Question

A square 12-turn coil with sides of length 40 cm carries a current of 3 A. It lies in the x-y plane as shown in Fig. in a uniform magnetic field.

(a) Find the magnetic moment of the coil.
(b) Find the torque exerted on the coil.
(c) Find the potential energy of the coil.

Answer 1

The magnetic moment of the loop is in the positive z-direction
(right hand thumb rule).
(a) The magnetic moment of the loop is given by
`vecM=NIAhatk=(12)(3)(0.40)^2hatk=5.76Am^2hatk`
(b) The torque on the current loop is given by
`vectau =vecMxxvecB= (5.76hatk)xx(0.3hati+0.4hatj)=1.73Nmhatj`
(c) The potential energy is the negative dot product of `vecmu and vecB`:
`U=-vecM.vecB=-(5.76hatk).(0.3hati+0.4hatk)=-2.30J`