A circular coil of 25 turns and radius 6.0 cm, carrying a current of 10 A, is suspended vertically in a uniform magnetic field of magnitude 1.2 T. The field lines run horizontally in the plane of the coil. Calculate the force and torque on coil due to the magnetic field. In which direction should a balancing torque be applied to prevent the coil from turning?

Solution

Given, a circular coil.
Number of turns on the coil, n = 25
Radius of the coil, r = 6 cm
Current carried by the coil, I = 10 A
Uniform magnetic field, B = 1.2 T

Consider any element

\vec{dl}

of the wire.
Force on this element is I

(\vec{dl} \times \vec{B}) .

For each element

\vec{dl},

there is another length element

- \vec{dl}

on the closed loop.

Since

\vec{B}

is uniform, therefore, the forces cancel for each pair of such elements.
Therefore, the net force on the coil is zero.

The torque

\vec{τ}

on a plane loop of any shape carrying a current I in a magnetic field B is given by

\vec{τ} = IA \hat{n} \times \vec{B}

where,

\hat{n}

is a unit vector normal to the plane of the loop (direction of motion of a right- handed screw rotating in the direction of current).

For a circular coil of radius r and N turns,