A rectangular coil of length $0.12 ext{ m}$ and width $0.1 ext{ m}$ having $50$ turns of wire is suspended

Question

A rectangular coil of length $0.12 	ext{ m}$ and width $0.1 	ext{ m}$ having $50$ turns of wire is suspended vertically in a uniform magnetic field of strength $0.2 	ext{ Wb/m}^2$. The coil carries a current of $2 	ext{ A}$. If the plane of the coil is inclined at an angle of $30^\circ$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be:

Accepted Answer

1. **Formula:** The torque ($	au$) exerted on a current-carrying coil in a magnetic field is given by $	au = NIAB \sin 	heta$, where:
   - $N$ is the number of turns.
   - $I$ is the current.
   - $A$ is the area of the coil.
   - $B$ is the magnetic field strength.
   - $	heta$ is the angle between the **normal to the coil's area** and the magnetic field .
2. **Identify Angle:** The problem states the **plane of the coil** makes an angle of $30^\circ$ with the field. Therefore, the angle between the normal to the coil and the field is $	heta = 90^\circ - 30^\circ = 60^\circ$ .
3. **Calculation:**
   - Area $A = 	ext{length} 	imes 	ext{width} = 0.12 	imes 0.1 = 0.012 	ext{ m}^2$.
   - Substituting values: $	au = 50 	imes 2 	imes 0.012 	imes 0.2 	imes \sin(60^\circ)$.
   - $	au = 0.24 	imes \frac{\sqrt{3}}{2} \approx 0.24 	imes 0.866$.
   - $	au \approx 0.2078 	ext{ N-m}$.
4. **Conclusion:** The value closest to the calculated result is $0.20 	ext{ N-m}$.

Step-by-Step Solution

Practice All PHYSICS Questions