Question

A current carrying coil is placed in an external uniform magnetic field. The coil is free to turn in the magnetic field. What is the net force acting on the coil? Obtain the orientation of the coil in stable equilibrium. Show that in this orientation the flux of the total field (field produced by the loop + external field) through the coil is maximum.

Answer 1

A current-carrying coil placed in an external uniform magnetic field receives torque but no net force if the field is uniform. The force on a current-carrying element dl of the coil is given by:

dF = Idl × B

In a closed loop, the forces on opposite sides may be equal and opposite, so canceling each other out in a uniform field. Thus, the net force acting on the coil is zero, but a torque is present.

The torque on a current loop in an external magnetic field is given by:

T = m × B

Where m is the magnetic dipole moment of the coil, m = NIA_n.

The torque tries to align the coil’s magnetic moment with the external magnetic field B, obtaining a stable equilibrium when the torque is zero, i.e., m × B = 0. This occurs when the magnetic field is parallel to B, showing that the plane of the coil is perpendicular to the external field. The system’s potential energy in an external field is:

u = −m.B

= −m B cosθ

The minimum potential energy (stable equilibrium) occurs at θ = 0°.

The magnetic flux through the coil is:

Φ = ∫B_total.dA

B_total is the sum of the external field B and the field generated by the coil. The field produced by the coil aligns with the direction of its magnetic moment. When the coil aligns with B, its own field increases the external field, resulting in a maximum total field.

Φ_max = (B_ext + B_coil)A

Thus, in the stable equilibrium position where m parallels B, the flux through the coil is increased.