Q78 / 120 · NEET 2010 (Mains) · Physics
A closely wound solenoid of 2000 turns and area of cross-section $1.5 \times 10^{-4} \mathrm{~m}^2$ carries a current of $2.0 \mathrm{~A}$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2} \mathrm{~T}$ making an angle of $30^{\circ}$ with the axis of the solenoid. The torque on the solenoid will be
- A$3 \times 10^{-3} \mathrm{~N}-\mathrm{m}$
- B$1.5 \times 10^{-3} \mathrm{~N}-\mathrm{m}$
- Correct answerC$1.5 \times 10^{-2} \mathrm{~N}$-m
- D$3 \times 10^{-2} \mathrm{~N}-\mathrm{m}$
Solution
Given, $\mathrm{N}=2000, \mathrm{~A}=1.5 \times 10^{-4} \mathrm{~m}^2$ $\mathrm{i}=2.0 \mathrm{AB}=5 \times 10^{-2} \mathrm{~T}$, and $\theta=30^{\circ}$
Torque, $\tau=$ NiBA $\sin \theta$
$$
\begin{aligned}
& =2000 \times 2 \times 5 \times 10^{-2} \times 1.5 \times 10^{-4} \times \sin 30^{\circ} \\
& =2000 \times 50 \times 10^{-6} \times \frac{1}{2} \\
& =1.5 \times 10^{-2} \mathrm{Nm}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.