Question

A bar magnet of moment $M$ gives a time period $T$ at a place in a vibration magnetometer. Four such similar bar magnets are placed in the frame one over the other out of which one magnet is placed with opposite polarity. The new time period is:
(A) $\dfrac{T}{{\sqrt 2 }}$
(B) $\sqrt 2 T$
(C) $T$
(D) $2T$

Answer 1

Hint:A magnetometer is a scientific device used to measure magnetic field in the vicinity of the instrument. First we calculate the torque of the magnet here and then find the time period of the magnetic field.

Complete step by step answer:
An object’s magnetic dipole moment is conveniently defined in terms of the torque encountered by the object in a given magnetic field. A magnetic dipole’s magnetic field is directly proportional to the moment on the magnetic dipole.
The two poles of a bar magnet encounter force when a bar magnet is put in a magnetic field. The forces are similar in magnitude and opposite on direction and don’t have the same action line. They form a couple of forces which generates a torque. The torque aims to spin the magnet in such a way that it is parallel to the field direction. The bar magnet also experiences maximum and minimum torque.
Since, the torque on a magnet is proportional to its magnetic moment, we get-
$\tau = - BM\sin \theta $
If $\theta $ is very very small $\sin \theta \approx \theta $
$\tau = - BM\theta $
               ...... (1)
The above equation is similar to the equation of simple harmonic motion
$\tau = - I{\omega ^2}\theta $
               ...... (2)
Also, we know that $\omega = \dfrac{{2\pi }}{T}$
               ...... (3)
From (1), (2) and (3), it can be concluded that
$T\alpha \dfrac{{\sqrt I }}{{\sqrt M }}$
In the second case, the net magnetic moment will be ${M_2} = 3M - M = 2M$ and the net moment of inertia will be $4I$.
$ \dfrac{{{T_2}}}{{{T_1}}} = \sqrt {\dfrac{{{I_2}{M_1}}}{{{M_2}{I_1}}}} = \sqrt 2 \\
{T_2} = \sqrt 2 T \\ $
So, the new time period is $\sqrt 2 T$.

Hence, option (B) is correct.

Note:Here the bar magnet experiences a torque not a force. Also for the second case we have to carefully observe which magnets and how magnets in the opposite poles.