Moving Charges and Magnetism

Answer: (d)

1. Axial field: $B = \frac{\mu_0 n I r^2}{2(r^2+x^2)^{3/2}}$
2. $x^2 = 8r^2 \Rightarrow (r^2+x^2)^{3/2} = (9r^2)^{3/2} = 27r^3$
3. $B = \frac{\mu_0 n I r^2}{2\times 27 r^3} = \frac{\mu_0 n I}{54 r}$
4. So the answer is (d).

Answer: (b)

1. Axial field: $B = \frac{\mu_0 n i R^2}{2(R^2+r^2)^{3/2}}$; for $r \gg R$, $B \approx \frac{\mu_0 n i R^2}{2r^3}$.
2. With $A = \pi R^2$: $B = \frac{\mu_0 n i A}{2\pi r^3} = \frac{2\mu_0 n i A}{4\pi r^3}$
3. So the answer is (b).

Answer: (c)

1. $B_c = \frac{\mu_0 I}{2R}$, $B_x = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$
2. $\frac{B_c}{B_x} = \frac{(R^2+x^2)^{3/2}}{R^3}$; with $x = R$: $\frac{(2R^2)^{3/2}}{R^3} = 2\sqrt{2} = \sqrt{8}$
3. So the answer is (c) $\sqrt{8}:1$.

Answer: (b)

1. $\frac{B_x}{B_c} = \frac{R^3}{(R^2+x^2)^{3/2}}$
2. $x = \sqrt{3}R \Rightarrow (4R^2)^{3/2} = 8R^3$, so $B_x = \frac{B_c}{8} = \frac{16}{8}$
3. So the answer is (b) $2\ \mu\text{T}$.

Answer: (b)

1. Arc of angle $\theta$: $B = \frac{\mu_0 I\theta}{4\pi R}$.
2. $\theta = \frac{\pi}{2}$: $B = \frac{\mu_0 I}{4\pi R}\cdot\frac{\pi}{2} = \frac{\mu_0 I}{8R}$
3. So the answer is (b).

Answer: (d)

1. Arc of angle $\theta$: $B = \frac{\mu_0 i\theta}{4\pi r}$.
2. $\theta = \frac{\pi}{5}$: $B = \frac{\mu_0 i}{4\pi r}\cdot\frac{\pi}{5} = \frac{\mu_0 i}{20r}$
3. So the answer is (d).

Answer: (b)

1. Single-turn coils: $B = \frac{\mu_0 I}{2r}$; equal $B$ needs $I \propto r$.
2. Same wire (same area, same material): $R_{wire} \propto \text{length} = 2\pi r$, so $R_{wire} \propto r$.
3. $V = IR_{wire} \propto r^2 \Rightarrow \frac{V_A}{V_B} = \left(\frac{2}{1}\right)^2 = 4$
4. So the answer is (b) $4 : 1$.

Answer: (b)

1. $B = \frac{\mu_0 I}{2r}$ same at both centres $\Rightarrow I \propto r$.
2. Same wire: resistance $\propto$ length $\propto r$.
3. $V = IR \propto r \cdot r = r^2 \Rightarrow V_A : V_B = 4 : 1$
4. So the answer is (b) $4 : 1$.

Answer: (a)

1. Fixed wire length $L$: $n$ turns gives $r = \frac{L}{2\pi n}$, so $B = \frac{\mu_0 n I}{2r} \propto n^2$.
2. $n$ goes from 1 to 2, so $B' = 2^2 B = 4B$.
3. So the answer is (a) $4B$.

Answer: (c)

1. Same length: $2\pi R = n(2\pi r) \Rightarrow r = R/n$.
2. $B_n = \frac{\mu_0 n I}{2r} = n^2 \frac{\mu_0 I}{2R} = n^2 B_1$
3. So the answer is (c) $1 : n^2$.

