Question

A magnet, freely suspended in a vibration magnetometer makes $40$ oscillations per minute at a place $A$ and $20$ oscillations per minute at $B$. If ${B_H}$​ at $B$ is $36 \times {10^{ - 6}}T$, then its value at $A$is :
A. $36 \times {10^{ - 6}}T$
B. $72 \times {10^{ - 6}}T$
C. $144 \times {10^{ - 6}}T$
D. $288 \times {10^{ - 6}}T$

Answer 1

Hint: In order to solve this problem, we are going to use the expression of time period of a bar magnet according to which the time period of magnet is directly proportional to the moment of inertia of magnet and inversely proportional to magnetic moment and magnetic field.

Formula used:
The time period of bar magnet is given by,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
Here, $T$ is time period of oscillation of bar magnet, $I$ is moment of inertia of bar magnet, $M$ is magnetic moment and $B$ is magnetic field intensity of bar magnet.

Complete step by step solution:
In the question, we have given the number of oscillations made by the magnet is $40$ and the magnetic field intensity is ${B_H} = 36 \times {10^{ - 6}}T$. In order for the model to be true, the net force exerted on the object at the pendulum's end must be commensurate to the displacement.

As we know, the motion of a simple pendulum may be described by the basic harmonic motion and simple harmonic motion can also be used to describe molecular vibration. The time period of bar magnet is given by,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
Similarly, the period of oscillation is given by,
$T = 2\pi \sqrt {\dfrac{I}{{M{B_H}}}} $
As we know that the frequency of an oscillation is directly proportional to the square root of the magnetic field, so we have $T \propto \dfrac{1}{{\sqrt {{B_H}} }}$.
Then, we have $\dfrac{{T_1^2}}{{T_2^2}} = \dfrac{{{B_{{H_A}}}}}{{{B_{{H_B}}}}}$

Now, substitute the given values in the above formula, then we obtain:
$\dfrac{{{{(\dfrac{{40}}{{60}})}^2}}}{{{{(\dfrac{{20}}{{60}})}^2}}} = \dfrac{{{B_{{H_A}}}}}{{36 \times {{10}^{ - 6}}}} \\$
$\Rightarrow \dfrac{{(\dfrac{{40}}{{60}})}}{{(\dfrac{{20}}{{60}})}} = \sqrt {\dfrac{{{B_{{H_B}}}}}{{36 \times {{10}^{ - 6}}}}} \\$
$\therefore {B_{{H_A}}} = 144 \times {10^{ - 6}}T \\ $
Therefore, the value of $A$ is $144 \times {10^{ - 6}}T$.

Thus, the correct option is C.

Note: The magnetic field strength or magnetic field intensity is the ratio of the magnetomotive force needed to create flux density per unit length of a certain material. The oscillation period and the magnet's moment of inertia are directly related, and the magnetic field of the planet is produced deep beneath the earth's core. The Earth's core generates an electric current because the movement of liquid iron generates magnetic fields.