Q176 / 200 · MHT CET - Medical 2026 (22 April Shift 2) · Physics
A closely wound solenoid of $1000$ turns and area of cross-section $2 \times 10^{-4}\ \text{m}^2$ carries a current of $2$ A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2}$ T making an angle of $60^\circ$ with the axis of solenoid. The torque on a solenoid will be nearly $\left(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\right)$
- Correct answerA$1.7 \times 10^{-2}$ Nm
- B$1.5 \times 10^{-2}$ Nm
- C$3.4 \times 10^{-2}$ Nm
- D$2 \times 10^{-2}$ Nm
Solution
The magnetic moment of the solenoid is given by $M = NIA$
Substituting the given values:
$M = 1000 \times 2 \times 2 \times 10^{-4} = 0.4\ \text{A m}^2$
The torque acting on the solenoid in a uniform magnetic field is given by $\tau = MB \sin \theta$
Substituting the values:
$\tau = 0.4 \times 5 \times 10^{-2} \times \sin 60^\circ$
$\tau = 2 \times 10^{-2} \times \dfrac{\sqrt{3}}{2}$
$\tau = \sqrt{3} \times 10^{-2} \approx 1.732 \times 10^{-2}\ \text{Nm}$
Answer: $1.7 \times 10^{-2}$ Nm
Substituting the given values:
$M = 1000 \times 2 \times 2 \times 10^{-4} = 0.4\ \text{A m}^2$
The torque acting on the solenoid in a uniform magnetic field is given by $\tau = MB \sin \theta$
Substituting the values:
$\tau = 0.4 \times 5 \times 10^{-2} \times \sin 60^\circ$
$\tau = 2 \times 10^{-2} \times \dfrac{\sqrt{3}}{2}$
$\tau = \sqrt{3} \times 10^{-2} \approx 1.732 \times 10^{-2}\ \text{Nm}$
Answer: $1.7 \times 10^{-2}$ Nm
Hence, the correct answer is option: 'A'.