Question

A circular coil of radius $r = 10\text{ cm}$ having $300$ turns carries a current of $2\text{ A}$. The coil is suspended vertically in a uniform magnetic field of strength $7\text{ T}$. If the plane of the coil makes an angle of $30^\circ$ with the magnetic field, the torque needed to prevent it from turning is:

• A. $22.84\text{ N m}$
• B. $1.1\text{ N m}$
• C. $5.71\text{ N m}$
• D. $11.42\text{ N m}$

Hint:

Always check the wording for angles in magnetic torque questions. If it says "angle with the normal to the plane", use $\sin\theta$. If it says "angle with the plane of the coil", use $\cos\alpha$ directly!

Answer 1

The Correct Option is D

Understanding the Concept: A current-carrying magnetic loop experiences a mechanical torque ($\tau$) when placed in an external magnetic field, given by: \[ \tau = NIAB\sin\theta \] where $\theta$ represents the angle between the normal vector (perpendicular) of the coil's area plane and the magnetic field lines. If the problem states the angle relative to the plane of the coil itself ($\alpha$), then $\theta = 90^\circ - \alpha$, converting the formula to $\tau = NIAB\cos\alpha$.
Step 1: Verify standard values and angles.
We are given:
Number of turns, $N = 300$
Current, $I = 2\text{ A}$
Radius, $r = 10\text{ cm} = 0.1\text{ m}$
Field strength, $B = 0.7\text{ T}$
Angle of plane with field, $\alpha = 30^\circ$
The angle $\theta$ for our standard formula is: \[ \theta = 90^\circ - 30^\circ = 60^\circ \]
Step 2: Calculate loop area ($A$) and evaluate torque magnitude.
\[ A = \pi r^2 = \pi \times (0.1)^2 = 0.01\pi\text{ m}^2 \] Plugging all values into the torque formula: \[ \tau = 300 \times 2 \times (0.01\pi) \times 0.7 \times \sin(60^\circ) \] \[ \tau = 6\pi \times 0.7 \times \frac{\sqrt{3}}{2} = 2.1 \times \sqrt{3} \times \pi \] Substitute $\sqrt{3} \approx 1.732$ and $\pi \approx 3.1416$: \[ \tau \approx 2.1 \times 1.732 \times 3.1416 = 3.6372 \times 3.1416 \approx 11.42\text{ N m} \]