Question

A magnet oscillating in a horizontal plane has a time period of 2 seconds at a place where the angle of dip is 30o and 3 seconds at another place where the angle of dip is 60o . The ratio of resultant magnetic fields at the two places is:

Answer 1

time period due to bar magnet is given by, $T=2\pi\sqrt{\frac{I}{MB_H}}$

where I is moment of inertia of bar magnet about its axis, M is magnetic moment of the bar magnet and $B_H$ is the horizontal component of bar magnet.

Let $B_1$ and $B_2$ are earth's magnetic fields at different locations.

$B_{H_1}=B_1cos30^{\circ}$ ; $T_1$= 2sec

so, $T_1=2=2\pi\sqrt{\frac{I}{MB_1cos30^{\circ}}}$.....(1)

$B_{H_2}=B_2cos60^{\circ}$ ; $T_2$= 3 sec

so, $T_2=3=2\pi\sqrt{\frac{I}{MB_2cos60^{\circ}}}$.....(2)

from equations (1) and (2),

$\frac{2}{3}=\sqrt{\frac{B_2cos60^{\circ}}{B_1cos30^{\circ}}}$

or, $\frac{4}{9}=\frac{B_2}{B_1\sqrt{3}}$

or, $\frac{B_1}{B_2}=\frac{3\sqrt{3}}{4}$

Answer 2

Answer:

Explanation:

Time period of magnet oscillation T = 2π√I/MB_H

Where,

I = The moment of inertia of the needle

B_H = horizontal component of Earth's magnetic field

M = magnetic moment of needle

Suppose, B1 and B2 is the magnetic field of the earth at particular location.

B_H1 = B_1 cos 30 = √3/2 B_1

B_H2 = B_2 cos 60 = 1/2 B_2

T_1 = 2 s

T_2 = 3 s

Thus, T_1 =  2π√I/MB_H1

as well as T_2 =  2π√I/MB_H2

∴ T_1/T_2 = √B_H2/B_H1

(T_1/T_2)^{2} = B_H2/B_H1

T_1^{2}/T_2^{2} = B_H2/B_H1  

T_1^{2}/T_2^{2} = 1/2B_2 / √3/2 B_1 = B_2/ √3B_1

4/9 = B_2/ √3B_1

B_2/B_1 = √3x4/9 = 1.73x4/9 = 0.77