Question

A bar magnet of magnetic dipole moment 10 Am2
is in stable equilibrium. The work done to rotate the
magnet through 60° in a magnetic field of 0.2T is​

Answer 1

Answer:

M=10Am2

B = 0.2 T

U = −MBcosθ

= −10×0.2×3–√2

=−3–√J

Explanation: