MHT CET · Physics · Magnetic Properties of Matter
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? \(\left[\sin \left(\frac{\pi}{2}\right)=1\right]\)
- A \(\sin ^{-1}(0 \cdot 8660)\)
- B \(\sin ^{-1}(0 \cdot 7071)\)
- C \(\sin ^{-1}(1)\)
- D \(\sin ^{-1}(0 \cdot 5)\)
Answer & Solution
Correct Answer
(D) \(\sin ^{-1}(0 \cdot 5)\)
Step-by-step Solution
Detailed explanation
(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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