Magnetic Properties of Materials PYQs: MHT CET, JEE Main, NEET

Q1MHT CET 2026 · 14 MAY · SHIFT 1Easy

Bar Magnet & Magnetic Moment · Magnetic moment of a magnet & a current loop

The angle made by electron's orbital angular momentum ( $L$ ) with the direction of the orbital magnetic moment ( $M$ ) is

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Answer: (c)

For an orbiting charge, $\vec M = \dfrac{q}{2m}\vec L$ .
The electron has $q = -e$ , so $\vec M = -\dfrac{e}{2m_e}\vec L$ : $\vec M$ is opposite to $\vec L$ .
So the answer is (c) $180^\circ$ .

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Q2JEE Main 2025 · 7 APR · SHIFT 2Easy

Bar Magnet & Magnetic Moment · Poles, monopoles & field lines

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Magnetic monopoles do not exist.

Reason (R): Magnetic field lines are continuous and form closed loops.

In the light of the above statements, choose the most appropriate answer from the options given below:

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Answer: (a)

No isolated N or S pole has ever been found, so (A) is correct.
Field lines leave N, return to S outside and go S to N inside, so they form closed loops; (R) is correct.
Closed loops have no start or end point, which is exactly why there is no source (monopole). So (R) explains (A).
So the answer is (a).

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Q3MHT CET 2024 · 4 MAY · SHIFT 1Medium

Bar Magnet & Magnetic Moment · Magnetic moment of a magnet & a current loop

The magnetic moments associated with two closely wound circular coils A and B of radius $r_A = 10\text{ cm}$ and $r_B = 20\text{ cm}$ respectively are equal if $(N_A, I_A$ and $N_B, I_B$ are number of turns and current of A and B respectively)

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Answer: (c)

Magnetic moment of a coil $M = NI\pi r^2$ .
$N_A I_A \pi (0.1)^2 = N_B I_B \pi (0.2)^2$
$N_A I_A = 4 N_B I_B$
So the answer is (c).

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Q4NEET 2024Medium

Bar Magnet & Magnetic Moment · Cutting & bending a magnet

The magnetic moment of an iron bar is $M$ . It is now bent in such a way that it forms an arc section of a circle subtending an angle of $60^\circ$ at the centre. The magnetic moment of this arc section is

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Answer: (a)

Pole strength $m = M/L$ is unchanged; $M' = m\times$ chord.
Arc $L$ subtends $60^\circ = \pi/3$ : $r = \dfrac{3L}{\pi}$ .
With $60^\circ$ at the centre the triangle is equilateral, so chord $= r = \dfrac{3L}{\pi}$ .
$M' = \dfrac{M}{L}\cdot\dfrac{3L}{\pi} = \dfrac{3M}{\pi}$
So the answer is (a) $\frac{3M}{\pi}$ .

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Q5NEET 2024Medium

Bar Magnet & Magnetic Moment · Cutting & bending a magnet

An iron bar of length $L$ has magnetic moment $M$ . It is bent at the middle of its length such that the two arms make an angle $60^\circ$ with each other. The magnetic moment of this new magnet is:

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Answer: (b)

Pole strength $m = M/L$ is unchanged.
Each arm is $L/2$ with $60^\circ$ between them, so the triangle formed is equilateral: end-to-end distance $= L/2$ .
$M' = \dfrac{M}{L}\cdot\dfrac{L}{2} = \dfrac{M}{2}$
So the answer is (b) $\frac{M}{2}$ .

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Q6MHT CET 2023 · 9 MAY · SHIFT 2Easy

Bar Magnet & Magnetic Moment · Cutting & bending a magnet

A thin rod of length $L$ has magnetic moment $M$ when magnetised. If rod is bent in a semicircular arc what is magnetic moment in new shape?

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Answer: (d)

Pole strength $m = M/L$ stays the same; the new moment is $m \times$ (straight distance between the ends).
Semicircle of length $L$ : $\pi r = L \Rightarrow r = L/\pi$ .
Distance between ends $= 2r = 2L/\pi$ .
$M' = \dfrac{M}{L}\cdot\dfrac{2L}{\pi} = \dfrac{2M}{\pi}$
So the answer is (d) $\frac{2M}{\pi}$ .

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Q7JEE Main 2021 · 18 MAR · SHIFT 2Medium

Bar Magnet & Magnetic Moment · Poles, monopoles & field lines

Which of the following statements are correct?

(A) Electric monopoles do not exist whereas magnetic monopoles exist.

(B) Magnetic field lines due to a solenoid at its ends and outside cannot be completely straight and confined.

(C) Magnetic field lines are completely confined within a toroid.

(D) Magnetic field lines inside a bar magnet are not parallel.

(E) $\chi = -$ 1 is the condition for a perfect diamagnetic material, where $\chi$ is its magnetic susceptibility. Choose the correct answer from the options given below:

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Answer: (c)

(A) Wrong: electric monopoles (single charges) exist; magnetic monopoles do not.
(B) Field lines spread out at the ends of a finite solenoid, so they are not straight and confined there.
(C) Correct: in an ideal toroid all field lines stay inside the core.
(D) Wrong: inside a bar magnet the lines are nearly parallel (S to N).
(E) Correct: perfect diamagnet (superconductor) has $\chi = -1$ , so $\mu_r = 0$ and $B = 0$ inside.
So the answer is (c) (C) and (E) only.

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Q8MHT CET 2021 · 24 SEP · SHIFT 1Medium

Bar Magnet & Magnetic Moment · Magnetic moment of a magnet & a current loop

A current ' $I$ ' is flowing in a conductor of length ' $L$ ' when it is bent in the form of a circular loop, its magnetic moment will be

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Answer: (d)

Loop of length $L$ : $2\pi r = L \Rightarrow r = \dfrac{L}{2\pi}$ .
Area $A = \pi r^2 = \dfrac{L^2}{4\pi}$ .
$M = IA = \dfrac{IL^2}{4\pi}$
So the answer is (d).

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Q9NEET 2021Medium

Bar Magnet & Magnetic Moment · Magnetic moment of a magnet & a current loop

A uniform conducting wire of length 12a and resistance 'R' is wound up as a current carrying coil in the shape of,

(i) an equilateral triangle of side 'a'

(ii) A square of side 'a'

The magnetic dipole moments of the coil in each case respectively are:

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Answer: (b)

Triangle of side $a$ : perimeter $3a$ , so turns $N = 12a/3a = 4$ ; area $= \dfrac{\sqrt3}{4}a^2$ .
$M_1 = 4\cdot I\cdot\dfrac{\sqrt3}{4}a^2 = \sqrt3\,Ia^2$
Square of side $a$ : perimeter $4a$ , so $N = 12a/4a = 3$ ; area $= a^2$ .
$M_2 = 3Ia^2$
So the answer is (b) $\sqrt3\,Ia^2$ and $3Ia^2$ .

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Q10MHT CET 2020 · 19 OCT · SHIFT 2Easy

Bar Magnet & Magnetic Moment · Cutting & bending a magnet

A steel wire of length / has a magnetic moment $M$ . It is then bent into a semicircular arc. The new magnetic moment is

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Answer: (c)

Pole strength $m = M/\ell$ is unchanged on bending.
Semicircle: $\pi r = \ell \Rightarrow r = \ell/\pi$ ; end-to-end distance $= 2r = 2\ell/\pi$ .
$M' = \dfrac{M}{\ell}\cdot\dfrac{2\ell}{\pi} = \dfrac{2M}{\pi}$
So the answer is (c) $\frac{2M}{\pi}$ .

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Q11NEET 2014 · AIPMTMedium

Bar Magnet & Magnetic Moment · Magnetic moment of a magnet & a current loop

Following figures show the arrangement of bar magnets in different configurations. Each magnet has magnetic dipole moment $\vec{m}.$ Which configuration has highest net magnetic dipole moment?

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Answer: (c)

Each moment points from S to N; add them as vectors: $M_{net} = 2m\cos(\theta/2)$ for angle $\theta$ between them.
(1) $\theta = 90^\circ$ : $\sqrt2\,m \approx 1.41m$ . (2) Antiparallel: $0$ .
(3) $\theta = 30^\circ$ : $2m\cos15^\circ \approx 1.93m$ .
(4) $\theta = 60^\circ$ : $2m\cos30^\circ = \sqrt3\,m \approx 1.73m$ .
So the answer is (c) configuration (3).

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Q12NEET 2013Easy

Bar Magnet & Magnetic Moment · Cutting & bending a magnet

A bar magnet of length $'l'$ and magnetic dipole moment 'M' is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be

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Answer: (d)

Pole strength $m = M/l$ is unchanged.
Arc of length $l$ subtends $60^\circ = \pi/3$ : $r\cdot\dfrac{\pi}{3} = l \Rightarrow r = \dfrac{3l}{\pi}$ .
The two radii and the chord form an equilateral triangle, so chord $= r = \dfrac{3l}{\pi}$ .
$M' = \dfrac{M}{l}\cdot\dfrac{3l}{\pi} = \dfrac{3M}{\pi}$
So the answer is (d) $\frac{3}{\pi}M$ .

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Q13NEET 2007 · AIPMTEasy

Bar Magnet & Magnetic Moment · Magnetic moment of a magnet & a current loop

A charged particle (charge q) is moving in a circle of radius R with uniform speed v. The associated magnetic moment $\mu$ is given by

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Answer: (d)

Equivalent current $I = \dfrac{q}{T} = \dfrac{qv}{2\pi R}$ .
Area of the orbit $A = \pi R^2$ .
$\mu = IA = \dfrac{qv}{2\pi R}\cdot \pi R^2 = \dfrac{qvR}{2}$
So the answer is (d) $qvR/2$ .

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Q14JEE Main 2003 · AIEEEEasy

Bar Magnet & Magnetic Moment · Poles, monopoles & field lines

The magnetic lines of force inside a bar magnet

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Answer: (d)

Magnetic field lines form closed loops.
Outside the magnet they go from N to S, so to close the loop they must go from S to N inside.
So the answer is (d) from south-pole to north-pole.

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Q15NEET 1997 · AIPMTEasy

Bar Magnet & Magnetic Moment · Cutting & bending a magnet

A bar magnet of magnetic moment $M$ is cut into two parts of equal length. The magnetic moment of each part will be

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Answer: (d)

Cutting into two equal lengths (perpendicular to the axis) keeps pole strength $m$ but halves the length.
$M' = m\cdot\dfrac{l}{2} = \dfrac{M}{2}$
So the answer is (d) $0.5\,M$ .

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Q16MHT CET 2026 · 15 MAY · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Permeability: B = μH

A magnetising field of $5000$ A/m produces a magnetic flux of $4 \times 10^{-5}$ Wb in an iron rod of cross- sectional area $0.4$ cm ${}^{2}$ . The permeability of the rod (in Wb/Am) is

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Answer: (c)

$B = \Phi/A$ and $\mu = B/H$ .
$B = \dfrac{4\times10^{-5}}{0.4\times10^{-4}} = 1\text{ T}$
$\mu = \dfrac{1}{5000} = 2\times10^{-4}\text{ Wb/Am}$
So the answer is (c).

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Q17MHT CET 2026 · 13 MAY · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Permeability: B = μH

An iron rod is placed parallel to magnetic field of intensity H. The magnetic flux through the rod is $\Phi$ and its cross-sectional area is A. The magnetic permeability ( $\mu$ ) of the rod is

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Answer: (c)

$B = \Phi/A$ and $B = \mu H$ .
$\mu = \dfrac{B}{H} = \dfrac{\Phi}{AH}$
So the answer is (c).

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Q18MHT CET 2026 · 15 APR · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Permeability: B = μH

An iron rod is placed parallel to magnetic field intensity 1000 A/m. The magnetic flux through the rod is $3 \times 10^{-4}$ Wb and its cross-sectional area is $1.5$ cm ${}^{2}$ . The magnetic permeability of rod in Wb/A-m is

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Answer: (b)

$B = \dfrac{\Phi}{A} = \dfrac{3\times10^{-4}}{1.5\times10^{-4}} = 2\text{ T}$
$\mu = \dfrac{B}{H} = \dfrac{2}{1000} = 2\times10^{-3}\text{ Wb/A-m}$
So the answer is (b).

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Q19MHT CET 2026 · 13 APR · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Permeability: B = μH

An iron rod is placed parallel to magnetic field intensity 2000 A/m. The magnetic flux through rod is $6 \times 10^{-4}$ wb and its cross sectional area is 3 cm ${}^{2}$ . The magnetic permeability of the rod in wb/A-m is

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Answer: (c)

$B = \dfrac{\Phi}{A} = \dfrac{6\times10^{-4}}{3\times10^{-4}} = 2\text{ T}$
$\mu = \dfrac{B}{H} = \dfrac{2}{2000} = 10^{-3}\text{ Wb/A-m}$
So the answer is (c).

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Q20MHT CET 2026 · 15 MAY · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The relationship between the magnetic susceptibility ( $\chi$ ) and magnetic permeability ( $\mu$ ) of a material is ( $\mu_0$ and $\mu_r$ is the permeability of free space and relative permeability respectively)

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Answer: (a)

$\mu_r = 1 + \chi$ and $\mu_r = \mu/\mu_0$ .
$\chi = \mu_r - 1 = \dfrac{\mu}{\mu_0} - 1$
So the answer is (a).

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Q21MHT CET 2026 · 12 MAY · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The relation between total magnetic field $(B)$ , magnetic intensity $(H)$ , permeability of free space $(\mu_0)$ and susceptibility $(\chi)$ is

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Answer: (a)

$B = \mu H = \mu_0\mu_r H$ and $\mu_r = 1+\chi$ .
$\dfrac{B}{H} = \mu_0(1+\chi)$
So the answer is (a).

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Q22JEE Main 2026 · 4 APR · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

A solenoid has a core made of material with relative permeability 400. The magnetic field produced in the interior of solenoid is 1.0 T. The magnetic intensity in SI units is $\alpha \times 10^{5}$ . The value of $\alpha$ is ______.

(Free space permeability $\mu_0 = 4\pi \times 10^{-7}$ SI units.)

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Answer: (b)

Inside the core $B = \mu_0 \mu_r H$ , so $H = \dfrac{B}{\mu_0 \mu_r}$ .
$H = \dfrac{1.0}{4\pi \times 10^{-7} \times 400} = \dfrac{10^{4}}{1.6\pi} = \dfrac{1}{16\pi} \times 10^{5}\ \text{A m}^{-1}$
So the answer is (b) $\alpha = \dfrac{1}{16\pi}$ .

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Q23MHT CET 2026 · 14 MAY · SHIFT 2Medium

Magnetisation: B, H, M, χ & μ · Intensity of magnetisation: M = m/V

A paramagnetic substance in the form of a cube with sides $1$ cm has a magnetic dipole moment $20 \times 10^{-6}$ J/T when a magnetic intensity $60 \times 10^{3}$ A/m is applied. Its magnetic susceptibility is

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Answer: (a)

Magnetisation $M = \dfrac{m}{V} = \dfrac{20 \times 10^{-6}}{(10^{-2})^3} = 20\ \text{A m}^{-1}$
Susceptibility $\chi = \dfrac{M}{H} = \dfrac{20}{60 \times 10^{3}} = 3.3 \times 10^{-4}$
So the answer is (a) $3.3 \times 10^{-4}$ .

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Q24MHT CET 2026 · 13 MAY · SHIFT 2Medium

Magnetisation: B, H, M, χ & μ · Intensity of magnetisation: M = m/V

A domain in a ferromagnetic substance is in the form of a cube of side ' $L$ '. If it contains ' $N$ ' atoms and each atomic dipole moment $M$ then the magnetisation of the domain is

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Answer: (b)

Total dipole moment of the domain $= NM$ ; volume $= L^3$ .
Magnetisation $= \dfrac{\text{moment}}{\text{volume}} = \dfrac{MN}{L^3}$
So the answer is (b).

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Q25MHT CET 2026 · 16 APR · SHIFT 1Medium

Magnetisation: B, H, M, χ & μ · Intensity of magnetisation: M = m/V

The magnetic moment produced in a sample of 2 gram is $8 \times 10^{-7}$ A/m ${}^{2}$ . If its density is $4$ g/cm ${}^{3}$ then the magnetization of the sample is

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Answer: (a)

Volume $V = \dfrac{\text{mass}}{\rho} = \dfrac{2}{4} = 0.5\ \text{cm}^3 = 5 \times 10^{-7}\ \text{m}^3$
$M_z = \dfrac{m}{V} = \dfrac{8 \times 10^{-7}}{5 \times 10^{-7}} = 1.6\ \text{A m}^{-1}$ (the moment's unit should read $\text{A m}^2$ ).
So the answer is (a) 1.6.

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Q26MHT CET 2026 · 18 APR · SHIFT 1Medium

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

The region inside a current carrying toroid is filled with a material (Niobium) having susceptibility $\chi = 2.6 \times 10^{-5}$ . The percentage increase in the magnetic field in the presence of Niobium over that without it, is

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Answer: (c)

With the material $B' = (1+\chi)B$ , so percentage increase $= \chi \times 100$ .
$= 2.6 \times 10^{-5} \times 100 = 2.6 \times 10^{-3}\%$
So the answer is (c).

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Q27MHT CET 2026 · 13 APR · SHIFT 2Difficult

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

A rod with a circular cross-section area $2$ cm ${}^{2}$ and length $40$ cm is wound uniformly with $400$ turns of an insulated wire. If a current $0.4$ A flows in the wire winding, the total magnetic flux produced inside the winding is $4\pi \times 10^{-6}$ Wb. The relative permeability of the rod is

(Given permeability of vacuum $\mu_0 = 4\pi \times 10^{-7}$ N m A ${}^{-2}$ )

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Answer: (c)

$B = \dfrac{\Phi}{A} = \dfrac{4\pi \times 10^{-6}}{2 \times 10^{-4}} = 2\pi \times 10^{-2}\ \text{T}$
$n = \dfrac{400}{0.4} = 1000\ \text{m}^{-1}$ , so $\mu_0 nI = 4\pi \times 10^{-7} \times 1000 \times 0.4 = 1.6\pi \times 10^{-4}\ \text{T}$
$\mu_r = \dfrac{B}{\mu_0 nI} = \dfrac{2\pi \times 10^{-2}}{1.6\pi \times 10^{-4}} = 125$
So the answer is (c) 125.

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Q28JEE Main 2025 · 29 JAN · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Definitions & units

Match List - I with List - II.

Choose the correct answer from the options given below:

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Answer: (a)

Magnetic induction $B$ : unit tesla or gauss, so (A)-(III).
Magnetic intensity $H$ : A/m $=$ A m $^{-1}$ , so (B)-(IV).
Magnetic flux: weber, so (C)-(II).
Magnetic moment: A m $^2$ , so (D)-(I).
So the answer is (a).

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Q29MHT CET 2025 · 20 APR · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The magnetic susceptibility of iron is 5499. The relative permeability of iron will be

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Answer: (c)

$\mu_r = 1 + \chi$
$\mu_r = 1 + 5499 = 5500$
So the answer is (c) 5500.

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Q30MHT CET 2025 · 19 APR · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

If $M$ is the magnetisation induced in the material, H is the magnetic field intensity, B is the net magnetic field inside the material then the correct relation between them is ( $\mu_0 =$ permeability of free space)

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Answer: (d)

Inside a material the field has two parts: the applied $\mu_0 H$ and the extra $\mu_0 M$ from magnetisation.
$B = \mu_0(H + M)$
So the answer is (d).

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Q31JEE Main 2025 · 2 APR · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The relationship between the magnetic susceptibility $(\chi)$ and the magnetic permeability $(\mu)$ is given by:

( $\mu_0$ is the permeability of free space and $\mu_r$ is relative permeability)

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Answer: (a)

$\mu_r = \mu/\mu_0 = 1 + \chi$ .
$\chi = \dfrac{\mu}{\mu_0} - 1$
So the answer is (a).

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Q32MHT CET 2025 · 5 MAY · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

The magnetic field (B) inside a long solenoid having ' $n$ ' turns per unit length and carrying current ' i ' when iron core is kept in it, is ( $\mu_0 =$ permeability of vacuum, $\chi =$ magnetic susceptibility)

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Answer: (a)

Without core $B_0 = \mu_0 n i$ ; the core multiplies it by $\mu_r$ .
$B = \mu_0\mu_r n i = \mu_0 n i(1+\chi)$
So the answer is (a).

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Q33JEE Main 2025 · 7 APR · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

The percentage increase in magnetic field (B) when space within a current carrying solenoid is filled with magnesium (magnetic susceptibility $\chi_{\text{Mg}} = 1.2 \times 10^{-5}$ ) is:

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Answer: (d)

With the material, $B' = \mu_0(1+\chi)nI$ , so $\dfrac{B' - B}{B} = \chi$ .
Percentage increase $= \chi \times 100 = 1.2 \times 10^{-5} \times 100 = 1.2 \times 10^{-3}\%$
So the answer is (d) $\frac{6}{5} \times 10^{-3}\%$ .

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Q34MHT CET 2025 · 26 APR · SHIFT 1Medium

Magnetisation: B, H, M, χ & μ · Intensity of magnetisation: M = m/V

A bar of iron having magnetic moment $2.4\text{ Am}^2$ weighs 66 g. If the density of the material of the bar is $7700\text{ kg/m}^3$ , the intensity of magnetisation in $\text{Am}^{-1}$ is

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Answer: (b)

Volume $V = \dfrac{0.066}{7700} = 8.57 \times 10^{-6}\ \text{m}^3$
Intensity of magnetisation $= \dfrac{m}{V} = \dfrac{2.4}{8.57 \times 10^{-6}} = 2.8 \times 10^{5}\ \text{A m}^{-1}$
So the answer is (b).

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Q35MHT CET 2025 · 21 APR · SHIFT 2Medium

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

The space within the current carrying toroid is filled with aluminium of susceptibility ' $\chi$ '. The percentage increase in the magnetic field ' $B$ ' will be

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Answer: (d)

$B' = \mu_0(1+\chi)nI = (1+\chi)B$
$\dfrac{B' - B}{B} \times 100 = \chi \times 100$
So the answer is (d) $\chi \times 100$ .

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Q36MHT CET 2024 · 11 MAY · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Intensity of magnetisation: M = m/V

A bar magnet has length 4 cm, cross-sectional area $2\text{ cm}^2$ and magnetic moment $6\text{ Am}^2$ . The intensity of magnetisation of bar magnet is

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Answer: (b)

Intensity of magnetisation $M_z = \dfrac{m}{V}$ .
$V = 4\times 2 = 8\text{ cm}^3 = 8\times10^{-6}\text{ m}^3$
$M_z = \dfrac{6}{8\times10^{-6}} = 7.5\times10^{5}\text{ A/m}$
So the answer is (b).

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Q37MHT CET 2024 · 4 MAY · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Permeability: B = μH

A magnetic intensity of $500\text{ A/m}$ , produces a magnetic flux of $2.4 \times 10^{-5}\text{ Wb}$ in an iron bar of cross- sectional area $0.4\text{ cm}^2$ . The magnetic permeability of the iron bar is

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Answer: (b)

$B = \dfrac{\Phi}{A} = \dfrac{2.4\times10^{-5}}{0.4\times10^{-4}} = 0.6\text{ T}$
$\mu = \dfrac{B}{H} = \dfrac{0.6}{500} = 1.2\times10^{-3}$ (SI unit T m/A, i.e. N/A $^2$ ; the 'N/m $^2$ ' in the options is a misprint).
So the answer is (b).

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Q38MHT CET 2024 · 9 MAY · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The correct relation between total magnetic field $(B)$ , magnetic intensity $(H)$ , permeability of free space $(\mu_0)$ and susceptibility $(\chi)$ is

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Answer: (c)

$B = \mu_0(H + M)$ with $M = \chi H$ .
$B = \mu_0 H(1+\chi) \Rightarrow \dfrac{B}{H} = \mu_0(1+\chi)$
So the answer is (c).

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Q39MHT CET 2024 · 3 MAY · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The relation between total magnetic field (B), magnetic intensity $(H)$ , permeability of free space $(\mu_0)$ and susceptibility $(\chi)$ is

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Answer: (b)

$B = \mu_0(H + \chi H) = \mu_0(1+\chi)H$
$\dfrac{B}{H} = \mu_0(1+\chi)$
So the answer is (b).

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Q40MHT CET 2024 · 3 MAY · SHIFT 2Medium

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The magnetic susceptibility of the material of a rod is 599. The absolute permeability of the material of the rod will be [ $\mu_0 = 4\pi \times 10^{-7}$ SI unit]

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Answer: (b)

$\mu = \mu_0 \mu_r = \mu_0(1+\chi)$
$\mu = 4\pi \times 10^{-7} \times 600 = 2400\pi \times 10^{-7} = 2.4\pi \times 10^{-4}$ SI units
So the answer is (b).

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Q41MHT CET 2024 · 16 MAY · SHIFT 2Medium

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

A solenoid having 400 turns per metre has a core of a material with relative permeability 400. When a current of 0.5 A is passed through it, the magnetization of the core material in $\text{Am}^{-1}$ is nearly

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Answer: (d)

$H = nI = 400 \times 0.5 = 200\ \text{A m}^{-1}$
$M = \chi H = (\mu_r - 1)H = 399 \times 200 \approx 8 \times 10^{4}\ \text{A m}^{-1}$
So the answer is (d).

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Q42JEE Main 2023 · 11 APR · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

The free space inside a current carrying toroid is filled with a material of susceptibility $2 \times 10^{-2}$ . The percentage increase in the value of magnetic field inside the toroid will be

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Answer: (c)

$B' = \mu_0(1+\chi)nI$ , so the fractional increase is $\chi$ .
Percentage increase $= 2 \times 10^{-2} \times 100 = 2\%$
So the answer is (c) 2%.

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Q43JEE Main 2023 · 25 JAN · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is:

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Answer: (c)

Magnetic intensity inside a long solenoid: $H = nI$ (independent of the core and the diameter).
$n = \dfrac{1200}{2} = 600\ \text{m}^{-1}$
$H = 600 \times 2 = 1.2 \times 10^{3}\ \text{A m}^{-1}$
So the answer is (c).

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Q44JEE Main 2023 · 8 APR · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

The magnetic intensity at the center of a long current carrying solenoid is found to be $1.6 \times 10^{3}\text{ Am}^{-1}$ . If the number of turns is 8 per cm, then the current flowing through the solenoid is __________ A.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

$H = nI$ with $n = 8\ \text{per cm} = 800\ \text{m}^{-1}$ .
$I = \dfrac{H}{n} = \dfrac{1.6 \times 10^{3}}{800} = 2$
So the current is 2 A.

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Q45MHT CET 2023 · 10 MAY · SHIFT 2Medium

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

A solenoid of 500 turns/m is carrying a current of 3 A. Its core is made of iron which has relative permeability 5001. The magnitude of magnetization is

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Answer: (c)

$H = nI = 500 \times 3 = 1500\ \text{A m}^{-1}$
$M = (\mu_r - 1)H = 5000 \times 1500 = 7.5 \times 10^{6}\ \text{A m}^{-1}$
So the answer is (c).

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Q46JEE Main 2022 · 27 JUN · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The susceptibility of a paramagnetic material is 99. The permeability of the material in Wb/A-m, is: [Permeability of free space $\mu$ 0 = 4 $\pi \times$ 10 $-$ 7 Wb/A-m]

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Answer: (c)

$\mu_r = 1 + \chi = 100$
$\mu = \mu_r\mu_0 = 100\times4\pi\times10^{-7} = 4\pi\times10^{-5}\text{ Wb/A-m}$
So the answer is (c).

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Q47JEE Main 2022 · 28 JUN · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to 1.2 $\times$ 10 $-$ 5. What is fractional increase in the magnetic field inside solenoid with respect to air as medium inside the solenoid?

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Answer: (a)

$B_{\text{air}} = \mu_0 nI$ , $B' = \mu_0(1+\chi)nI$ .
Fractional increase $= \dfrac{B' - B}{B} = \chi = 1.2 \times 10^{-5}$
So the answer is (a) $1.2 \times 10^{-5}$ .

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Q48MHT CET 2022 · 11 AUG · SHIFT 2Medium

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The magnetic susceptibility of the material of a rod is 349 and permeability of vacuum $\mu_0$ is $4\pi \times 10^{-7}$ SI units. Absolute permeability of the material of the rod in SI units is

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Answer: (a)

$\mu = \mu_0(1+\chi) = 4\pi \times 10^{-7} \times 350$
$= 1400\pi \times 10^{-7} \approx 4400 \times 10^{-7}$ SI units
So the answer is (a).

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Q49MHT CET 2021 · 23 SEP · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Intensity of magnetisation: M = m/V

A bar magnet has length $3\text{ cm}$ , cross-sectional area $2\text{ cm}^2$ and magnetic moment $3\text{ Am}^2$ . The intensity of magnetisation of bar magnet is

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Answer: (d)

$M_z = \dfrac{m}{V}$ , $V = 3\times2 = 6\text{ cm}^3 = 6\times10^{-6}\text{ m}^3$ .
$M_z = \dfrac{3}{6\times10^{-6}} = 5\times10^{5}\text{ A/m}$
So the answer is (d).

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Q50MHT CET 2021 · 22 SEP · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The relative permeability of iron is 2000. Its absolute permeability in SI unit will be $(\frac{\mu_0}{4\pi} = 10^{-7}$ SI unit $)$

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Answer: (c)

$\mu = \mu_r\mu_0$ , with $\mu_0 = 4\pi\times10^{-7}$ .
$\mu = 2000\times4\pi\times10^{-7} = 8\pi\times10^{-4}$ SI unit
So the answer is (c).

