Question

Work done in rotating a bar magnet of magnetic moment m from its unstable equilibrium position to stable equilibrium position in a uniform magnetic field b is

Answer 1

when a bar magnet is in unstable equilibrium then its angle with the magnetic field will be 180 degree while when bar magnet is in stable equilibrium then its angle with magnetic field will be 0 degree

Now in order to find the work done we can say

$W = U_f - U_i$

here U = potential energy of bar magnet

so we will have

$U = - M.B$

$U = - MBcos\theta$

now by plug in the value

$U = -MBcos\theta_f + MBcos\theta_i$

also we know that

$\theta_f = 0 degree$

$\theta_i = 180 degree$

now we have

$W = -MBcos0 + MBcos180$

$W = -2MB$

so work done to change the orientation of magnet is - 2MB

Answer 2

Explanation:

hope this will help you