A magnet of magnetic moment \(M\) is rotated through \(360^{\circ}\) in a magnetic field \(H\), the work…
A magnet of magnetic moment \(M\) is rotated through \(360^{\circ}\) in a magnetic field \(H\), the work done will be
\(\mathrm{MH}\)
\(2 M H\)
\(2 \pi M H\)
0
Solution
Work done to rotate magnetic dipole in magnetic field from angle \(\theta_1\) to \(\theta_2\) is given as
\(W=M B\left(\cos \theta_1-\cos \theta_2\right)\)
Here, \(\theta_1=0^{\circ}\) and \(\theta_2=360^{\circ}\)
\(\therefore \quad W=M B\left(\cos 0^{\circ}-\cos 360^{\circ}\right)=M B(1-1)=0\)