Question

A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of \[0.1weber/{m^2}\] . The amount of work done in rotating it through ${180^0}$ from its equilibrium position will be-
a) $0.1J$
b) $0.2J$
c) $0.4J$
d) $0.8J$

Answer 1

Hint: In order to solve this question, we should know that a current carrying conductor always experience a torque when placed in magnetic field and in order to change its position we need to do work against the torque here, we will use general formula of torque and work done to rotate a conductor in magnetic field with given conditions.

Complete answer:
A coil of current-carrying wire has a magnetic moment. This magnetic moment is a vector, with a magnitude and direction. The magnitude of the magnetic moment is proportional to the current in the coil and the number of turns in the coil.

The magnetic moment of a circular current carrying coil is also proportional to the area of the coil. The direction of the magnetic moment is perpendicular to the plane of the coil and points in the direction of the magnetic field that the coil creates.

From here we get a formula of the magnetic moment of the circular coil carrying current.
$M = iAn$
here, $M = {\text{ magnetic moment, i = current, A = area of the coil, n = number of turns}}{\text{.}}$

So now if we now try rotate the coil, we have to do a work done. We can find out the work done by change in energy of two states. So first we are considering that the coil is in a horizontal plane.
Energy in that state is $ - MB\cos {0^0}$, now when we rotate the coil by ${180^0}$ . Then energy now is $ - MB\cos {180^0}$ .

Now we are very close to the answer. Let’s calculate the change in energy. $MB\left( {\cos {0^0} - \cos {{180}^0}} \right)$ is the change in energy.

We can write $i(\pi {r^2})nB(\cos {0^0} - \cos {180^0})$ .
or, $2 \times 50 \times \pi \times 4 \times {10^{ - 2}} \times 0.1 \times (1 + 1) = 0.1J$ . We get our answer.

Hence option a is the correct option.

Note: While solving such questions, always remember that the equilibrium position is when conductor makes zero angle with magnetic field and make sure the proper signs are used in cosine terms to avoid any such minor calculation mistakes.