Question

A magnet oscillating in a horizontal plane has a time period of $2$ second at a place where the angle of dip is $30^{\circ}$ and $3$ seconds at another place where the angle of dip is $60^{\circ}$. The ratio of resultant magnetic field at the two places is
A. $\dfrac{4\sqrt 3}{7}$
B. $\dfrac{4}{9\sqrt 3}$
C. $\dfrac{9}{4\sqrt 3}$
D. $\dfrac{9}{\sqrt 3}$

Answer 1

A magnet is metal that can produce a magnetic field. It has two poles, namely the North and the South Pole on two sides of the metal. We also know that the earth has a magnetic field. Thus when a magnet is suspended in the air, due the earth’s magnetic field, the magnet undergoes oscillation and finally rests at the geographical north and south.
Formula used:
$T\propto \dfrac{1}{\sqrt{B_H}}$
$B_{H} =B\cos\theta$

Complete step by step answer:
Since the magnet is oscillating, we can say that there is a time period for this oscillation, say, $T$. Angle of dip or the magnetic dip or magnetic inclination is the angle made by the earth’s magnetic field and the plane of horizon at any given location. Clearly this value varies at different places on the surface of the earth. Since earth’s magnetic field $B$ is a vector quantity, it has two components,as shown in the figure below:

Horizontal component of earth’s magnetic field $B$ : $B_{H} =B\cos\theta$
Vertical component of earth’s magnetic field $B$: $B_{V} =B\sin\theta$
Let us assume that the oscillation in happening in the xy- plane, then the time period $T$ depends on the horizontal component of the magnetic field, then, we have
$T=2\pi\sqrt{\dfrac{I}{\mu B_H}}$
$T\propto \dfrac{1}{\sqrt{B_H}}$
Taking the ratio, we have
$\dfrac{T_1}{T_2}=\sqrt{\dfrac{B_2\cos\theta_2}{B_1\cos\theta_1}}$
Given that $T_{1}= 2s$ when $\theta_1=30^{\circ}$ and $T_2=3s$ when $\theta_2=60^{\circ}$, substituting the values we have,
$\dfrac{2}{3}=\sqrt{\dfrac{B_2 \cos 60^{\circ}}{B_1\cos 30^{\circ}}}$
$\dfrac{4}{9}=\dfrac{B_2 \cos 60^{\circ}}{B_1\cos 30^{\circ}}$
$\implies \dfrac{B_{2}}{B_{1}}=\dfrac{4\cos 30^{\circ}}{9\cos 60^{\circ}}$
$\implies \dfrac{B_{2}}{B_{1}}=\dfrac{4\times \dfrac{\sqrt{3}}{2}}{9\times \dfrac{1}{2}}$
$\implies\dfrac{B_{2}}{B_{1}}=\dfrac{4\sqrt{3}}{9}$
Thus the ratio is $\dfrac{B_{1}}{B_{2}}=\dfrac{9}{4\sqrt{3}}$

So, the correct answer is “Option C”.

Note: We know that the magnetic field is a vector quality, and hence has both direction and magnitude. We can also say that the direction of the magnetic field is given from north to south outside the magnet and south to north inside the metal.