WBJEE · Physics · Magnetic Effects of Current
A bar magnet has a magnetic moment of \(200 \mathrm{Am}^{2}\). The magnet is suspended in a magnetic field of \(0.30 \mathrm{NA}^{-1} \mathrm{m}^{-1}\). The torque required to rotate the magnet from its equilibrium position through an angle of \(30^{\circ},\) will be
- A \(30 \mathrm{N}\) -m
- B \(30 \sqrt{3} \mathrm{N}-\mathrm{m}\)
- C \(60 \mathrm{N}\) -m
- D \(60 \sqrt{3} \mathrm{N} \cdot \mathrm{m}\)
Answer & Solution
Correct Answer
(A) \(30 \mathrm{N}\) -m
Step-by-step Solution
Detailed explanation
Given, \(\mathrm{M}=200 \mathrm{A} \mathrm{m}^{2}\) \(\mathrm{B}=0.30 \mathrm{NA}^{-} \mathrm{M}^{-1}\) and \(\theta=30^{\circ}\) We know that the Torque,…
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