Question

The time period of vibration of magnetic dipole is T. On bringing a bar magnet closer long the axis of vibration, the new time period T’ will be
A. T’ > T
B. T’ < T
C. T’ = T
D. None of these

Answer 1

Hint: A time period which is denoted by 'T' is the time taken for one complete cycle of vibration to pass a given point. As the frequency of a given wave increases, the time period of the wave decreases. The unit for time period is 'seconds'. We have to consider the effect an external magnetic field will have on the magnetic dipole to solve these types of questions.

Complete step-by-step answer:
A bar magnet is a rectangular piece of an object which is made up of iron, steel or any other ferromagnetic substance or ferromagnetic composite, that shows permanent magnetic properties. It has two poles, a north and a south pole such that when suspended freely, the magnet aligns itself so that the northern pole points towards the magnetic north pole of the earth.
When we bring a bar magnet near a magnetic dipole placed in a magnetic field then it will tend to reduce the restoring torque acting on the dipole while doing SHM.
Therefore, when restoring torque is lesser than the new time period T′ will be lesser than the previous time period T.
Hence the correct option will be option B which is T’ < T

Note: When the particle or the body undergoes small angular displacement about mean position it results that the body under stable equilibrium is disturbed by a small external torque. In turn, the rotating system tries to oppose the change and generates a restoring torque, which tries to restore equilibrium.