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A square 12-turn coil with sides of leng...

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.

Text Solution

Verified by Experts

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`
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