Question

1. A bar magnet is lying in the magnetic meridian. How much work need to be done to rotate it through angle 0 ? Given that magnetic moment of the magnet is Pmand horizontal component of Earth's magnetic BH a) P.Bicos e b) P.B(1 - cos O) c) P.Busino d) PMB (1 - sin o)​

Answer 1

Answer:

Potential energy of a magnetic moment at an angle θ with the magnetic field = −MBcosθ

Thus work done in rotating from angle θ

1

to θ

2

=MB(cosθ

1

−cosθ

2

)

=MB(cos0

∘

−cos90

∘

)

=MB(1−0)

=MB