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Q51MHT CET 2021 · 22 SEP · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The permeability of a metal is $0.1256\text{ TmA}^{-1}$ . Its relative permeability will be $\left(\frac{\mu_0}{4\pi} = 10^{-7}\text{ SI unit}\right)$ ( $\pi = 3.14$ )

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Answer: (a)

$\mu_r = \dfrac{\mu}{\mu_0}$ , with $\mu_0 = 4\pi\times10^{-7} = 1.256\times10^{-6}$ .
$\mu_r = \dfrac{0.1256}{1.256\times10^{-6}} = 10^{5}$
So the answer is (a) $10^5$ .

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Q52MHT CET 2021 · 21 SEP · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

What is susceptibility of a medium, if its relative permeability is 0.85?

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Answer: (c)

$\chi = \mu_r - 1$
$\chi = 0.85 - 1 = -0.15$ (diamagnetic)
So the answer is (c) $-0.15$ .

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Q53JEE Main 2021 · 20 JUL · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

The magnetic susceptibility of a material of a rod is 499. Permeability in vacuum is 4 $\pi \times$ 10 $-$ 7 H/m. Absolute permeability of the material of the rod is:

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Answer: (b)

$\mu_r = 1 + \chi = 500$
$\mu = 500\times4\pi\times10^{-7} = 2000\pi\times10^{-7} = 2\pi\times10^{-4}\text{ H/m}$
So the answer is (b).

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Q54JEE Main 2021 · 25 JUL · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Solenoid & toroid with a core

The value of aluminium susceptibility is 2.2 $\times$ 10 $-$ 5. The percentage increase in the magnetic field if space within a current carrying toroid is filled with aluminium is $\frac{x}{10^4}$ . Then the value of x is _________________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 22

Filling the toroid raises $B$ by the factor $(1+\chi)$ , so percentage increase $= \chi \times 100$ .
$= 2.2 \times 10^{-5} \times 100 = 2.2 \times 10^{-3}\% = \dfrac{22}{10^{4}}\%$
So $x = 22$ .

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Q55MHT CET 2020 · 16 OCT · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Permeability: B = μH

An iron rod is placed parallel to magnetic field of intensity $2000\text{ A/m}$ . The magnetic flux through the rod is $6 \times 10^{-4}\text{ Wb}$ and its cross-sectional area is $3\text{ cm}^2$ . The magnetic permeability of the rod in $\frac{\text{Wb}}{\text{A-m}}$ is

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Answer: (d)

$B = \dfrac{\Phi}{A} = \dfrac{6\times10^{-4}}{3\times10^{-4}} = 2\text{ T}$
$\mu = \dfrac{B}{H} = \dfrac{2}{2000} = 10^{-3}\text{ Wb/A-m}$
So the answer is (d) $10^{-3}$ .

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Q56NEET 2020 · PHASE 1Easy

Magnetisation: B, H, M, χ & μ · Relations: B = μ0(H + M), μr = 1 + χ

An iron rod of susceptibility 599 is subjected to a magnetizing field of 1200 A m-1. The permeability of the material of the rod is:

$\left(\mu_0 = 4\pi \times 10^{-7}\text{ TmA}^{-1}\right)$

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Answer: (d)

$\mu_r = 1 + \chi = 600$ (the field 1200 A/m is not needed).
$\mu = \mu_r\mu_0 = 600\times4\pi\times10^{-7} = 2400\pi\times10^{-7}$
$\mu = 2.4\pi\times10^{-4}\text{ TmA}^{-1}$
So the answer is (d).

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Q57MHT CET 2020 · 19 OCT · SHIFT 2Medium

Magnetisation: B, H, M, χ & μ · Intensity of magnetisation: M = m/V

A domain in a ferromagnetic substance is in the form of a cube of side $1\,\mu\text{m}$ . If it contains $8 \times 10^{10}$ atoms and each atomic dipole has a dipole moment of $9 \times 10^{-24}\text{ Am}^2$ , then the magnetisation of the domain is

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Answer: (a)

Total moment $= 8 \times 10^{10} \times 9 \times 10^{-24} = 7.2 \times 10^{-13}\ \text{A m}^2$
Volume $= (10^{-6})^3 = 10^{-18}\ \text{m}^3$
Magnetisation $= \dfrac{7.2 \times 10^{-13}}{10^{-18}} = 7.2 \times 10^{5}\ \text{A m}^{-1}$
So the answer is (a).

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Q58MHT CET 2019 · 2 MAY · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ · Intensity of magnetisation: M = m/V

The magnetization of bar magnet of length 5 cm, cross sectional area $2\text{ cm}^2$ and net magnetic moment $1\text{ Am}^2$ is

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Answer: (d)

$M_z = \dfrac{m}{V}$ , $V = 5\times2 = 10\text{ cm}^3 = 10^{-5}\text{ m}^3$ .
$M_z = \dfrac{1}{10^{-5}} = 1\times10^{5}\text{ A/m}$
So the answer is (d).

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Q59MHT CET 2019 · 3 MAY · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ · Permeability: B = μH

A magnetizing field of $5000\text{ A/m}$ produces a magnetic flux of $4 \times 10^{-5}\text{ Wb}$ in an iron rod of cross- sectional area $0.4\text{ cm}^2$ . The permeabilit of the rod in Wb/A-m, is

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Answer: (c)

$B = \dfrac{\Phi}{A} = \dfrac{4\times10^{-5}}{0.4\times10^{-4}} = 1\text{ T}$
$\mu = \dfrac{B}{H} = \dfrac{1}{5000} = 2\times10^{-4}\text{ Wb/A-m}$
So the answer is (c).

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Q60JEE Main 2019 · 11 JAN · SHIFT 2Medium

Magnetisation: B, H, M, χ & μ · Intensity of magnetisation: M = m/V

A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of 20 $\times$ 10−6 J/ T when a magnetic intensity of 60 $\times$ 103 A/m is applied. Its magnetic susceptibility is

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Answer: (a)

$M = \dfrac{m}{V} = \dfrac{20 \times 10^{-6}}{10^{-6}} = 20\ \text{A m}^{-1}$
$\chi = \dfrac{M}{H} = \dfrac{20}{60 \times 10^{3}} = 3.3 \times 10^{-4}$
So the answer is (a) $3.3 \times 10^{-4}$ .

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Q61JEE Main 2026 · 4 APR · SHIFT 2Medium

Field of a Magnet & Magnet in a Uniform Field · Field of a bar magnet

Two identical small bar magnets each of dipole moment $3\sqrt{5}\text{ J/T}$ are placed at a center to center separation of 10 cm, with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point P midway between the magnets is $\alpha \times 10^{-3}\text{ T}$ . The value of $\alpha$ is $\underline{\qquad}\ \left(\mu_0 = 4\pi \times 10^{-7}\text{ Tm/A}\right)$

Numerical answer type. Work it out, then open the answer.

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Answer: 12

P is 5 cm from each centre: on the axis of the left magnet and on the equatorial line of the right magnet.
$B_1 = \dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$ (horizontal), $B_2 = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}$ (vertical); they are perpendicular.
$B = \dfrac{\mu_0}{4\pi}\dfrac{\sqrt{5}M}{r^3} = 10^{-7} \times \dfrac{\sqrt{5} \times 3\sqrt{5}}{(0.05)^3} = \dfrac{15 \times 10^{-7}}{1.25 \times 10^{-4}}$
$B = 1.2 \times 10^{-2}\ \text{T} = 12 \times 10^{-3}\ \text{T}$
So $\alpha = 12$ .

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Q62MHT CET 2026 · 19 MAY · SHIFT 1Medium

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A magnetic dipole lying parallel to a magnetic field requires ' $W$ ' units of work to turn it through $30^\circ$ . The torque required to keep this needle in this position is

$\left[\sin 30^\circ = \cos 60^\circ = 0.5, \sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}, \sin 90^\circ = \cos 0^\circ = 1\right]$

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Answer: (d)

Work from $0^\circ$ to $30^\circ$ : $W = MB(1 - \cos 30^\circ) = MB\dfrac{2 - \sqrt{3}}{2}$ , so $MB = \dfrac{2W}{2 - \sqrt{3}}$ .
Torque at $30^\circ$ : $\tau = MB\sin 30^\circ = \dfrac{MB}{2}$
$\tau = \dfrac{W}{2 - \sqrt{3}}$
So the answer is (d).

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Q63JEE Main 2026 · 24 JAN · SHIFT 1Medium

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A short bar magnet placed with its axis at $30^\circ$ with an external field of 800 Gauss, experiences a torque of $0.016\text{ N.m}$ . The work done in moving it from most stable to most unstable position is $\alpha \times 10^{-3}\text{ J}$ . The value of $\alpha$ is ______.

Numerical answer type. Work it out, then open the answer.

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Answer: 64

$B = 800\ \text{G} = 0.08\ \text{T}$ ; $\tau = MB\sin 30^\circ \Rightarrow MB = \frac{0.016}{0.5} = 0.032\ \text{J}$
Most stable to most unstable: $W = 2MB = 0.064\ \text{J} = 64 \times 10^{-3}\ \text{J}$
So $\alpha = 64$ .

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Q64MHT CET 2025 · 26 APR · SHIFT 1Medium

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

The work done in turning a magnet of magnetic moment ' $M$ ' by an angle of $90^\circ$ from the meridian is ' $n$ ' times the corresponding Work done to turn it through an angle of $60^\circ$ where the value of ' $n$ ' is $(\cos 90^\circ = 0, \cos 60^\circ = 0.5)$

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Answer: (b)

$W = MB(1 - \cos\theta)$ starting from the meridian.
$W_{90} = MB(1 - 0) = MB$ , $W_{60} = MB(1 - 0.5) = \dfrac{MB}{2}$
$n = \dfrac{W_{90}}{W_{60}} = 2$
So the answer is (b) 2.

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Q65JEE Main 2025 · 3 APR · SHIFT 2Medium

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A magnetic dipole experiences a torque of $80\sqrt{3}\text{ N m}$ when placed in uniform magnetic field in such a way that dipole moment makes angle of $60^\circ$ with magnetic field. The potential energy of the dipole is:

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Answer: (b)

$\tau = MB\sin 60^\circ = 80\sqrt{3} \Rightarrow MB\frac{\sqrt{3}}{2} = 80\sqrt{3} \Rightarrow MB = 160$
$U = -MB\cos 60^\circ = -160 \times \frac{1}{2}$
So the answer is (b) $-80\ \text{J}$ .

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Q66JEE Main 2025 · 28 JAN · SHIFT 2 (One or more correct)Difficult

Field of a Magnet & Magnet in a Uniform Field · Field of a bar magnet

A bar magnet has total length $2l = 20$ units and the field point P is at a distance $d = 10$ units from the centre of the magnet. If the relative uncertainty of length measurement is $1\%$ , then uncertainty of the magnetic field at point P is:

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Answer: (c)

P is on the equatorial line: $B = \frac{\mu_0}{4\pi}\frac{M}{(d^2+l^2)^{3/2}}$ (treat $M$ as given).
$\frac{\Delta B}{B} = \frac{3}{2}\cdot\frac{2d\,\Delta d + 2l\,\Delta l}{d^2+l^2}$
$d = l = 10$ , $\Delta d = \Delta l = 1\%$ of 10 $= 0.1$ : $\frac{\Delta B}{B} = \frac{3}{2}\cdot\frac{2 + 2}{200} = 0.03$
So the answer is (c) 3%.

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Q67NEET 2024Easy

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

The magnetic potential energy, when a magnetic bar of magnetic moment $\vec{m}$ is placed perpendicular to the magnetic field $\vec{B}$ is

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Answer: (b)

$U = -mB\cos\theta$ with $\theta = 90^\circ$ .
$U = -mB\cos 90^\circ = 0$
So the answer is (b) zero.

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Q68JEE Main 2024 · 4 APR · SHIFT 2Easy

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

The magnetic moment of a bar magnet is $0.5\text{ Am}^2$ . It is suspended in a uniform magnetic field of $8 \times 10^{-2}\text{ T}$ . The work done in rotating it from its most stable to most unstable position is:

Show answer and solution

Answer: (c)

Most stable $\theta = 0^\circ$ , most unstable $\theta = 180^\circ$ : $W = MB(1 - \cos 180^\circ) = 2MB$ .
$W = 2 \times 0.5 \times 8 \times 10^{-2} = 8 \times 10^{-2}\ \text{J}$
So the answer is (c).

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Q69JEE Main 2024 · 29 JAN · SHIFT 1Medium

Field of a Magnet & Magnet in a Uniform Field · Field of a bar magnet

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $20\text{ cm}$ from its center is $1.5 \times 10^{-5}\text{ T m}$ . The magnetic moment of the dipole is _________ $\text{A m}^2$ . (Given: $\frac{\mu_0}{4\pi} = 10^{-7}\text{ TmA}^{-1}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 6

Magnetic potential on the axis: $V = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^2}$ .
$M = \dfrac{V r^2}{\mu_0/4\pi} = \dfrac{1.5 \times 10^{-5} \times (0.2)^2}{10^{-7}} = 6$
So the magnetic moment is $6\ \text{A m}^2$ .

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Q70JEE Main 2023 · 31 JAN · SHIFT 1Easy

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A bar magnet with a magnetic moment $5.0\text{ Am}^2$ is placed in parallel position relative to a magnetic field of $0.4\text{ T}$ . The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is _____________.

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Answer: (d)

Parallel to antiparallel: $W = MB(\cos 0^\circ - \cos 180^\circ) = 2MB$ .
$W = 2 \times 5.0 \times 0.4 = 4\ \text{J}$
So the answer is (d) 4 J.

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Q71JEE Main 2022 · 26 JUN · SHIFT 2Easy

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A bar magnet having a magnetic moment of 2.0 $\times$ 105 JT ${}^{-1}$ , is placed along the direction of uniform magnetic field of magnitude B = 14 $\times$ 10 $-$ 5 T. The work done in rotating the magnet slowly through 60 ${}^\circ$ from the direction of field is:

Show answer and solution

Answer: (a)

$W = MB(1 - \cos 60^\circ)$
$W = 2.0 \times 10^{5} \times 14 \times 10^{-5} \times 0.5 = 14\ \text{J}$
So the answer is (a) 14 J.

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Q72MHT CET 2021 · 24 SEP · SHIFT 2Easy

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A magnetic dipole of magnetic moment $M$ , is freely suspended in a magnetic field of induction B. The minimum and maximum values of potential energy of the dipole, respectively are

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Answer: (a)

$U = -MB\cos\theta$
Minimum at $\theta = 0^\circ$ : $U = -MB$ ; maximum at $\theta = 180^\circ$ : $U = +MB$ .
So the answer is (a) $-MB, +MB$ .

