Earth's Magnetism

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 $\tan 37^\circ = \frac{3}{4})$

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:

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

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)

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

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

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)

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:

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 -

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

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.)

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:

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)

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

Tangent galvanometer is used to measure