Q151 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Physics
A short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800 \mathrm{G}$ experiences a torque of $0.016 \mathrm{Nm}$. The magnetic moment of the bar magnet is
- Correct answerA$0.4 \mathrm{Am}^2$
- B$0.5 \mathrm{Am}^2$
- C$0.6 \mathrm{Am}^2$
- D$0.7 \mathrm{Am}^2$
Solution
$\because \tau=M B \sin \theta$
where, $\tau=$ torque experienced by bar magnet
$=0.016 \mathrm{~N}-\mathrm{m}$
$M=$ magnetic moment of bar magnet
$B=$ external magnetic field intensity
$=800 \mathrm{G}=800 \times 10^{-4} \mathrm{~T}=8 \times 10^{-2} \mathrm{~T}$
$\theta=$ angle made by the axis of bar magnet with the external magnetic field $=30^{\circ}$
$\begin{aligned}
& \text { So, } 0.016=M \times 8 \times 10^{-2} \times \sin \left(30^{\circ}\right) \\
& \Rightarrow \quad 0.016=M \times 8 \times 10^{-2} \times \frac{1}{2} \\
& \Rightarrow \quad 1.6 \times 10^{-2}=4 M \times 10^{-2} \Rightarrow M=\frac{1.6}{4} \\
& \Rightarrow \quad M=0.4 \mathrm{~A}-\mathrm{m}^2
\end{aligned}$
Hence, the correct answer is option: 'A'.