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Q73JEE Main 2021 · 26 AUG · SHIFT 1Difficult

Field of a Magnet & Magnet in a Uniform Field · Field of a bar magnet

Two short magnetic dipoles m1 and m2 each having magnetic moment of 1 Am2 are placed at point O and P respectively. The distance between OP is 1 meter. The torque experienced by the magnetic dipole m2 due to the presence of m1 is ........... $\times$ 10 $-$ 7 Nm.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1

P is on the equatorial line of $m_1$ (which points up), so $B_1 = \frac{\mu_0}{4\pi}\frac{m_1}{r^3}$ , pointing down.
$B_1 = 10^{-7} \times \frac{1}{1^3} = 10^{-7}\ \text{T}$
$m_2$ points right, perpendicular to $B_1$ : $\tau = m_2B_1\sin 90^\circ = 1 \times 10^{-7}\ \text{N m}$
So the answer is 1.

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Q74MHT CET 2020 · 16 OCT · SHIFT 1Medium

Field of a Magnet & Magnet in a Uniform Field · Field of a bar magnet

Two identical bar magnets each of magnetic moment $M$ , separated by some distance are kept perpendicular to each other. The magnetic induction at a point at the same distance $d$ from the centre of magnets, is ( $\mu_0 =$ permeability of free space)

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Answer: (c)

The point lies on the axis of one magnet and on the equatorial line of the other, at distance $d$ from both.
$B_{\text{axial}} = \dfrac{\mu_0}{4\pi}\dfrac{2M}{d^3}$ , $B_{\text{eq}} = \dfrac{\mu_0}{4\pi}\dfrac{M}{d^3}$ , and the two fields are perpendicular.
$B = \sqrt{B_{\text{axial}}^2 + B_{\text{eq}}^2} = \dfrac{\mu_0}{4\pi}\dfrac{M}{d^3}\sqrt{4+1}$
So the answer is (c) $\frac{\mu_0}{4\pi}(\sqrt{5})\frac{M}{d^3}$ .

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Q75JEE Main 2020 · 4 SEP · SHIFT 1Medium

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A small bar magnet placed with its axis

at 30° with an external field of 0.06 T

experiences a torque of 0.018 Nm. The

minimum work required to rotate it from its

stable to unstable equilibrium position is:

Show answer and solution

Answer: (c)

$\tau = MB\sin 30^\circ \Rightarrow MB = \frac{0.018}{0.5} = 0.036\ \text{J}$
Stable ( $0^\circ$ ) to unstable ( $180^\circ$ ): $W = MB(\cos 0^\circ - \cos 180^\circ) = 2MB$
$W = 2 \times 0.036 = 0.072\ \text{J}$
So the answer is (c) $7.2 \times 10^{-2}\ \text{J}$ .

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Q76JEE Main 2019 · 8 APR · SHIFT 2Difficult

Field of a Magnet & Magnet in a Uniform Field · Field of a bar magnet

Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A particle of charge q is passing, through their midpoint P, at angle $\theta$ = 45° with the horizontal line, as shown in figure. What would be the magnitude of force on the particle at that instant?

(d is much larger than the dimensions of the dipole)

Show answer and solution

Answer: (d)

P lies on the axis of X: $B_X = \frac{\mu_0}{4\pi}\frac{2M}{(d/2)^3}$ , along X's moment (S to N), i.e. to the right.
P lies on the equatorial line of Y: $B_Y = \frac{\mu_0}{4\pi}\frac{2M}{(d/2)^3}$ , opposite to Y's moment (Y points down), i.e. upward.
$B_X = B_Y$ , so the resultant $\vec{B}$ points at $45^\circ$ to the horizontal.
The velocity is also at $45^\circ$ , so $\vec{v} \parallel \vec{B}$ and $F = qvB\sin 0^\circ = 0$ .
So the answer is (d) 0.

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Q77JEE Main 2019 · 10 JAN · SHIFT 1Difficult

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A magnet of total magnetic moment 10−2 $\hat{i}$ A-m2 is placed in a time varying magnetic field, B $\hat{i}$ (cos $\omega t$ ) where B = 1 Tesla and $\omega$ = 0.125 rad/s. The work done for reversing the direction of the magnetic moment at t = 1 second, is -

Show answer and solution

Answer: (a)

Work to reverse a dipole in field $B'$ : $W = MB'(\cos 0^\circ - \cos 180^\circ) = 2MB'$ .
At $t = 1$ s: $B' = B\cos(\omega t) = \cos(0.125\ \text{rad}) \approx 0.992\ \text{T}$
$W = 2 \times 10^{-2} \times 0.992 \approx 0.0198\ \text{J}$
The keyed value $0.014\ \text{J} = 2MB\cos 45^\circ$ corresponds to $\omega t = \pi/4$ . Key: (a) 0.014 J.

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Q78NEET 2017Medium

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A 250-turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 $\mu$ A and subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by 180° against the torque is

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Answer: (d)

Rotating through $180^\circ$ from stable position: $W = 2NIAB$ .
$A = 2.1 \times 10^{-2} \times 1.25 \times 10^{-2} = 2.625 \times 10^{-4}\ \text{m}^2$
$W = 2 \times 250 \times 85 \times 10^{-6} \times 2.625 \times 10^{-4} \times 0.85 \approx 9.5 \times 10^{-6}\ \text{J}$
The nearest option is 9.1 $\mu$ J, so the answer is (d).

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Q79NEET 2016 · PHASE 2Medium

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is

Show answer and solution

Answer: (b)

Work to rotate from equilibrium: $W = MB(1-\cos 60^\circ) = \frac{MB}{2}$ , so $MB = 2W$ .
Torque at $60^\circ$ : $\tau = MB\sin 60^\circ = 2W \cdot \frac{\sqrt{3}}{2}$
So the answer is (b) $\sqrt{3}W$ .

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Q80NEET 2015 · AIPMTMedium

Field of a Magnet & Magnet in a Uniform Field · Torque on a magnet or current loop

A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 weber/m2. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be

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Answer: (d)

$NIAB = 50 \times 2 \times (0.12 \times 0.1) \times 0.2 = 0.24\ \text{N m}$
The plane makes $30^\circ$ with $B$ , so the normal (moment) makes $60^\circ$ : $\tau = NIAB\sin 60^\circ = NIAB\cos 30^\circ$ .
$\tau = 0.24 \times 0.866 \approx 0.21\ \text{N m}$ , nearest option 0.20 N m.
So the answer is (d).

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Q81JEE Main 2014 · 9 APR · SHIFT 1Medium

Field of a Magnet & Magnet in a Uniform Field · Field of a bar magnet

The mid points of two small magnetic dipoles of length d in end-on positions, are separated by a distance x, (x > > d). The force between them is proportional to x $-$ n where n is:

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Answer: (d)

Field of the first dipole on its axis: $B = \dfrac{\mu_0}{4\pi}\dfrac{2m_1}{x^3}$ .
Force on the second dipole along the axis: $F = m_2\dfrac{dB}{dx} = \dfrac{\mu_0}{4\pi}\dfrac{6m_1m_2}{x^4}$
So $F \propto x^{-4}$ and the answer is (d) $n = 4$ .

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Q82NEET 2013Easy

Field of a Magnet & Magnet in a Uniform Field · Torque on a magnet or current loop

A bar magnet of magnetic moment M is placed at right angles to a magnetic induction B. If a force F is experienced by each pole of the magnet, the length of the magnet will be

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Answer: (a)

Perpendicular to $B$ , the two pole forces form a couple: $\tau = F \times l$ .
Also $\tau = MB\sin 90^\circ = MB$ .
$Fl = MB \Rightarrow l = \dfrac{MB}{F}$
So the answer is (a).

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Q83NEET 2013Easy

Field of a Magnet & Magnet in a Uniform Field · Torque on a magnet or current loop

A current loop in a magnetic field

Show answer and solution

Answer: (b)

$\tau = mB\sin\theta = 0$ at $\theta = 0^\circ$ and $\theta = 180^\circ$ : two equilibrium positions.
$U = -mB\cos\theta$ is minimum at $0^\circ$ (stable) and maximum at $180^\circ$ (unstable).
So the answer is (b).

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Q84JEE Main 2013Difficult

Field of a Magnet & Magnet in a Uniform Field · Field of a bar magnet

Two short bar magnets of length $1\text{ cm}$ each have magnetic moments $1.20\text{ Am}^{2}$ and $1.00\text{ Am}^{2}$ respectively. They are placed on a horizontal table parallel to each other with their $N$ poles pointing towards the South. They have a common magnetic equator and are separated by a distance of $20.0\text{ cm.}$ The value of the resultant horizontal magnetic induction at the mid-point $O$ of the line joining their centres is close to $($ Horizontal component of earth's magnetic induction is $3.6 \times 10^{-5}\text{ Wb/m}^{2})$

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Answer: (b)

O lies on the common equatorial line of both magnets. Their moments point south, so each equatorial field at O points north (opposite to $M$ ), same as $B_H$ .
$r = 0.1\ \text{m}$ : $B_1 + B_2 = \frac{\mu_0}{4\pi}\frac{M_1+M_2}{r^3} = 10^{-7} \times \frac{2.2}{10^{-3}} = 2.2 \times 10^{-4}\ \text{T}$
$B = 2.2 \times 10^{-4} + 0.36 \times 10^{-4} = 2.56 \times 10^{-4}\ \text{T}$
So the answer is (b) $2.56 \times 10^{-4}\ \text{Wb/m}^2$ .

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Q85NEET 2012 · AIPMT · MAINSMedium

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}J$ of work to turn it through 60°. The torque needed to maintain the needle in this position will be

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Answer: (b)

$W = MB(1-\cos 60^\circ) = \frac{MB}{2} = \sqrt{3}\ \text{J} \Rightarrow MB = 2\sqrt{3}\ \text{J}$
$\tau = MB\sin 60^\circ = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3$
So the answer is (b) $3\ \text{J}$ (i.e. 3 N m).

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Q86NEET 2011 · AIPMT · MAINSEasy

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A short bar magnet of magnetic moment 0.4 J T ${}^{-1}$ is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy is

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Answer: (b)

Stable equilibrium: $\theta = 0^\circ$ , so $U = -MB$ .
$U = -0.4 \times 0.16 = -0.064\ \text{J}$
So the answer is (b).

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Q87NEET 2010 · AIPMT · MAINSEasy

Field of a Magnet & Magnet in a Uniform Field · Field of a bar magnet

Two identical bar magnets are fixed with their centers at a distance d apart. A stationary charge Q is placed at P in between the gap of the two magnets at a distance D from the centre O as shown in the figure

The force on the charge Q is

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Answer: (a)

Magnetic force on a charge: $\vec F = Q\,\vec v \times \vec B$ .
The charge is stationary, $v = 0$ , so $F = 0$ whatever the field at P.
So the answer is (a) zero.

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Q88NEET 2009 · AIPMTEasy

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A bar magnet having a magnetic moment of 2 $\times$ 104 J T ${}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 $\times$ 10 $-$ 4 T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is

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Answer: (b)

$W = MB(\cos\theta_1 - \cos\theta_2) = MB(1 - \cos 60^\circ)$
$W = 2 \times 10^{4} \times 6 \times 10^{-4} \times 0.5 = 6\ \text{J}$
So the answer is (b) 6 J.

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Q89JEE Main 2005 · AIEEEEasy

Field of a Magnet & Magnet in a Uniform Field · Torque on a magnet or current loop

A magnetic needle is kept in a non-uniform magnetic field. It experiences:

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Answer: (d)

In a non-uniform field the forces on the two poles are unequal, so a net force acts.
The pole forces are also not along the same line in general, so a torque acts too.
So the answer is (d) a force and a torque.

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Q90NEET 2005 · AIPMTDifficult

Field of a Magnet & Magnet in a Uniform Field · Torque on a magnet or current loop

A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $\vec{B}$ is in plane of the coil. If due to a current i in the triangle a torque $\tau$ acts on it, the side $l$ of the triangle is

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Answer: (b)

$\vec{B}$ in the plane of the coil, so $\vec{m} \perp \vec{B}$ : $\tau = iAB$ .
Area of equilateral triangle $A = \frac{\sqrt{3}}{4}l^2 \Rightarrow \tau = \frac{\sqrt{3}}{4}l^2 iB$
$l^2 = \frac{4\tau}{\sqrt{3}Bi} \Rightarrow l = 2\left(\frac{\tau}{\sqrt{3}Bi}\right)^{1/2}$
So the answer is (b).

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Q91JEE Main 2003 · AIEEEMedium

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^\circ.$ The torque needed to maintain the needle in this position will be:

Show answer and solution

Answer: (a)

$W = MB(1-\cos 60^\circ) = \frac{MB}{2} \Rightarrow MB = 2W$
$\tau = MB\sin 60^\circ = 2W \cdot \frac{\sqrt{3}}{2} = \sqrt{3}W$
So the answer is (a) $\sqrt{3}\,W$ .

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Q92NEET 1999 · AIPMTEasy

Field of a Magnet & Magnet in a Uniform Field · Torque on a magnet or current loop

A bar magnet of magnetic moment $\vec{M}$ is placed in a magnetic field of induction $\vec{B}$ . The torque exerted on it is

Show answer and solution

Answer: (c)

Torque on a dipole in a field: $\vec\tau = \vec M \times \vec B$ , magnitude $MB\sin\theta$ .
So the answer is (c) $\vec M \times \vec B$ .

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Q93NEET 1995 · AIPMTMedium

Field of a Magnet & Magnet in a Uniform Field · Work & potential energy

The work done in turning a magnet of magnetic moment $M$ by an angle of $90^\circ$ from the meridian, is $n$ times the corresponding work done to turn it through an angle of $60^\circ$ . The value of $n$ is given by

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Answer: (c)

Work from the meridian: $W = MB(1-\cos\theta)$ .
$W_{90} = MB(1-0) = MB$ , $W_{60} = MB(1-\frac{1}{2}) = \frac{MB}{2}$
$n = \frac{W_{90}}{W_{60}} = 2$
So the answer is (c) $2$ .

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Q94JEE Main 2022 · 29 JUL · SHIFT 2EasyOld syllabus

Earth's Magnetism · Dip, BH, BV & Earth as a dipole

The vertical component of the earth's magnetic field is $6 \times 10^{-5}\text{ T}$ at any place where the angle of dip is $37^\circ$ . The earth's resultant magnetic field at that place will be $($ Given $\tan 37^\circ = \frac{3}{4})$

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Answer: (d)

$V = B_E\sin\delta$ ; $\tan 37^\circ = \frac{3}{4} \Rightarrow \sin 37^\circ = \frac{3}{5}$
$B_E = \frac{V}{\sin 37^\circ} = \frac{6 \times 10^{-5}}{0.6} = 1 \times 10^{-4}\ \text{T}$
So the answer is (d) $1 \times 10^{-4}\ \text{T}$ .