Answer: (a)

1. Field at the centre: $B = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \times 2}{0.2} = 4\pi \times 10^{-6}\text{ T}$.
2. Energy density: $u = \frac{B^2}{2\mu_0} = \frac{16\pi^2 \times 10^{-12}}{8\pi \times 10^{-7}} = 2\pi \times 10^{-5}\text{ J/m}^3$.
3. Cube volume $= (10^{-3})^3 = 10^{-9}\text{ m}^3$, so $U = 2\pi \times 10^{-14} = 6.28 \times 10^{-14}\text{ J}$.
4. So the answer is (a) $\alpha = 6.28$.

Answer: (b)

1. Equivalent current $I = qf = 100 \times 1.6 \times 10^{-19} \times 1 = 1.6 \times 10^{-17}\text{ A}$.
2. $B = \frac{\mu_0 I}{2r} = \frac{\mu_0 \times 1.6 \times 10^{-17}}{1.6} = \mu_0 \times 10^{-17}$
3. So the answer is (b) $\mu_0 10^{-17}$.

Answer: (c)

1. Rotating ring acts as current $I = qf$.
2. $B = \frac{\mu_0 I}{2R} = \frac{\mu_0 q f}{2R}$
3. So the answer is (c) $\frac{\mu_0 q f}{2R}$.

Answer: (c)

1. Field on the axis of a loop: $B = \frac{\mu_0 I R^2}{2(z^2 + R^2)^{3/2}}$.
2. Comparing with $\frac{x R^2}{(z^2 + R^2)^{3/2}}$ gives $x = \frac{\mu_0 I}{2}$.
3. So the answer is (c) $\frac{\mu_0 I}{2}$.

Answer: (b)

1. On the axis $B = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ and at the centre $B_c = \dfrac{\mu_0 I}{2R}$, so $\dfrac{B_2}{B_1} = \dfrac{(R^2+x^2)^{3/2}}{R^3}$.
2. $8 = \dfrac{(R^2+x^2)^{3/2}}{R^3} \Rightarrow R^2+x^2 = 4R^2$
3. $x^2 = 3R^2 \Rightarrow x = \sqrt{3}R$
4. So the answer is (b) $\sqrt{3}\,R$.

Answer: (b)

1. Each loop is at axial distance $x = r$ from O: $B = \dfrac{\mu_0 I r^2}{2(2r^2)^{3/2}} = \dfrac{\mu_0 I}{4\sqrt{2}r}$ for current $I$.
2. $B_P = \dfrac{\mu_0 I}{4\sqrt{2}r}$, $B_Q = \dfrac{4\mu_0 I}{4\sqrt{2}r}$
3. Clockwise as seen from O means each field at O points away from O: $B_P$ towards P, $B_Q$ towards Q (opposite).
4. $B_{net} = B_Q - B_P = \dfrac{3\mu_0 I}{4\sqrt{2}r}$ towards Q
5. So the answer is (b).

Answer: (b)

1. The two halves are in parallel; $R_{ABC} = \frac{1}{2}R_{ADC}$, so $I_{ABC} = \dfrac{2I_0}{3}$, $I_{ADC} = \dfrac{I_0}{3}$.
2. Both currents go from A to C, so at O their fields are opposite. The straight leads pass through O and give zero.
3. $B = \dfrac{\mu_0}{4r}\left(\dfrac{2I_0}{3}-\dfrac{I_0}{3}\right) = \dfrac{\mu_0 I_0}{12r}$
4. So the answer is (b).

Answer: (c)

1. On the axis $B = \dfrac{\mu_0 I r^2}{2(r^2+x^2)^{3/2}} \approx \dfrac{\mu_0 I r^2}{2x^3}$ for $x \gg r$.
2. Coil X: $B_x = \dfrac{\mu_0 I r^2}{2(d/2)^3} = \dfrac{4\mu_0 I r^2}{d^3}$
3. Coil Y: $B_y = \dfrac{\mu_0 I (2r)^2}{2d^3} = \dfrac{2\mu_0 I r^2}{d^3}$
4. So $B_x = 2B_y$, answer (c).

