Question

The strength of the magnetic field in which the magnet of a vibration magnetometer is oscillating is increased 4 times its original value. The frequency of oscillation would then become
A. Twice its original value
B. Four times its original value
C. Half of its original value
D. One-fourth of its original value

Answer 1

Hint: A vibration magnetometer is a device used to measure the horizontal component of the Earth's magnetic field or to compare the magnetic moments of two magnets. When a magnet hanging in a consistent magnetic field (such the one caused by the earth's magnetic field) is moved out of its equilibrium position and starts to vibrate merely harmonically about the direction of the field.

Formula used:
The expression of oscillation frequency of magnetometer is,
$v={\displaystyle \frac{1}{2\pi }}\sqrt{{\displaystyle \frac{mB}{I}}}$
Here, $I$ is the moment of inertia, $B$ is the magnetic field and m$ is the magnetic moment.

Complete step by step solution:
When a bar magnet is positioned in a uniform magnetic field, the force operating in the field's direction causes the bar magnet to spin and make an effort to align itself with the magnetic field. The bar magnet will move in a straightforward harmonic fashion because of the rotational inertia brought on by the torque, which will drive the magnet to rotate and try to align with the field.

The torque acting on a bar magnet and rotational inertia are the basic operating principles of a vibration magnetometer. The formula of oscillating frequency is,
$v={\displaystyle \frac{1}{2\pi }}\sqrt{{\displaystyle \frac{mB}{I}}}$
From the above expression it is clear that frequency is directly proportional to the square root of the magnetic field.
$v\propto \sqrt{B}$
$\Rightarrow {\displaystyle \frac{v_1}{v_2}}=\sqrt{{\displaystyle \frac{B_1}{B_2}}}$
We are given that the magnetic field in which the magnet of a vibration magnetometer is oscillating is increased 4 times its original value. So we can say that $B_2=4B_1$
$\Rightarrow {\displaystyle \frac{v_1}{v_2}}=\sqrt{{\displaystyle \frac{B_1}{4B_1}}}$
$\Rightarrow {\displaystyle \frac{v_1}{v_2}}={\displaystyle \frac{1}{2}}$
$\therefore v_2=2v_1$
The oscillation frequency would then double from its initial value.

Therefore, option A is the correct answer.

Note: The torque and rotation in a uniform magnetic field is the basis for the operation of vibration magnetometers. The magnet will not oscillate and will stop when aligned with the field if the field is weak or not uniform. Therefore, a homogeneous, weak magnetic field is required for the vibration magnetometer to function.