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Q95JEE Main 2022 · 29 JUN · SHIFT 1EasyOld syllabus

Earth's Magnetism · Dip, BH, BV & Earth as a dipole

At a certain place the angle of dip is 30 $^\circ$ and the horizontal component of earth's magnetic field is 0.5 G. The earth's total magnetic field (in G), at that certain place, is:

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Answer: (a)

$H = B_E\cos\delta \Rightarrow B_E = \frac{H}{\cos 30^\circ}$
$B_E = \frac{0.5}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}\ \text{G}$
So the answer is (a) $\frac{1}{\sqrt{3}}$ .

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Q96JEE Main 2022 · 27 JUL · SHIFT 1MediumOld syllabus

Earth's Magnetism · Apparent dip

A magnet hung at $45^\circ$ with magnetic meridian makes an angle of $60^\circ$ with the horizontal. The actual value of the angle of dip is -

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Answer: (a)

Apparent dip: $\tan\delta' = \frac{\tan\delta}{\cos\alpha} \Rightarrow \tan\delta = \tan\delta'\cos\alpha$
$\tan\delta = \tan 60^\circ\cos 45^\circ = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}}$
So the answer is (a) $\tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$ .

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Q97JEE Main 2021 · 22 JUL · SHIFT 2EasyOld syllabus

Earth's Magnetism · Apparent dip

Choose the correct option

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Answer: (c)

In a vertical plane at angle $\theta$ to the meridian only $H\cos\theta$ acts horizontally, so $\tan\delta' = \frac{\tan\delta}{\cos\theta}$ .
$\cos\theta \le 1 \Rightarrow \tan\delta' \ge \tan\delta$ , i.e. apparent dip $\ge$ true dip.
So true dip is less than the apparent dip: answer (c).

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Q98JEE Main 2021 · 20 JUL · SHIFT 2MediumOld syllabus

Earth's Magnetism · Apparent dip

At an angle of 30 $^\circ$ to the magnetic meridian, the apparent dip is 45 $^\circ$ . Find the true dip:

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Answer: (d)

$\tan\delta = \tan\delta'\cos\alpha$
$\tan\delta = \tan 45^\circ \cos 30^\circ = \frac{\sqrt{3}}{2}$
So the answer is (d) $\tan^{-1}\frac{\sqrt{3}}{2}$ .

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Q99JEE Main 2021 · 16 MAR · SHIFT 1MediumOld syllabus

Earth's Magnetism · Neutral points

A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the center of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is: (1G = 10 $-$ 4 T)

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Answer: (b)

N pole towards north: neutral point on the equatorial line, $\frac{\mu_0}{4\pi}\frac{M}{(r^2+l^2)^{3/2}} = B_H$ .
$l = 7$ cm, $r = 18$ cm: $r^2 + l^2 = 373\ \text{cm}^2$ , $(373)^{3/2} \approx 7203\ \text{cm}^3 = 7.2 \times 10^{-3}\ \text{m}^3$
$M = \frac{0.4 \times 10^{-4} \times 7.2 \times 10^{-3}}{10^{-7}} \approx 2.88\ \text{J T}^{-1}$
So the answer is (b) 2.880 J T $^{-1}$ .

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Q100MHT CET 2020 · 16 OCT · SHIFT 2Medium

Earth's Magnetism · Dip, BH, BV & Earth as a dipole

If the angle of dip at places $A$ and $B$ are $30^\circ$ and $45^\circ$ respectively, then the ratio of horizontal component of earth's magnetic field at $A$ to that at $B$ will be

$\left[\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}, \sin\frac{\pi}{6} = \frac{1}{2}, \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\right]$

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Answer: (c)

$H = B_E\cos\delta$ ; taking the same total field $B_E$ at both places,
$\frac{H_A}{H_B} = \frac{\cos 30^\circ}{\cos 45^\circ} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}$
So the answer is (c) $\sqrt{3} : \sqrt{2}$ .

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Q101NEET 2019Easy

Earth's Magnetism · Dip, BH, BV & Earth as a dipole

The relations amongst the three elements of earth's magnetic field, namely horizontal component $H$ , vertical component $V$ and dip $\delta$ are, ( $B_E$ = total magnetic field)

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Answer: (c)

$B_E$ makes angle $\delta$ with the horizontal.
Horizontal component $H = B_E\cos\delta$ , vertical component $V = B_E\sin\delta$ .
So the answer is (c).

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Q102NEET 2019EasyOld syllabus

Earth's Magnetism · Dip, BH, BV & Earth as a dipole

At a point A on the earth’s surface the angle of dip, $\delta$ = + 25°. At a point B on the earth's surface the angle of dip, $\delta$ = – 25°. We can interpret that:

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Answer: (b)

Positive dip: the north end of the needle dips below horizontal, which happens in the northern hemisphere.
Negative dip: the north end points upward (south end dips), which happens in the southern hemisphere.
So the answer is (b) A is in the northern and B in the southern hemisphere.

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Q103JEE Main 2019 · 10 JAN · SHIFT 2DifficultOld syllabus

Earth's Magnetism · Dip, BH, BV & Earth as a dipole

At some location on earth, the horizontal component of earth’s magnetic field is 18 × 10−6 T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid- point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is -

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Answer: (d)

Needle rests at $45^\circ$ (the dip), so $B_V = B_H\tan 45^\circ = 18 \times 10^{-6}\ \text{T}$ .
When horizontal, $B_V$ gives a couple $\tau = mB_V L$ about the midpoint.
Force $F$ at one end (arm $L/2$ ): $F\frac{L}{2} = mB_V L \Rightarrow F = 2mB_V$
$F = 2 \times 1.8 \times 18 \times 10^{-6} = 6.48 \times 10^{-5}\ \text{N}$
So the answer is (d) $6.5 \times 10^{-5}$ N.

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Q104NEET 2017Medium

Earth's Magnetism · Apparent dip

If $\theta$ 1 and $\theta$ 2 be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $\theta$ is given by

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Answer: (d)

In perpendicular planes: $\tan\theta_1 = \frac{V}{H\cos\alpha}$ , $\tan\theta_2 = \frac{V}{H\sin\alpha}$ .
$\cot^2\theta_1 + \cot^2\theta_2 = \frac{H^2(\cos^2\alpha + \sin^2\alpha)}{V^2} = \frac{H^2}{V^2} = \cot^2\theta$
So the answer is (d) $\cot^2\theta = \cot^2\theta_1 + \cot^2\theta_2$ .

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Q105JEE Main 2015 · 11 APR · SHIFT 1Medium

Earth's Magnetism · Neutral points

A short bar magnet is placed in the magnetic meridian of the earth with north pole pointing north. Neutral points are found at a distance of 30 cm from the magnet on the east - West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in Am2 is close to: (Given $\frac{\mu_0}{4\pi}$ = 10 $-$ 7 in SI units and BH = Horizontal component of earth's magnetic field = 3.6 $\times$ 10 $-$ 5 Tesla.)

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Answer: (a)

N pole pointing north: neutral points lie on the equatorial (east-west) line, where the magnet's field points south and cancels $B_H$ .
$\frac{\mu_0}{4\pi}\frac{M}{r^3} = B_H \Rightarrow M = \frac{B_H r^3}{10^{-7}}$
$M = \frac{3.6 \times 10^{-5} \times (0.3)^3}{10^{-7}} = 3.6 \times 10^{-5} \times 2.7 \times 10^{5} = 9.72$
So the answer is (a) 9.7 A m $^2$ .

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Q106JEE Main 2014 · 9 APR · SHIFT 1MediumOld syllabus

Earth's Magnetism · Dip, BH, BV & Earth as a dipole

The magnetic field of earth at the equator is approximately 4 $\times$ 10 $-$ 5 T. The radius of earth is 6.4 $\times$ 106 m. Then the dipole moment of the earth will be nearly of the order of:

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Answer: (a)

Equatorial field of earth's dipole: $B = \frac{\mu_0}{4\pi}\frac{M}{R^3}$
$M = \frac{BR^3}{10^{-7}} = \frac{4 \times 10^{-5} \times (6.4 \times 10^6)^3}{10^{-7}}$
$(6.4 \times 10^6)^3 \approx 2.6 \times 10^{20} \Rightarrow M \approx 1.05 \times 10^{23}\ \text{A m}^2$
So the answer is (a) $10^{23}\ \text{A m}^2$ .

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Q107JEE Main 2013 · 25 APR · SHIFT 1MediumOld syllabus

Earth's Magnetism · Dip, BH, BV & Earth as a dipole

The earth's magnetic field lines resemble that of a dipole at the centre of the earth. If the magnetic moment of this dipole is close to $8 \times 10^{22}$ Am $^{2}$ , the value of earth's magnetic field near the equator is close to (radius of the earth = $6.4 \times 10^{6}$ m)

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Answer: (d)

$B_{eq} = \frac{\mu_0}{4\pi}\frac{M}{R^3}$
$B = 10^{-7} \times \frac{8 \times 10^{22}}{(6.4 \times 10^6)^3} = \frac{8 \times 10^{15}}{2.62 \times 10^{20}} \approx 3.05 \times 10^{-5}\ \text{T}$
$1\ \text{G} = 10^{-4}\ \text{T} \Rightarrow B \approx 0.3\ \text{G}$
So the answer is (d) 0.32 Gauss.

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Q108NEET 2012 · AIPMT · PRELIMSEasy

Earth's Magnetism · Dip, BH, BV & Earth as a dipole

A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It

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Answer: (b)

At a magnetic pole the earth's field is vertical, so $H = B_E\cos 90^\circ = 0$ .
A needle free only in the horizontal plane feels no horizontal torque, so it rests in any direction.
So the answer is (b) it will stay in any position.

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Q109NEET 2001 · AIPMTEasy

Earth's Magnetism · Tangent galvanometer

Tangent galvanometer is used to measure

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Answer: (b)

In a tangent galvanometer $I = \frac{2rB_H}{\mu_0 N}\tan\theta$ , i.e. $I \propto \tan\theta$ .
The deflection gives the current directly.
So the answer is (b) current.

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Q110MHT CET 2026 · 20 APR · SHIFT 1Easy

Vibration Magnetometer · Time period: T = 2π√(I/MB)

The periodic time of a magnet in vibration magnetometer is $T_0$ . If a magnet is replaced by another magnet whose moment of inertia is three times and magnetic moment is one-third of the original magnet, then time period of new magnet will be

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Answer: (c)

$T = 2\pi\sqrt{\frac{I}{MB}} \Rightarrow T \propto \sqrt{\frac{I}{M}}$
$\frac{T}{T_0} = \sqrt{\frac{3I}{I} \cdot \frac{M}{M/3}} = \sqrt{9} = 3$
So the answer is (c) $3T_0$ .

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Q111MHT CET 2026 · 19 APR · SHIFT 1Medium

Vibration Magnetometer · Comparing fields at two places

The frequency of oscillation of a small magnet in a magnetic field of induction 'B' is ' $n$ '. If the frequency of oscillations of the same magnet in a field of induction 'X' falls to $\left(\frac{n}{3}\right)$ , the value of 'X' is

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Answer: (d)

$n = \frac{1}{2\pi}\sqrt{\frac{MB}{I}} \Rightarrow n \propto \sqrt{B}$
$\frac{n/3}{n} = \sqrt{\frac{X}{B}} \Rightarrow \frac{X}{B} = \frac{1}{9}$
So the answer is (d) $\frac{B}{9}$ .

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Q112MHT CET 2025 · 20 APR · SHIFT 2Medium

Vibration Magnetometer · Sum & difference positions

Two bar magnets A and B are geometrically similar but the magnetic moment of $A$ is twice that of $B.\ T_1$ is the time period of oscillation when their like poles are kept together. When unlike poles are kept together, the time period of oscillation is $T_2$ . The ratio $T_1 : T_2$ will be

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Answer: (c)

Both arrangements have the same moment of inertia $2I$ , so $T \propto \dfrac{1}{\sqrt{M_{net}}}$ .
Let $M_B = M$ , $M_A = 2M$ . Like poles together (sum position): $M_{net} = 2M + M = 3M$ .
Unlike poles together (difference position): $M_{net} = 2M - M = M$ .
$\dfrac{T_1}{T_2} = \sqrt{\dfrac{M}{3M}} = \dfrac{1}{\sqrt{3}}$
So the answer is (c) $1 : \sqrt{3}$ .

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Q113MHT CET 2024 · 3 MAY · SHIFT 1Medium

Vibration Magnetometer · Time period: T = 2π√(I/MB)

A magnetic needle of magnetic moment $6 \times 10^{-2}\text{ Am}^{2}$ and moment of inertia $9.6 \times 10^{-5}\text{ kg m}^{2}$ performs simple harmonic motion in a magnetic field of 0.01 T. Time taken to complete 10 oscillations is [Take $\pi = 3.14$ ]

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Answer: (c)

$T = 2\pi\sqrt{\frac{I}{MB}} = 2\pi\sqrt{\frac{9.6 \times 10^{-5}}{6 \times 10^{-2} \times 0.01}} = 2\pi\sqrt{0.16}$
$T = 2 \times 3.14 \times 0.4 = 2.512\ \text{s}$
10 oscillations: $10T = 25.12\ \text{s}$
So the answer is (c) 25.12 s.

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Q114NEET 2024Medium

Vibration Magnetometer · Time period: T = 2π√(I/MB)

The magnetic moment and moment of inertia of a magnetic needle as shown are, respectively, $1.0 \times 10^{-2}\text{ A m}^{2}$ and $\frac{10^{-6}}{\pi^{2}}\text{ kg m}^{2}$ . If it completes 10 oscillations in $10\text{ s}$ , the magnitude of the magnetic field is

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Answer: (c)

10 oscillations in 10 s: $T = 1$ s.
$T = 2\pi\sqrt{\frac{I}{MB}} \Rightarrow B = \frac{4\pi^2 I}{MT^2}$
$B = \frac{4\pi^2 \times 10^{-6}/\pi^2}{10^{-2} \times 1} = 4 \times 10^{-4}\ \text{T} = 0.4\ \text{mT}$
So the answer is (c) 0.4 mT.

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Q115NEET 2024Medium

Vibration Magnetometer · Time period: T = 2π√(I/MB)

In a uniform magnetic field of $0.049\text{ T}$ , a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is $9.8 \times 10^{-6}\text{ kg m}^{2}$ . If the magnitude of magnetic moment of the needle is $x \times 10^{-5}\text{ Am}^{2}$ , then the value of ' $x$ ' is:

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Answer: (d)

20 oscillations in 5 s: $T = 0.25$ s.
$M = \frac{4\pi^2 I}{BT^2} = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{0.049 \times 0.0625}$
$M = 4\pi^2 \times 3.2 \times 10^{-3} = 1280\pi^2 \times 10^{-5}\ \text{A m}^2$
So $x = 1280\pi^2$ , answer (d).

