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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? $\left[\sin \left(\frac{\pi}{2}\right)=1\right]$
Options:
on the magnet is to be halved by rotating it. Through what angle it should be
rotated? $\left[\sin \left(\frac{\pi}{2}\right)=1\right]$
Solution:
1492 Upvotes
Verified Answer
The correct answer is:
$\sin ^{-1}(0 \cdot 5)$
(D)
$T=M B \sin \theta$
When $\theta=\frac{\pi}{2}, \quad \mathrm{~T}=\mathrm{MB} \sin \frac{\pi}{2}=\mathrm{MB}$
Let $\quad T^{\prime}=\frac{T}{2}=M B \sin \theta^{\prime}=\frac{M B}{2}=M B \sin \theta^{\prime}$
$\therefore \sin \theta^{\prime}=\frac{1}{2}=0.5$
$\theta^{\prime}=\sin ^{-1} 0.5$
$T=M B \sin \theta$
When $\theta=\frac{\pi}{2}, \quad \mathrm{~T}=\mathrm{MB} \sin \frac{\pi}{2}=\mathrm{MB}$
Let $\quad T^{\prime}=\frac{T}{2}=M B \sin \theta^{\prime}=\frac{M B}{2}=M B \sin \theta^{\prime}$
$\therefore \sin \theta^{\prime}=\frac{1}{2}=0.5$
$\theta^{\prime}=\sin ^{-1} 0.5$
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