Answer: (c)

1. Field at centre: $B = \frac{\mu_0 n I}{2r}$, so $n = \frac{2rB}{\mu_0 I}$.
2. $n = \frac{2 \times 0.08 \times 3.14\times10^{-4}}{12.56\times10^{-7}\times 0.4} = \frac{5.024\times10^{-5}}{5.024\times10^{-7}}$
3. So the answer is (c) $n = 100$.

Answer: (c)

1. Axial field: $B = \frac{\mu_0 n I r^2}{2(r^2+x^2)^{3/2}}$
2. $x = 2\sqrt{2}\,r \Rightarrow r^2+x^2 = 9r^2$, $(9r^2)^{3/2} = 27r^3$
3. $B = \frac{\mu_0 n I}{54 r}$
4. So the answer is (c).

Answer: (c)

1. Biot-Savart law: $dB = \frac{\mu_0}{4\pi}\frac{I\,\Delta x \sin\theta}{r^2}$, with $\theta = 90^\circ$ for a point on the y-axis.
2. $dB = 10^{-7} \times \frac{10 \times 0.01}{(0.5)^2} = 10^{-7} \times 0.4$
3. So the answer is (c) $4 \times 10^{-8}\text{ T}$.

Answer: (a)

1. For a fixed length of wire, $B = \frac{\mu_0 n I}{2r}$ with $r \propto \frac{1}{n}$, so $B \propto n^2$.
2. $\frac{B_2}{B_1} = \left(\frac{3}{9}\right)^2 = \frac{1}{9}$
3. So the answer is (a) $\frac{B_1}{9}$.

Answer: (c)

1. Same length: $2\pi R = n \cdot 2\pi r_1 \Rightarrow r_1 = R/n$.
2. $B' = \frac{\mu_0 n I}{2r_1} = n^2 \frac{\mu_0 I}{2R} = n^2 B$
3. So the answer is (c) $n^2 B$.

Answer: (d)

1. $B = \frac{\mu_0 I}{2r}$ with $I = \frac{ev}{2\pi r}$, so $B \propto \frac{v}{r^2}$.
2. Bohr model: $v \propto \frac{1}{n}$, $r \propto n^2$.
3. $B \propto \frac{n^{-1}}{n^4} = n^{-5}$
4. So the answer is (d) $n^{-5}$.

Answer: (d)

1. $I = qf = 1000 \times 1.6 \times 10^{-19} \times 1 = 1.6 \times 10^{-16}\text{ A}$.
2. $B = \frac{\mu_0 I}{2r} = x\mu_0 \Rightarrow r = \frac{I}{2x} = \frac{1.6 \times 10^{-16}}{4 \times 10^{-16}}$
3. So the answer is (d) $0.4\text{ m}$.

Answer: (c)

1. $\dfrac{B_{axis}}{B_c} = \dfrac{R^3}{(R^2+x^2)^{3/2}} = \dfrac{1}{27}$
2. $(R^2+x^2)^{3/2} = 27R^3 \Rightarrow R^2+x^2 = 9R^2$
3. $x^2 = 8R^2 \Rightarrow x = 2\sqrt{2}R$
4. So the answer is (c) $2\sqrt{2}R$.

Answer: (a)

1. $\dfrac{B_2}{B_1} = \dfrac{R^3}{(R^2+x^2)^{3/2}}$ with $x = \dfrac{3}{4}R$.
2. $R^2+x^2 = \dfrac{25}{16}R^2 \Rightarrow (R^2+x^2)^{3/2} = \dfrac{125}{64}R^3$
3. $\dfrac{B_2}{B_1} = \dfrac{64}{125}$
4. So the answer is (a) $64 : 125$.

Answer: (b)

1. Straight parts PQ and RS point towards O, so they give no field at O.
2. Outer semicircle ($R_1$, clockwise): $\dfrac{\mu_0 I}{4R_1}$ into the page.
3. Inner semicircle ($R_2$, anticlockwise): $\dfrac{\mu_0 I}{4R_2}$ out of the page.
4. $B = \dfrac{\mu_0 I}{4}\left[\dfrac{1}{R_2}-\dfrac{1}{R_1}\right]$ (since $R_2 < R_1$)
5. So the answer is (b).

