Q124 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Physics
A bar magnet of length $10 \mathrm{~cm}$ and having the pole strength equal to $10^{-3} \mathrm{~A}-\mathrm{m}$ is kept in a magnetic field having magnetic induction $B$ equal to $4 \pi \times 10^{-3} \mathrm{~T}$. It makes an angle of $30^{\circ}$ with the direction of magnetic induction. The value of the torque acting on the magnet is
- Correct answerA$2 \pi \times 10^{-7} \mathrm{Nm}$
- B$2 \pi \times 10^{-5} \mathrm{Nm}$
- C$0.5 \mathrm{Nm}$
- D$0.5 \times 10^2 \mathrm{Nm}$
Solution
Given, length of bar magnet,
$$
l=10 \mathrm{~cm}=10^{-1} \mathrm{~m}
$$
Pole strength, $m=10^{-3} \mathrm{~A}-\mathrm{m}$
$$
\begin{aligned}
& B=4 \pi \times 10^{-3} \mathrm{~T} \\
& \theta=30^{\circ}
\end{aligned}
$$
$\therefore$ Magnetic dipole moment,
$$
M=m l=10^{-3} \times 10^{-1}=10^{-4} \mathrm{~A}-\mathrm{m}^2
$$
$\therefore$ Torque, $\tau=M B \sin \theta$
$$
\begin{aligned}
& =10^{-4} \times 4 \pi \times 10^{-3} \times \sin 30^{\circ} \\
& =4 \pi \times 10^{-7} \times \frac{1}{2}=2 \pi \times 10^{-7} \mathrm{~N}-\mathrm{m}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.