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Q116MHT CET 2024 · 2 MAY · SHIFT 1Medium

Vibration Magnetometer · Comparing fields at two places

At certain place a magnet makes 30 oscillations per minute. At another place if the magnetic induction is increased by two times the magnetic induction at first place, then the time period of same magnet will be

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Answer: (a)

30 oscillations per minute: $T_1 = \frac{60}{30} = 2$ s.
Field increased by two times the original: $B_2 = B + 2B = 3B$ .
$T \propto \frac{1}{\sqrt{B}} \Rightarrow T_2 = \frac{2}{\sqrt{3}}\ \text{s}$
So the answer is (a) $\frac{2}{\sqrt{3}}$ s.

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Q117JEE Main 2023 · 12 APR · SHIFT 1DifficultOld syllabus

Vibration Magnetometer · Comparing fields at two places

A compass needle oscillates 20 times per minute at a place where the dip is $30^\circ$ and 30 times per minute where the dip is $60^\circ$ . The ratio of total magnetic field due to the earth at two places respectively is $\frac{4}{\sqrt{x}}$ . The value of $x$ is

Numerical answer type. Work it out, then open the answer.

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Answer: 243

Frequency $n \propto \sqrt{B\cos\delta}$ , so $n^2 \propto B\cos\delta$ .
$\dfrac{20^2}{30^2} = \dfrac{B_1\cos 30^\circ}{B_2\cos 60^\circ} \Rightarrow \dfrac{4}{9} = \dfrac{B_1}{B_2}\cdot\sqrt{3}$
$\dfrac{B_1}{B_2} = \dfrac{4}{9\sqrt{3}} = \dfrac{4}{\sqrt{81\times 3}} = \dfrac{4}{\sqrt{243}}$
So $x = 243$ .

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Q118JEE Main 2022 · 27 JUL · SHIFT 1MediumOld syllabus

Vibration Magnetometer · Time period: T = 2π√(I/MB)

Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of $3\text{ s}$ and $4\text{ s}$ respectively. If their moments of inertia are in the ratio of $3 : 2$ , then the ratio of their magnetic moments will be:

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Answer: (b)

$T = 2\pi\sqrt{\frac{I}{MB_H}} \Rightarrow M \propto \frac{I}{T^2}$
$\frac{M_1}{M_2} = \frac{I_1}{I_2}\cdot\frac{T_2^2}{T_1^2} = \frac{3}{2} \times \frac{16}{9} = \frac{8}{3}$
So the answer is (b) 8 : 3.

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Q119JEE Main 2022 · 27 JUL · SHIFT 2DifficultOld syllabus

Vibration Magnetometer · Comparing fields at two places

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place $\mathrm{P}$ of $\text{dip } 30^\circ$ . The number of oscillations per minute become 10 at another place $\mathrm{Q}$ of $60^\circ$ dip. The ratio of the total magnetic field at the two places $(B_Q : B_P)$ is:

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Answer: (a)

Frequency $n \propto \sqrt{B_H}$ and $B_H = B\cos\delta$ , so $n^2 \propto B\cos\delta$ .
$\dfrac{n_P^2}{n_Q^2} = \dfrac{B_P\cos 30^\circ}{B_Q\cos 60^\circ} \Rightarrow \dfrac{400}{100} = \dfrac{B_P}{B_Q}\cdot\dfrac{\sqrt{3}/2}{1/2}$
$4 = \sqrt{3}\,\dfrac{B_P}{B_Q} \Rightarrow \dfrac{B_Q}{B_P} = \dfrac{\sqrt{3}}{4}$
So the answer is (a) $\sqrt{3} : 4$ .

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Q120MHT CET 2021 · 24 SEP · SHIFT 2Medium

Vibration Magnetometer · Time period: T = 2π√(I/MB)

Two bar magnets ' $\mathrm{P}$ ' and ' $\mathrm{Q}$ ' are kept in uniform magnetic field ' $\mathrm{B}$ ' with magnetic moments ' $\mathrm{M}_{\mathrm{P}}$ ' and ' $\mathrm{M}_{\mathrm{Q}}$ ' respectively. Magnet 'P' is oscillating with frequency twice that of magnet 'Q'. If the moment of inertia of the magnet 'P' is twice that of magnet 'Q' then

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Answer: (c)

$f = \frac{1}{2\pi}\sqrt{\frac{MB}{I}} \Rightarrow M \propto f^2 I$ (same $B$ )
$\frac{M_P}{M_Q} = \left(\frac{f_P}{f_Q}\right)^2\frac{I_P}{I_Q} = 4 \times 2 = 8$
So the answer is (c) $M_P = 8M_Q$ .

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Q121JEE Main 2021 · 27 JUL · SHIFT 1Medium

Vibration Magnetometer · Time period: T = 2π√(I/MB)

In a uniform magnetic field, the magnetic needle has a magnetic moment 9.85 $\times$ 10 $-$ 2 A m2 and moment of inertia 5 $\times$ 10 $-$ 6 kg m2. If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is _______________ mT. [Take $\pi$ 2 as 9.85]

Numerical answer type. Work it out, then open the answer.

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Answer: 8

10 oscillations in 5 s: $T = 0.5$ s.
$B = \frac{4\pi^2 I}{MT^2} = \frac{4 \times 9.85 \times 5 \times 10^{-6}}{9.85 \times 10^{-2} \times 0.25}$
$B = \frac{20 \times 10^{-6}}{2.5 \times 10^{-3}} = 8 \times 10^{-3}\ \text{T}$
So the answer is 8 mT.

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Q122JEE Main 2019 · 12 APR · SHIFT 1DifficultOld syllabus

Vibration Magnetometer · Comparing fields at two places

A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45°, and 40 times per minute where the dip is 30°. If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio $\frac{B_1}{B_2}$ is best given by:

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Answer: (c)

Frequency $n \propto \sqrt{B_H}$ with $B_H = B\cos\delta$ , so $n^2 \propto B\cos\delta$ .
$\dfrac{30^2}{40^2} = \dfrac{B_1\cos 45^\circ}{B_2\cos 30^\circ} \Rightarrow \dfrac{9}{16} = \dfrac{B_1}{B_2}\cdot\dfrac{0.707}{0.866}$
$\dfrac{B_1}{B_2} = 0.5625 \times 1.225 \approx 0.69$
So the answer is (c) 0.7.

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Q123JEE Main 2017Medium

Vibration Magnetometer · Time period: T = 2π√(I/MB)

A magnetic needle of magnetic moment 6.7 $\times$ 10−2 A m2 and moment of inertia 7.5 $\times$ 10−6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:

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Answer: (b)

$T = 2\pi\sqrt{\frac{I}{MB}} = 2\pi\sqrt{\frac{7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times 0.01}}$
$\frac{I}{MB} = 0.0112 \Rightarrow T = 2\pi \times 0.1058 \approx 0.665\ \text{s}$
10 oscillations: $10T \approx 6.65\ \text{s}$
So the answer is (b) 6.65 s.

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Q124NEET 2010 · AIPMT · PRELIMSMedium

Vibration Magnetometer · Comparing fields at two places

A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in earth's horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be

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Answer: (d)

For a vibration magnetometer $T = 2\pi\sqrt{\dfrac{I}{MB_H}}$ , so $T \propto \dfrac{1}{\sqrt{B_H}}$ .
Net horizontal field $= 24 - 18 = 6\ \mu\text{T}$ (wire's field is opposite to earth's field).
$\dfrac{T'}{T} = \sqrt{\dfrac{24}{6}} = 2 \Rightarrow T' = 2 \times 2 = 4\ \text{s}$
So the answer is (d) 4 s.

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Q125JEE Main 2004 · AIEEEDifficult

Vibration Magnetometer · Cut, stacked or heavier magnets

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2\text{ s.}$ The magnet is cut along its length into three equal parts and these parts are then placed on each other with their like poles together. The time period of this combination will be

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Answer: (b)

Taking the three pieces as each of length $\dfrac{l}{3}$ : mass $\dfrac{m}{3}$ , moment $\dfrac{M}{3}$ , and $I_1 = \dfrac{(m/3)(l/3)^2}{12} = \dfrac{I}{27}$ .
Stacked with like poles together: $I' = 3\cdot\dfrac{I}{27} = \dfrac{I}{9}$ , $M' = 3\cdot\dfrac{M}{3} = M$ .
$T' = 2\pi\sqrt{\dfrac{I'}{M'B_H}} = \dfrac{T}{3} = \dfrac{2}{3}\ \text{s}$
So the answer is (b) $\dfrac{2}{3}$ s.

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Q126NEET 2003 · AIPMTMedium

Vibration Magnetometer · Cut, stacked or heavier magnets

A bar magnet is oscillating in the Earth's magnetic field with a period T. What happens to its period and motion if its mass is quadrupled?

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Answer: (b)

$T = 2\pi\sqrt{\dfrac{I}{MB_H}}$ and $I \propto m$ (same shape), so $T \propto \sqrt{m}$ .
The restoring torque is still $\propto \theta$ for small angles, so the motion stays SHM.
$m' = 4m \Rightarrow T' = \sqrt{4}\,T = 2T$
So the answer is (b) SHM with period $2T$ .

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Q127JEE Main 2003 · AIEEEMediumOld syllabus

Vibration Magnetometer · Cut, stacked or heavier magnets

A thin rectangular magnet suspended freely has a period of oscillation equal to $T.$ Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T',$ the ratio $\frac{T'}{T}$ is

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Answer: (b)

$T = 2\pi\sqrt{\dfrac{I}{MB_H}}$ with $I = \dfrac{m l^2}{12}$ for a thin magnet.
Each half: mass $\dfrac{m}{2}$ , length $\dfrac{l}{2}$ , so $I' = \dfrac{(m/2)(l/2)^2}{12} = \dfrac{I}{8}$ and $M' = \dfrac{M}{2}$ .
$\dfrac{T'}{T} = \sqrt{\dfrac{I'}{I}\cdot\dfrac{M}{M'}} = \sqrt{\dfrac{1}{8}\times 2} = \dfrac{1}{2}$
So the answer is (b) $\dfrac{1}{2}$ .

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Q128NEET 2002 · AIPMTEasy

Vibration Magnetometer · Sum & difference positions

Two bar magnets having same geometry with magnetic moments M and 2M, are firstly placed in such a way that their similar poles are same side then its time period of oscillation is T1. Now the polarity of one of the magnet is reversed then time period of oscillation is T2, then

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Answer: (a)

Like poles together (sum position): moment $3M$ , inertia $2I$ : $T_1 = 2\pi\sqrt{\frac{2I}{3MB}}$
One reversed (difference position): moment $M$ : $T_2 = 2\pi\sqrt{\frac{2I}{MB}}$
$\frac{T_1}{T_2} = \frac{1}{\sqrt{3}} < 1$
So the answer is (a) $T_1 < T_2$ .

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Q129NEET 1994 · AIPMTMedium

Vibration Magnetometer · Cut, stacked or heavier magnets

A bar magnet is oscillating in the Earth's magnetic field with a period $T$ . What happens to its period and motion if its mass is quadrupled?

Show answer and solution

Answer: (b)

$T = 2\pi\sqrt{\dfrac{I}{MB_H}}$ and $I \propto m$ (same shape), so $T \propto \sqrt{m}$ .
The restoring torque is still $\propto \theta$ for small angles, so the motion stays SHM.
$m' = 4m \Rightarrow T' = \sqrt{4}\,T = 2T$
So the answer is (b) SHM with period $2T$ .

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Q130MHT CET 2026 · 19 APR · SHIFT 1Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

A diamagnetic liquid is filled in a U-tube. One arm of a U-tube is placed in an external magnetic field with the meniscus in line with the field. The level of the liquid in that arm will

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Answer: (b)

A diamagnetic liquid is pushed from the stronger field region towards the weaker field region.
So the liquid in the arm inside the field is pushed out and its level goes down.
So the answer is (b) fall.

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Q131MHT CET 2026 · 15 APR · SHIFT 1Easy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

The product of magnetic susceptibility ( $\chi$ ) and absolute temperature (T) is constant for a

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Answer: (b)

Curie law: $\chi = \dfrac{C}{T} \Rightarrow \chi T = C$ (constant).
It holds for paramagnetic substances (ferromagnets follow $\chi = \dfrac{C}{T - T_C}$ only above $T_C$ ).
So the answer is (b) paramagnetic substance.

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Q132MHT CET 2026 · 11 APR · SHIFT 1Easy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

Which of the following figure best represents the variation of magnetic susceptibility ( $\chi$ ) with temperature for a diamagnetic substance?

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Answer: (c)

For a diamagnetic substance $\chi$ is negative and independent of temperature.
So the $\chi$ – $T$ graph is a horizontal line below the $T$ -axis.
So the answer is (c).

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Q133JEE Main 2026 · 23 JAN · SHIFT 1Medium

Magnetic Materials & Curie Law · Domains

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Consider a ferromagnetic material:

Assertion (A): The individual atoms in a ferromagnetic material possess a magnetic dipole moment and interact with one another in such a way that they spontaneously align themselves forming domains.

Reason (R): At high enough temperature, the domain structure of ferromagnetic material disintegrates. Thus, magnetization will disappear at high enough temperature known as Curie temperature.

In the light of the above statements, choose the correct answer from the options given below:

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Answer: (d)

(A) is true. Neighbouring atomic dipoles in a ferromagnet interact (exchange interaction) and align spontaneously, forming domains.
(R) is true. Above the Curie temperature, thermal agitation breaks up the domain structure and the magnetisation disappears.
(R) explains how domains are destroyed at high temperature. It does not explain why they form, which is what (A) states.
So the answer is (d): both are true, but (R) is not the correct explanation of (A).

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Q134MHT CET 2025 · 22 APR · SHIFT 1Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

Susceptibility of a paramagnetic substance is

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Answer: (d)

Paramagnetic dipoles align weakly along the field, so $\chi$ is positive but small (about $10^{-5}$ to $10^{-3}$ ).
So the answer is (d) positive and small.

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Q135MHT CET 2025 · 21 APR · SHIFT 1Easy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

The variation of magnetic susceptibility $(\chi)$ with temperature for a diamagnetic substance is shown correctly by graph

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Answer: (c)

Diamagnetic $\chi$ is negative and independent of $T$ : a horizontal line below the $T$ -axis.
Graph A shows exactly this (B is positive constant, C is ferromagnetic, D is paramagnetic).
So the answer is (c) A.

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Q136MHT CET 2024 · 16 MAY · SHIFT 1Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

The materials having negative magnetic susceptibility are

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Answer: (c)

Diamagnetic materials get magnetized opposite to the applied field, so $M$ is opposite to $H$ and $\chi = M/H < 0$ .
Para- and ferromagnetic materials have $\chi > 0$ .
So the answer is (c) diamagnetic.

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Q137NEET 2024Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

The incorrect relation for a diamagnetic material (all the symbols carry their usual meaning and $\varepsilon$ is a small positive number) is

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Answer: (d)

For a diamagnetic material $-1 \le \chi < 0$ , so $\mu_r = 1 + \chi$ lies in $0 \le \mu_r < 1$ and $\mu < \mu_0$ .
So (a), (b), (c) are correct relations.
$1 < \mu_r < 1 + \varepsilon$ is true for a paramagnetic material, not a diamagnetic one.
So the answer is (d).

