Question

A magnet of total magnetic moment $ {10^{ - 2}}\hat iA - {m^2} $ is placed in a time varying magnetic field, $ B\hat i\left( {\cos \omega t} \right) $ , where $ B = 1Tesla $ and $ \omega = 0.125rad/s $ . The work done for reversing the direction of the magnetic moment at $ t = 1s $ second, is
(A) $ 0.007J $
(B) $ 0.014J $
(C) $ 0.01J $
(D) $ 0.028J $

Answer 1

Hint
The magnetic moment of a magnet is a measure of the magnetic strength, expressed as torque per unit magnetic flux density. The work done for reversing the direction of the magnetic moment is given by the formula $ W = (\Delta \vec u).\vec B $ , using which we can calculate the value.
Formula used: In this solution we will be using the following formula,
 $ W = (\Delta \vec u).\vec B $
where $ W $ is the work done and $ B $ is the magnetic field strength.

Complete step by step answer
The relation for calculating the work done in reversing the direction of the magnetic moment is given as,
 $ W = (\Delta \vec u).\vec B $
In the question we are given the values, $ \vec B = B\hat i\left( {\cos \omega t} \right) $ , $ B = 1Tesla $ and $ \omega = 0.125rad/s $
While substituting the value in the formula, we get,
 $ W = 2[{10^{ - 2}} \times 1 \times \cos [0.125]] $
On calculating we get,
 $ W = 1.98 \times {10^{ - 2}} $
That is,
 $ W = 0.0198J $
Thus, we get the work done for reversing the direction of the magnetic moment is approximately 0.02 J.
Hence, the correct answer is option (D).

Note
The magnetic moment is the magnetic strength and orientation of a magnet or other object that produces a magnetic field. Objects that experience magnetic moments are loops of electric current, permanent magnets, moving elementary particles, various molecules, and many astronomical objects.