Question

A rectangular coil of length $0.12m$ and width $0.1m$ having 50 turns of wire is suspended vertically in a uniform magnetic field of strength $0.2\text{ Weber/}{{\text{m}}^{2}}$. The coil carries a current of $2A$. If the plane of the coil is inclined at an angle of ${{30}^{0}}$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be
(A). $0.12Nm$
(B). $0.15Nm$
(C). $0.20Nm$
(D). $0.24Nm$

Answer 1

Hint: This problem can be solved 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 and hence, the magnitude of the torque that is required to keep it in static equilibrium.

Formula used:
The magnitude of the torque $\left( \tau \right)$ experienced by a coil of $N$ turns 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 answer:
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 same magnitude of opposing torque will have to be applied to keep the coil in stable equilibrium.
The magnitude of torque $\left( \tau \right)$ experienced by a coil of $N$ turns carrying a current $I$ and having cross sectional area $A$ when kept in a magnetic field $B$ is given by
$\tau =NAIB\sin \theta $ --(1)
Where $\theta $ is the angle made by the perpendicular to the plane with the direction of the magnetic field.
This torque has a direction such that it tries to make the plane of the coil perpendicular to the magnetic field.

Let us now analyze the question.
Number of turns of the coil $N=50$.
Length of the coil $=0.12m$
Breadth of the coil $=0.1m$
Therefore, cross sectional area of the rectangular coil $A=\text{length}\times \text{width = }0.12\times 0.1=0.012{{m}^{2}}$
Current in the coil $I=2A$
Strength of magnetic field $B=0.2Wb/{{m}^{2}}$
Since the plane of the coil is inclined at an angle of ${{30}^{0}}$ with the field, the perpendicular to the plane will be inclined at ${{90}^{0}}-{{30}^{0}}={{60}^{0}}$ with the field.
Hence, $\theta ={{60}^{0}}$.
Using (1), the torque $\left( \tau \right)$ experienced by the coil will be
$\tau =50\times 0.012\times 2\times 0.2\times \sin {{60}^{0}}=0.12\times 2\times \dfrac{\sqrt{3}}{2}\approx 0.20Nm$ $\left( \sin {{60}^{0}}=\dfrac{\sqrt{3}}{2} \right)$
Therefore, the magnitude of torque exerted by the magnetic field on the coil (and the magnitude of the torque required to keep it in equilibrium) is $0.20Nm$ . Hence, the correct answer is C) $0.20Nm$.

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.