The square loop in Fig. has sides of length 20 cm. It has 5 turns and carries a current of 2 A. The normal to the loop is at `37^@` to a uniform field `vec B = 0.5 hat j T`.
a. Find the magnetic moment of the loop.
b. Find the work needed to rotate the loop from its position of minimum energy to the given orientation.

Explanation

(a) From Fig. we see that the unit vector normal to loop
`hatn=-sin37^@hati+cos37^@hatj=-0.6hati+0.8hatj`

The magnetic moment is
`vecM=NIAvecn=(5)(2)(0.2)^2(-0.6hati+0.8hatj)`
`=-0.24hati+0.32hatjAm^2`
(b) The torque, `hattau=vecMxxvecB=(-0.24hati+0.32hatj)xx(0.5hatj)`
`=-0.12hatk Nm`
(c) The potential energy of the loop is `U=-MB cos theta` where
`M=NIA=0.4Am^2` and the position of minimum energy
is `theta=0`.
Thus, the external work, `W_(ext)=+DeltaU`, needed to rotate it to
the given orientation, is given by `W_(ext)=U_f-U_i=(-MBcos 37^@)-(-MB cos0^@)`
`=(0.4)(0.5)(1-0.8)=0.04J`
The external work is postive since the dipole moment is rotated away from alignment with the field.