A bar magnet is held perpendicular to a uniform magnetic field. The couple acting on the magnet is to be…
A bar magnet is held perpendicular to a uniform magnetic field. The couple acting
on the magnet is to be halved by rotating it. Through what angle it should be
rotated? $\left[\sin \left(\frac{\pi}{2}\right)=1\right]$
$\sin ^{-1}(0 \cdot 8660)$
$\sin ^{-1}(0 \cdot 7071)$
$\sin ^{-1}(1)$
$\sin ^{-1}(0 \cdot 5)$
Solution
$T=M B \sin \theta$
When $\theta=\frac{\pi}{2}, \quad \mathrm{~T}=\mathrm{MB} \sin \frac{\pi}{2}=\mathrm{MB}$
Let $\quad T^{\prime}=\frac{T}{2}=M B \sin \theta^{\prime}=\frac{M B}{2}=M B \sin \theta^{\prime}$
$\therefore \sin \theta^{\prime}=\frac{1}{2}=0.5$
$\theta^{\prime}=\sin ^{-1} 0.5$