A circular coil with area\(A\)and\(N\)turns is free to rotate about a diameter that coincides with the\(x\)axis. Current\(I\)is circulating in the coil. There is a uniform magnetic field\(\vec B\)in the positive\(y\)direction. Calculate the magnitude and direction of the torque\(\vec \tau \) and the value of the potential energy\(U\)as given in Eq. (27.27), when the coil is oriented as shown in parts (a) through (d) of Fig. E27.45.

Area of each coil is\(A\).

Number of turns of each coil is \(N\).

Current in each coil is \(I\).

Direction of magnetic field \(\vec B\) is \( + y\).

The torque on a coil of area\(A\), carrying total current\(I\)in the presence of a magnetic field\(B\)is

\(\tau = IAB{\rm{sin}}\theta \) .....(i)

Here,\(\theta \)is the angle between the magnetic moment and the magnetic field. The direction of the magnetic moment is obtained from the right hand thumb rule.

The direction of the torque is the direction of the cross product between magnetic moment and magnetic field.

The potential energy of the coil is

\(U = - IAB{\rm{cos}}\theta \) .....(ii)

From right hand thumb rule, the magnetic moment of the coil is in the\( + z\)direction. Thus, angle between magnetic moment and magnetic field is

\(\theta = 90^\circ \).

The total current in the coil is\(NI\).

From equation (i), the magnitude of the torque is

\(\begin{aligned}\tau = NIAB{\rm{sin90}}^\circ \\ = NIAB\end{aligned}\)

The cross product of magnetic moment and field is along the cross product of\( + z\)and\( + y\), that is along\( - x\)axis.

Thus, the torque is\(NIAB\)along the\( - x\)axis.

From equation (ii), the potential energy is

\(\begin{aligned}U = - NIAB{\rm{cos90}}^\circ \\ = 0\end{aligned}\)

Thus, the potential energy is \(0\).