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Q138NEET 2024Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

Match List I with List II

Choose the correct answer from the options given below.

	List I (Material)		List II (Susceptibility ($$\chi))
A.	Diamagnetic	I.	$\chi = 0$
B.	Ferromagnetic	II.	$0 > \chi \geq -1$
C.	Paramagnetic	III.	$\chi \gg 1$
D.	Non-magnetic	IV.	$0 < \chi < \varepsilon$ (a small positive number)

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Answer: (a)

Diamagnetic: small negative $\chi$ , $0 > \chi \ge -1$ (A-II).
Ferromagnetic: $\chi \gg 1$ (B-III).
Paramagnetic: small positive $\chi$ , $0 < \chi < \varepsilon$ (C-IV).
Non-magnetic: $\chi = 0$ (D-I).
So the answer is (a) A-II, B-III, C-IV, D-I.

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Q139JEE Main 2024 · 8 APR · SHIFT 1Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

Paramagnetic substances:

A. align themselves along the directions of external magnetic field.

B. attract strongly towards external magnetic field.

C. has susceptibility little more than zero.

D. move from a region of strong magnetic field to weak magnetic field.

Choose the most appropriate answer from the options given below:

Show answer and solution

Answer: (b)

A: true, the dipoles (and a paramagnetic rod) align along the external field.
B: false, the attraction is weak, not strong.
C: true, $\chi$ is small and positive (slightly more than zero).
D: false, they move from weak to strong field; strong to weak is diamagnetic behaviour.
So the answer is (b) A, C only.

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Q140MHT CET 2024 · 2 MAY · SHIFT 2Easy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

A ferromagnetic material is heated above its curie temperature. The correct statement from the following is that

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Answer: (b)

Above the Curie temperature thermal agitation breaks the alignment inside domains.
The domain structure becomes random and the material behaves as paramagnetic.
So the answer is (b) ferromagnetic domains become random.

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Q141JEE Main 2023 · 12 APR · SHIFT 1Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

Given below are two statements:

Statement I: The diamagnetic property depends on temperature.

Statement II: The induced magnetic dipole moment in a diamagnetic sample is always opposite to the magnetizing field.

In the light of given statements, choose the correct answer from the options given below.

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Answer: (a)

Statement I: diamagnetism is independent of temperature, so it is incorrect.
Statement II: by Lenz's law the induced moment always opposes the magnetizing field, so it is true.
So the answer is (a).

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Q142JEE Main 2023 · 10 APR · SHIFT 2Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

Given below are two statements:

Statement I: For diamagnetic substance, $-1 \le \chi < 0$ , where $\chi$ is the magnetic susceptibility. Statement II: Diamagnetic substances when placed in an external magnetic field, tend to move from stronger to weaker part of the field.

In the light of the above statements, choose the correct answer from the options given below

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Answer: (a)

Statement I: diamagnetic $\chi$ is negative, with $\chi = -1$ for a perfect diamagnet, so $-1 \le \chi < 0$ is true.
Statement II: diamagnetic substances move from stronger to weaker field, true.
So the answer is (a) both statements are true.

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Q143MHT CET 2023 · 13 MAY · SHIFT 1Easy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

For which of the following substances, the magnetic susceptibility is independent of temperature?

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Answer: (a)

Paramagnetic: $\chi \propto \dfrac{1}{T}$ . Ferromagnetic: $\chi$ depends strongly on $T$ .
Diamagnetism comes from induced orbital motion, unaffected by thermal agitation, so its $\chi$ is independent of $T$ .
So the answer is (a) diamagnetic only.

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Q144MHT CET 2023 · 10 MAY · SHIFT 1Easy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

According to Curie's law in magnetism, the correct relation is ( $\mathrm{M} =$ magnetization in paramagnetic sample, $\mathrm{B} =$ applied magnetic field, $\mathrm{T} =$ absolute temperature of the material, $\mathrm{C} =$ curie's constant)

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Answer: (b)

Curie law: magnetization $M \propto \dfrac{B}{T}$ .
$M = C\,\dfrac{B}{T}$
So the answer is (b) $M = \dfrac{CB}{T}$ .

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Q145NEET 2023Easy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

The variation of susceptibility $(\chi)$ with absolute temperature ( $\mathrm{T}$ ) for a paramagnetic material is represented as:

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Answer: (a)

Curie law: $\chi = \dfrac{C}{T} = C\cdot\dfrac{1}{T}$ .
So $\chi$ versus $\dfrac{1}{T}$ is a straight line through the origin with positive slope $C$ .
So the answer is (a).

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Q146JEE Main 2022 · 25 JUN · SHIFT 2Medium

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

Given below are two statements:

Statement I: Susceptibilities of paramagnetic and ferromagnetic substances increase with decrease in temperature.

Statement II: Diamagnetism is a result of orbital motions of electrons developing magnetic moments opposite to the applied magnetic field.

Choose the correct answer from the options given below:

Show answer and solution

Answer: (a)

Statement I is true. For a paramagnet $\chi = C/T$ , and for a ferromagnet above $T_C$ , $\chi = C/(T - T_C)$ . Both increase as $T$ decreases.
Statement II is true. An applied field changes the orbital motion of electrons and induces magnetic moments opposite to the field (Lenz's law). This is diamagnetism.
So the answer is (a): both statements are true.

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Q147JEE Main 2021 · 25 FEB · SHIFT 2Easy

Magnetic Materials & Curie Law · Domains

In a ferromagnetic material, below the curie temperature, a domain is defined as:

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Answer: (b)

Inside a domain, exchange coupling aligns all atomic dipoles in the same direction.
So each domain is a macroscopic region magnetized to saturation; domains differ only in direction.
So the answer is (b).

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Q148JEE Main 2021 · 24 FEB · SHIFT 2Easy

Magnetic Materials & Curie Law · Domains

A soft ferromagnetic material is placed in an external magnetic field. The magnetic domains:

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Answer: (a)

In a field, domains aligned with the field grow and others shrink (domain wall motion).
In stronger fields the domain magnetization also rotates towards the field.
So domains may increase or decrease in size and change orientation: the answer is (a).

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Q149JEE Main 2021 · 1 SEP · SHIFT 2Medium

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

Following plots show magnetization (M) vs magnetizing field (H) and magnetic susceptibility ( $\chi$ ) vs temperature (T) graph:

Which of the following combination will be represented by a diamagnetic material?

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Answer: (a)

For a diamagnet, $\chi$ is small and negative, so $M = \chi H$ points opposite to $H$ . The $M$ – $H$ graph is a straight line with negative slope: graph (a).
Diamagnetic $\chi$ does not depend on temperature, so the $\chi$ – $T$ graph is a horizontal line below the $T$ -axis: graph (c).
So the answer is (a): graphs (a) and (c).

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Q150JEE Main 2021 · 22 JUL · SHIFT 2Medium

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

Statement I: The ferromagnetic property depends on temperature. At high temperature, ferromagnet becomes paramagnet.

Statement II: At high temperature, the domain wall area of a ferromagnetic substance increases. In the light of the above statements, choose the most appropriate answer from the options given below:

Show answer and solution

Answer: (b)

Statement I is true. A ferromagnet becomes paramagnetic above its Curie temperature.
Statement II is false. At high temperature the domain structure is destroyed by thermal agitation. The domain wall area does not increase.
So the answer is (b): Statement I is true but Statement II is false.

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Q151MHT CET 2020 · 19 OCT · SHIFT 2Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

For a paramagnetic substance, the magnetic susceptibility is

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Answer: (b)

Paramagnetic: weak magnetization along the field, so $\chi$ is small and positive.
So the answer is (b) small and positive.

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Q152MHT CET 2020 · 16 OCT · SHIFT 2Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

A thin light weight rod of diamagnetic substance such as silver is suspended in uniform external magnetic field. It will align itself with its length

Show answer and solution

Answer: (b)

A diamagnetic rod tends to move from stronger to weaker field regions.
It reduces its energy by keeping as little of its length along the field as possible, so it sets perpendicular to the field.
So the answer is (b) perpendicular to the magnetic field.

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Q153JEE Main 2020 · 3 SEP · SHIFT 2Medium

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

A perfectly diamagnetic sphere has a small spherical cavity at its centre, which is filled with a paramagnetic substance. The whole system is placed in a uniform magnetic field $\vec{B}$ . Then the field inside the paramagnetic substance is:

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Answer: (b)

A perfect diamagnet has $\chi = -1$ , so $\mu_r = 0$ and $B = \mu_0 \mu_r H = 0$ inside it. It expels the field completely, as a superconductor does.
No field lines enter the diamagnetic shell, so none reach the cavity at its centre.
With no field applied to it, the paramagnetic substance has no magnetisation, so the field there is zero.
So the answer is (b) zero.

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Q154MHT CET 2020 · 16 OCT · SHIFT 2Medium

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

The magnetic susceptibility of a paramagnetic material at $-73^\circ\text{C}$ is 0.0075. Its value at $-173^\circ\text{C}$ will be

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Answer: (a)

Curie's law: $\chi \propto \dfrac{1}{T}$ , so $\dfrac{\chi_2}{\chi_1} = \dfrac{T_1}{T_2}$ .
$T_1 = -73 + 273 = 200\text{ K}$ and $T_2 = -173 + 273 = 100\text{ K}$ .
$\chi_2 = 0.0075 \times \dfrac{200}{100} = 0.0150$
So the answer is (a) 0.0150.

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Q155JEE Main 2020 · 4 SEP · SHIFT 2Medium

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

A paramagnetic sample shows a net magnetisation of 6 A/m when it is placed in an external magnetic field of 0.4 T at a temperature of 4 K. When the sample is placed in an external magnetic field of 0.3 T at a temperature of 24 K, then the magnetisation will be:

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Answer: (c)

Curie's law: $M = \chi H$ with $\chi \propto \dfrac{1}{T}$ , so $M \propto \dfrac{B}{T}$ .
$\dfrac{M_2}{M_1} = \dfrac{B_2}{B_1} \cdot \dfrac{T_1}{T_2} = \dfrac{0.3}{0.4} \times \dfrac{4}{24} = \dfrac{1}{8}$
$M_2 = \dfrac{6}{8} = 0.75\text{ A/m}$
So the answer is (c) 0.75 A/m.

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Q156MHT CET 2019 · 2 MAY · SHIFT 2Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

Magnetic susceptibility of a paramagnetic substance is

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Answer: (b)

Paramagnetic: weak magnetization along the field, so $\chi$ is small and positive.
So the answer is (b) small and positive.

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Q157JEE Main 2019 · 12 JAN · SHIFT 2Medium

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

A paramagnetic material has 1028 atoms/m3. Its magnetic susceptibility at temperature 350 K is 2.8 $\times$ 10–4. Its susceptibility at 300 K is:

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Answer: (d)

Curie's law: $\chi \propto \dfrac{1}{T}$ , so $\chi_2 = \chi_1 \dfrac{T_1}{T_2}$ . The number of atoms per $\text{m}^3$ is not needed.
$\chi_2 = 2.8 \times 10^{-4} \times \dfrac{350}{300}$
$\chi_2 = 3.267 \times 10^{-4}$
So the answer is (d) $3.267 \times 10^{-4}$ .

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Q158NEET 2018Medium

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from

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Answer: (a)

A diamagnetic rod is pushed from a strong field region to a weaker one, so it gains potential energy.
When the current is switched on, the changing flux induces an emf in the electromagnet coil. The source does extra work against this emf.
That extra energy from the current source becomes the rod's gain in potential energy.
So the answer is (a) the current source.

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Q159NEET 2016 · PHASE 1Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

The magnetic susceptibility is negative for

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Answer: (c)

Only diamagnetic materials are magnetized opposite to the field, giving $\chi < 0$ .
Para- and ferromagnetic materials have $\chi > 0$ .
So the answer is (c) diamagnetic material only.

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Q160JEE Main 2014 · 19 APR · SHIFT 1Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

An example of a perfect diamagnet is a superconductor. This implies that when a superconductor is put in a magnetic field of intensity B, the magnetic field Bs inside the superconductor will be such that:

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Answer: (b)

Perfect diamagnet: $\chi = -1$ , so $\mu_r = 1 + \chi = 0$ .
$B_s = \mu_0\mu_r H = 0$ : the field is completely expelled (Meissner effect).
So the answer is (b) $B_s = 0$ .

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Q161JEE Main 2014 · 11 APR · SHIFT 1Easy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

Three identical bars A, B and C are made of different magnetic materials. When kept in a uniform magnetic field, the field lines around them look as follows:

Make the correspondence of these bars with their material being diamagnetic (D), ferromagnetic (F) and paramagnetic (P):

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Answer: (b)

A: field lines crowd strongly into the bar, so it is ferromagnetic (F).
B: field lines bend away and avoid the bar, so it is diamagnetic (D).
C: lines are drawn in only slightly, so it is paramagnetic (P).
So the answer is (b) A-F, B-D, C-P.

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Q162NEET 2011 · AIPMT · PRELIMSEasy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

There are four light-weight-rod samples A, B, C, D separately suspended by threads. A bar magnet is slowly brought near each sample and the following observations are noted

(i) A is feebly repelled

(ii) B is feebly attracted

(iii) C is strongly attracted

(iv) D remains unaffected

Which one of the following is true?

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Answer: (a)

Feebly repelled: diamagnetic (A). Feebly attracted: paramagnetic (B).
Strongly attracted: ferromagnetic (C). Unaffected: non-magnetic (D).
Only statement (a) matches.
So the answer is (a) B is of a paramagnetic material.

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Q163NEET 2010 · AIPMT · MAINSEasy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

The magnetic moment of a diamagnetic atom is

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Answer: (d)

In a diamagnetic atom the orbital and spin moments of the paired electrons cancel completely.
So the net (permanent) magnetic moment of the atom is zero.
So the answer is (d) equal to zero.

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Q164NEET 2009 · AIPMTEasy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is

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Answer: (d)

A diamagnetic substance is magnetized opposite to the applied field whichever pole is near.
So the near face always has the same polarity as the magnet's pole and it moves to the weaker field region.
So the answer is (d) repelled by both the poles.

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Q165JEE Main 2008 · AIEEEEasy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

Relative permittivity and permeability of a material $\varepsilon_r$ and $\mu_r,$ respectively. Which of the following values of these quantities are allowed for a diamagnetic material?

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Answer: (b)

Diamagnetic: $\chi < 0$ , so $\mu_r < 1$ .
For any material medium $\varepsilon_r > 1$ (vacuum has 1).
Only $\varepsilon_r = 1.5$ , $\mu_r = 0.5$ satisfies both.
So the answer is (b).

