Question

Twists of suspension fibre should be removed in vibration magnetometer so that
A. Time period be less
B. Time period be more
C. Magnet may vibrate freely
D. Cannot be said with certainty

Answer 1

Hint: Magnetic moments and magnetic fields are compared using a vibration magnetometer. The basis for this device's operation is the fact that anytime a freely suspended magnet in a uniform magnetic field is moved from its equilibrium point, it begins to vibrate about the mean location. In a nutshell, torque operating on the bar magnet is the basis for how vibration magnetometers operate.

Formula used:
The expression of time period of a magnet in a vibration magnetometer is,
$\mathrm{T}=2 \pi \sqrt{\dfrac{\mathrm{I}}{\mathrm{MB}}}$
Here, $I$ is the moment of inertia, $M$ is the magnetic moment and $B$ is the magnetic field.

Complete step by step solution:
The amount of time it takes for any string element to complete one oscillation is known as the wave's time period.
Magnet is suspended like a free body in a vibration magnetometer.
$\mathrm{T}=2 \pi \sqrt{\dfrac{1}{\mathrm{~g}}}$
Twist causes a reduction in length and from the above expression we can say that if length is reduced then the time period will decrease. Therefore, if the twists in the suspension fibre are removed, the time period increases.

Hence,the correct answer is option B.

Additional information: A bar magnet held freely in a uniform magnetic field begins to move in a straightforward harmonic manner about the equilibrium position when it is moved from that location. The sign indicating that the torque works in the direction of decreasing $\tau=-M B_{H} \theta$ is $\tau=-m B_{H}(2 \ell \sin \theta)=-M B_{H} \sin \theta$ .If $\tau=-M B_{H} \theta$ is small , then$\tau=-M B_{H} \theta$

Note: It is a tool for measuring the horizontal component of the earth's magnetic field as well as comparing the magnetic moments of two bar magnets and the horizontal components of the earth's magnetic field at two locations.