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Q. A bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction $B$. If it is rotated through an angle if $90^{\circ},$ then the work is $(M=$ magnetic dipole moment of bar magnet $)$

Magnetism and Matter

Solution:

Work done, $W=M B\left(\cos \theta_{1}-\cos \theta_{2}\right)$
Then, $W=M B\left(\cos 0^{\circ}-\cos 90^{\circ}\right)=M B$