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Q166NEET 2008 · AIPMTEasy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

Curie temperature above which

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Answer: (c)

Below the Curie temperature $T_C$ , a ferromagnet has aligned domains and strong magnetisation.
Above $T_C$ , thermal agitation destroys the domain alignment, and the material obeys the Curie–Weiss law $\chi = \dfrac{C}{T - T_C}$ like a paramagnet.
So the answer is (c): a ferromagnetic material becomes paramagnetic.

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Q167NEET 2007 · AIPMTEasy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show

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Answer: (d)

Nickel is ferromagnetic below its Curie temperature.
Above the Curie temperature the domains break up and the dipoles are only weakly aligned by the field, so $\chi = \dfrac{C}{T - T_C}$ is small and positive.
So the answer is (d) paramagnetism.

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Q168JEE Main 2006 · AIEEEEasy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

Needles $N_1, N_2$ and $N_3$ are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will

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Answer: (b)

Ferromagnetic $N_1$ : large positive $\chi$ , strongly attracted.
Paramagnetic $N_2$ : small positive $\chi$ , weakly attracted.
Diamagnetic $N_3$ : small negative $\chi$ , weakly repelled.
So the answer is (b).

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Q169NEET 2006 · AIPMTEasy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

Curie temperature above which

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Answer: (c)

Above the Curie temperature $T_C$ , thermal motion destroys the domain alignment of a ferromagnet.
The material then behaves like a paramagnet with $\chi = \dfrac{C}{T - T_C}$ .
So the answer is (c): a ferromagnetic material becomes paramagnetic.

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Q170NEET 2005 · AIPMTEasy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material are denoted by $\mu$ d, $\mu$ p and $\mu$ f respectively, then

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Answer: (a)

Diamagnetic atom: all electrons paired, net moment $\mu_d = 0$ .
Paramagnetic and ferromagnetic atoms have unpaired electrons, so $\mu_p \ne 0$ and $\mu_f \ne 0$ .
So the answer is (a) $\mu_d = 0$ and $\mu_p \ne 0$ .

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Q171JEE Main 2004 · AIEEEEasy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

Curie temperature is the temperature above which

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Answer: (a)

Above the Curie temperature, thermal agitation destroys the domain structure of a ferromagnet.
It then behaves as a paramagnet with $\chi = \dfrac{C}{T - T_C}$ .
So the answer is (a): a ferromagnetic material becomes paramagnetic.

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Q172NEET 2003 · AIPMTEasy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

A diamagnetic material in a magnetic field moves

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Answer: (a)

Induced moment is opposite to the field, so the energy $-\vec{m}\cdot\vec{B}$ is lower where $B$ is weaker.
So the material moves from stronger to weaker parts of the field.
So the answer is (a).

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Q173NEET 2003 · AIPMTEasy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

According to Curie's law, the magnetic susceptibility of a substance at an absolute temperature $T$ is proportional to

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Answer: (a)

Curie's law for a paramagnet: $\chi = \dfrac{C}{T}$ .
So $\chi \propto \dfrac{1}{T}$ .
So the answer is (a) $1/T$ .

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Q174NEET 2001 · AIPMTEasy

Magnetic Materials & Curie Law · Curie law, χ–T graphs & Curie temperature

Among which the magnetic susceptibility does not depend on the temperature?

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Answer: (a)

Diamagnetism comes from induced orbital moments that oppose the field. Thermal motion does not change these induced moments.
Para- and ferromagnetism depend on aligning permanent dipoles, and thermal agitation disturbs that alignment, so their $\chi$ depends on $T$ .
So the answer is (a) diamagnetism.

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Q175NEET 1999 · AIPMTEasy

Magnetic Materials & Curie Law · Dia, para & ferromagnetic properties

A diamagnetic substance is brought near a strong magnet, then it is

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Answer: (b)

A diamagnetic substance is magnetized opposite to the magnet's field, so it is pushed towards the weaker field.
This happens for both poles.
So the answer is (b) repelled by a magnet.

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Q176MHT CET 2026 · 19 MAY · SHIFT 2Easy

Hysteresis, Soft Iron vs Steel & Shielding · Soft iron vs steel: choice of material

Soft iron is used in making an electromagnet because

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Answer: (d)

An electromagnet must lose its magnetism as soon as the current is switched off.
Soft iron has low retentivity, low coercivity and high permeability, so it does this well.
So the answer is (d) it has low retentivity.

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Q177MHT CET 2025 · 22 APR · SHIFT 2Easy

Hysteresis, Soft Iron vs Steel & Shielding · Hysteresis curve: retentivity & coercivity

Magnetic hysteresis is exhibited by magnetic materials which are

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Answer: (c)

Hysteresis means $B$ lags behind $H$ and some magnetisation remains after $H$ is removed. This needs domains.
Only ferromagnetic materials have domains. Para- and diamagnets have no retentivity.
So the answer is (c) only ferromagnetic.

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Q178MHT CET 2025 · 25 APR · SHIFT 2Easy

Hysteresis, Soft Iron vs Steel & Shielding · Magnetic shielding

To protect the instrument from magnetic field, it is completely surrounded by

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Answer: (a)

A soft ferromagnetic material has very high permeability, so field lines crowd into the shell and pass around the inside.
The space enclosed by the shell is left almost field free. Dia- and paramagnets have $\mu_r \approx 1$ , so they cannot do this.
So the answer is (a) soft ferromagnetic substance.

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Q179MHT CET 2024 · 10 MAY · SHIFT 1Easy

Hysteresis, Soft Iron vs Steel & Shielding · Hysteresis curve: retentivity & coercivity

In the following graph of flux density versus magnetizing force, coercivity and retentivity are respectively represented by the points

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Answer: (a)

Retentivity is the value of $B$ left when $H = 0$ . It is the point where the loop cuts the $B$ -axis: point b.
Coercivity is the reverse $H$ needed to make $B = 0$ . It is the point where the loop cuts the negative $H$ -axis: point c.
So the answer is (a): c and b.

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Q180JEE Main 2024 · 8 APR · SHIFT 2Easy

Hysteresis, Soft Iron vs Steel & Shielding · Demagnetising current

The coercivity of a magnet is $5 \times 10^{3}\text{ A/m}$ . The amount of current required to be passed in a solenoid of length $30\text{ cm}$ and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is ________ A.

Numerical answer type. Work it out, then open the answer.

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Answer: 10

The solenoid's $H = nI$ must equal the coercivity: $I = \dfrac{H}{n}$ .
$n = \dfrac{150}{0.30} = 500\text{ m}^{-1}$
$I = \dfrac{5 \times 10^{3}}{500} = 10\text{ A}$
So the answer is 10 A.

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Q181MHT CET 2023 · 14 MAY · SHIFT 2Easy

Hysteresis, Soft Iron vs Steel & Shielding · Hysteresis curve: retentivity & coercivity

In the hysteresis curve the value of magnetization (B) which will be present in a substance when value of magnetizing force $(\mathrm{H})$ is made zero $(\mathrm{H} = 0)$ is called as

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Answer: (b)

When $H$ is reduced to zero, some magnetisation $B$ remains in the material.
This remaining value is the retentivity, also called remanence.
So the answer is (b) retentivity.

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Q182JEE Main 2023 · 10 APR · SHIFT 1Easy

Hysteresis, Soft Iron vs Steel & Shielding · Demagnetising current

The current required to be passed through a solenoid of 15 cm length and 60 turns in order of demagnetise a bar magnet of magnetic intensity $2.4 \times 10^{3}\text{ Am}^{-1}$ is ___________ A.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 6

The solenoid's $H = nI$ must equal the given magnetic intensity (the coercivity): $I = \dfrac{H}{n}$ .
$n = \dfrac{60}{0.15} = 400\text{ m}^{-1}$
$I = \dfrac{2.4 \times 10^{3}}{400} = 6\text{ A}$
So the answer is 6 A.

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Q183MHT CET 2023 · 9 MAY · SHIFT 1Easy

Hysteresis, Soft Iron vs Steel & Shielding · Soft iron vs steel: choice of material

The materials suitable for making electromagnets should have

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Answer: (b)

An electromagnet must lose its magnetism when the current stops, which needs low retentivity.
Low coercivity means it is easily demagnetised, and a narrow loop means little hysteresis loss.
So the answer is (b) low retentivity and low coercivity.

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Q184JEE Main 2023 · 8 APR · SHIFT 2Easy

Hysteresis, Soft Iron vs Steel & Shielding · Soft iron vs steel: choice of material

Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$

Assertion A: Electromagnets are made of soft iron.

Reason R: Soft iron has high permeability and low retentivity.

In the light of above, statements, choose the most appropriate answer from the options given below.

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Answer: (a)

A is true. Electromagnet cores are made of soft iron.
R is true. High permeability gives a strong field for a small current, and low retentivity means the core loses its magnetism when the current is switched off.
These are the reasons soft iron is used, so R explains A.
So the answer is (a).

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Q185MHT CET 2023 · 13 MAY · SHIFT 2Easy

Hysteresis, Soft Iron vs Steel & Shielding · Magnetic shielding

Magnetic shielding is done by surrounding the instrument to be protected from magnetic field by

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Answer: (a)

Shielding needs a material of very high permeability, so field lines run through the shell and not into the space inside.
Soft iron is such a material. Copper, aluminium and tantalum have $\mu_r \approx 1$ .
So the answer is (a) soft ferromagnetic substance soft iron.

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Q186JEE Main 2022 · 24 JUN · SHIFT 2Easy

Hysteresis, Soft Iron vs Steel & Shielding · Soft iron vs steel: choice of material

The soft-iron is a suitable material for making an electromagnet. This is because soft-iron has

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Answer: (c)

High permeability gives a strong magnetic field for a small current.
Low retentivity lets the magnet lose its magnetism when the current is switched off.
So the answer is (c) high permeability and low retentivity.

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Q187JEE Main 2020 · 7 JAN · SHIFT 2EasyOld syllabus

Hysteresis, Soft Iron vs Steel & Shielding · Hysteresis curve: retentivity & coercivity

The figure gives experimentally measured B vs H variation in a ferromagnetic material. The retentivity, coercivity and saturation, respectively, of the material are:

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Answer: (a)

Retentivity is $B$ at $H = 0$ , where the loop cuts the $B$ -axis: $1.0\text{ T}$ .
Coercivity is the $|H|$ at which $B = 0$ , where the loop cuts the $H$ -axis: $50\text{ A/m}$ .
Saturation is the maximum $B$ , at the tip of the loop: $1.5\text{ T}$ .
So the answer is (a) 1.0 T, 50 A/m and 1.5 T.

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Q188JEE Main 2020 · 2 SEP · SHIFT 1EasyOld syllabus

Hysteresis, Soft Iron vs Steel & Shielding · Soft iron vs steel: choice of material

Magnetic materials used for making permanent magnets (P) and magnets in a transformer (T) have different properties of the following, which property best matches for the type of magnet required?

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Answer: (b)

A permanent magnet must keep its magnetism (large retentivity) and resist demagnetisation (large coercivity).
A transformer core needs small coercivity and a narrow loop for low hysteresis loss. That rules out options (a) and (d).
So the answer is (b) P: large retentivity, large coercivity.

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Q189JEE Main 2019 · 9 JAN · SHIFT 1Easy

Hysteresis, Soft Iron vs Steel & Shielding · Demagnetising current

A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 A. The coercivity of the bar magnet is:

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Answer: (b)

To demagnetise the magnet, the solenoid field must equal the coercivity: $H = nI = \dfrac{N}{L} I$ .
$H = \dfrac{100}{0.2} \times 5.2 = 500 \times 5.2$
$H = 2600\text{ A/m}$
So the answer is (b) 2600 A/m.

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Q190JEE Main 2018 · 15 APR · SHIFT 1Medium

Hysteresis, Soft Iron vs Steel & Shielding · Demagnetising current

The $B\text{-}H$ curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a long solenoid with $1000$ turns/cm. The current that should be passed in the solenoid to demagnetise the ferromagnet completely is:

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Answer: (a)

From the graph, the loop cuts the $H$ -axis at $H = 100\text{ A/m}$ . This is the coercivity, the field needed to demagnetise the ferromagnet.
$n = 1000\text{ turns/cm} = 10^{5}\text{ m}^{-1}$
$I = \dfrac{H}{n} = \dfrac{100}{10^{5}} = 10^{-3}\text{ A}$
So the answer is (a) 1 mA.

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Q191JEE Main 2016MediumOld syllabus

Hysteresis, Soft Iron vs Steel & Shielding · Soft iron vs steel: choice of material

Hysteresis loops for two magnetic materials $A$ and $B$ are given below:

These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:

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Answer: (b)

Loop A is wide, so it has high retentivity and high coercivity. This suits permanent magnets, such as those in electric generators.
Loop B is narrow, so it has low coercivity, small hysteresis loss and is easily demagnetised. This suits transformer cores and electromagnets.
So the answer is (b): B for electromagnets and transformers.

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Q192JEE Main 2014Easy

Hysteresis, Soft Iron vs Steel & Shielding · Demagnetising current

The coercivity of a small magnet where the ferromagnet gets demagnetized is $3 \times 10^{3}\text{ Am}^{-1}.$ The current required to be passed in a solenoid of length $10\text{ cm}$ and number of turns $100,$ so that the magnet gets demagnetized when inside the solenoid, is:

Show answer and solution

Answer: (c)

The solenoid's $H = nI$ must equal the coercivity: $I = \dfrac{H}{n}$ .
$n = \dfrac{N}{L} = \dfrac{100}{0.1} = 1000\text{ m}^{-1}$
$I = \dfrac{3 \times 10^{3}}{1000} = 3\text{ A}$
So the answer is (c) 3 A.

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Q193NEET 2010 · AIPMT · PRELIMSEasy

Hysteresis, Soft Iron vs Steel & Shielding · Soft iron vs steel: choice of material

Electromagnets are made of soft iron because soft iron has

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Answer: (c)

An electromagnet should become magnetised strongly and lose that magnetism quickly when the current stops.
Soft iron has low retentivity and low coercive force, so it has a narrow hysteresis loop.
So the answer is (c) low retentivity and low coercive force.

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Q194JEE Main 2004 · AIEEEEasy

Hysteresis, Soft Iron vs Steel & Shielding · Soft iron vs steel: choice of material

The materials suitable for making electromagnets should have

Show answer and solution

Answer: (b)

An electromagnet must lose its magnetism when the current is switched off, so it needs low retentivity.
Low coercivity gives a narrow hysteresis loop, so it is easy to demagnetise and has little energy loss.
So the answer is (b) low retentivity and low coercivity.

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Q195NEET 1998 · AIPMTEasy

Hysteresis, Soft Iron vs Steel & Shielding · Magnetic shielding

For protecting a sensitive equipment from the external magnetic field, it should be

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Answer: (b)

Iron has very high permeability, so field lines are drawn into the iron walls and pass around the inside.
The inside of an iron can is nearly field free. Aluminium and copper have $\mu_r \approx 1$ and give no shielding.
So the answer is (b) placed inside an iron can.

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