Question

A current carrying loop, free to turn, is placed in a magnetic field b. What will be its orientation relative to b in the equilibrium state​

Answer 1

Answer:

The current carrying circular loop behaves as a magnetic dipole of magnetic moment

M

acting perpendicular to its plane. The torque on the current loop of magnetic dipole moment M in the magnetic field B is

τ=MBsinα=IA×Bsinα,(∵M=AI)

where α is the angle between

M

and

B

. The system will be in stable equilibrium if torque is zero, which is so if α=0

∘

. This is possible if

B

is parallel to

A

i.e.

B

is perpendicular to the plane of the loop. In this orientation, the magnetic field produced by the loop is in the same direction as that of external field, both normal to the plane of loop. It is due to this fact, the magnetic flux due to total field is maximum.