Question

A square coil of 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle 30 with the direction of a uniform horizontal magnetic field of magnitude 0.08 T. What is the magnitude of torque experienced by the coil?

Answer 1

This problem can be solved by making use of the fact that when a current carrying coil is placed in a magnetic field, it experiences a torque by the magnetic field. This torque tries to rotate the coil such that the magnetic field becomes perpendicular to the plane of the coil. By using this concept, we can find out the torque exerted by the magnetic field on the coil.

Formula Used:
The magnitude of the torque $\left( \tau \right)$ experienced by a coil of turns $N$ when it is carrying a current $I$ and has a cross sectional area $A$ and it is kept in a magnetic field $B$ is given by
$\tau = NAIB\sin \theta$
Here $\theta$ is the angle made by the perpendicular to the plane with the magnetic field direction.
This torque has a direction such that it tries to make the plane of the coil perpendicular to the magnetic field.

Complete step by step solution:
Whenever a current carrying coil is placed in a magnetic field, it experiences a torque that tries to make the plane of the coil perpendicular to the direction of the magnetic field.
The magnitude of the torque $\left( \tau \right)$ experienced by a coil of turns $N$ when it is carrying a current $I$ and has a cross sectional area $A$ and it is kept in a magnetic field $B$ is given by
$\tau = NAIB\sin \theta$
Let us now analyse the question,
Length of a side of the square coil, $l = 10{\text{ cm}}$
$\Rightarrow l = 0.1{\text{ m}}$
Current flowing in the coil, $I = 12{\text{ A}}$
Number of turns on the coil, $n = 20$
Angle made by the plane of the coil with magnetic field, $\theta {\text{ = 3}}{{\text{0}}^ \circ }$
Strength of magnetic field, $B = 0.80{\text{ T}}$
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
$\tau = NAIB\sin \theta$
Now, substitute the known values
$\Rightarrow l \times l = 0.1 \times 0.1 = 0.01{\text{ }}{{\text{m}}^2}$
$\Rightarrow \tau = 20 \times 0.8 \times 12 \times 0.01 \times \sin {30^ \circ }$
We know
$\sin {30^ \circ } = \dfrac{1}{2}$
$\Rightarrow \tau = 0.96{\text{ N m}}$
Therefore, the magnitude of torque exerted by the magnetic field on the coil is $0.96{\text{ N m}}$.

Note:
Students must take care while determining the value of $\theta$. It is the angle made by the perpendicular to the plane and the magnetic field. In many questions like this, the angle is explicitly not given and the angle between the plane and the magnetic field is instead given. In such situations, one must proceed to find the angle as done above. The wrong value of $\theta$ will lead to a completely wrong answer and a huge waste of time especially in competitive exams.