Answer: 1

1. Straight parts AB and CD lie along a line through O: no field.
2. Inner arc ($R_2$) is clockwise, outer arc ($R_1$) is anticlockwise, so the fields are opposite.
3. $B = \dfrac{\mu_0 I}{4}\left(\dfrac{1}{R_2}-\dfrac{1}{R_1}\right) = \dfrac{4\pi\times10^{-7}\times12}{4}\left(\dfrac{1}{4\pi}-\dfrac{1}{6\pi}\right)$
4. $B = 12\pi\times10^{-7}\times\dfrac{1}{12\pi} = 1\times10^{-7}$ T
5. So $k = 1$.

Answer: (b)

1. Each coil gives $B = \dfrac{\mu_0 NI}{2R}$ at O.
2. The coils are perpendicular, so the two fields are perpendicular.
3. $B_{net} = \sqrt{2}\cdot\dfrac{\mu_0 NI}{2R} = \dfrac{\mu_0 NI}{\sqrt{2}R}$
4. So the answer is (b).

Answer: (d)

1. Statement I: with time-varying currents, the magnetic forces between elements need not be equal and opposite; momentum balance holds only when the field's momentum is included. True.
2. Statement II: Ampere's circuital law is derived from (and equivalent to) the Biot-Savart law for steady currents, so it does depend on it. False.
3. So the answer is (d) Statement I is true but Statement II is false.

Answer: (c)

1. $B = \frac{\mu_0 n I}{2r}$
2. $B = \frac{4\pi\times10^{-7}\times 100\times 7}{2\times 0.1} = \frac{8.8\times10^{-4}}{0.2} = 4.4\times10^{-3}\text{ T}$
3. So the answer is (c) 4.4 mT.

Answer: (b)

1. Equivalent current $I = qf = 100\times1.6\times10^{-19}\times 1 = 1.6\times10^{-17}\text{ A}$.
2. $B = \frac{\mu_0 I}{2r} = \frac{\mu_0 \times 1.6\times10^{-17}}{2\times 0.8}$
3. So the answer is (b) $10^{-17}\mu_0$.

Answer: (b)

1. Arc of angle $\theta$: $B = \frac{\mu_0 I\theta}{4\pi R}$.
2. $\theta = \frac{\pi}{2}$: $B = \frac{\mu_0 I}{8R}$
3. So the answer is (b).

Answer: (d)

1. Arc field = (fraction of circle) $\times \frac{\mu_0 I}{2R}$.
2. $B = \frac{3}{4}\cdot\frac{\mu_0 I}{2R} = \frac{3\mu_0 I}{8R}$
3. Anticlockwise current (seen from above): by the right-hand rule the field at the centre points towards the viewer, i.e. upward.
4. So the answer is (d) $\frac{3\mu_0 I}{8R}$ upward.

Answer: (b)

1. Arc of angle $\theta$: $B = \frac{\mu_0 I\theta}{4\pi r}$.
2. $\theta = \frac{\pi}{8}$: $B = \frac{\mu_0 I}{4\pi r}\cdot\frac{\pi}{8} = \frac{\mu_0 I}{32r}$
3. So the answer is (b).

Answer: (b)

1. $B_P = \frac{\mu_0 I}{2R}$, $B_Q = \frac{\sqrt{8}\,\mu_0 I}{2R}$; the coils are in perpendicular planes, so the fields are perpendicular.
2. $B = \sqrt{B_P^2+B_Q^2} = \frac{\mu_0 I}{2R}\sqrt{1+8} = \frac{3\mu_0 I}{2R}$
3. So the answer is (b).

Answer: (c)

1. For zero net field: $\frac{\mu_0 I_A}{2r_A} = \frac{\mu_0 I_B}{2r_B}$, with fields opposite.
2. $I_B = I_A\frac{r_B}{r_A} = 0.5\times\frac{10}{20} = 0.25\text{ A}$
3. Opposite field needs opposite current, i.e. clockwise.
4. So the answer is (c) 0.25 A clockwise.