JEE Mains · Physics · STD 12 - 5. Magnetism and matter
A small bar magnet placed with its axis at \(30^{\circ}\) with an external field of \(0.06\, T\) experiences a torque of \(0.018\, Nm\). The minimum work required to rotate it from its stable to unstable equilibrium position is
- A \(9.2 \times 10^{-3} J\)
- B \(11.7 \times 10^{-3} J\)
- C \(6.4 \times 10^{-2} J\)
- D \(7.2 \times 10^{-2} J\)
Answer & Solution
Correct Answer
(D) \(7.2 \times 10^{-2} J\)
Step-by-step Solution
Detailed explanation
Torque on a bar magnet: \(I = MB\) sin \(\theta\) Here, \(\theta=30^{\circ}, I =0.018\, N-m , B =0.06 T\) \(\Rightarrow 0.018= M \times 0.06 \times \sin 30^{\circ}\) \(\Rightarrow 0.018=M \times 0.06 \times \frac{1}{2}\) \(\Rightarrow M =0.6 A - m ^{2}\) Now…
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