JEE Mains · Physics · STD 12 - 5. Magnetism and matter
A bar magnet having a magnetic moment of \(2.0 \times\) \(10^{5} \; JT ^{-1}\), is placed along the direction of uniform magnetic field of magnitude \(B =14 \times 10^{-5}\; T\). The work done in rotating the magnet slowly through \(60^{\circ}\) from the direction of field is .............. \(J\)
- A \(14\)
- B \(8.4\)
- C \(4\)
- D \(1.4\)
Answer & Solution
Correct Answer
(A) \(14\)
Step-by-step Solution
Detailed explanation
Work done \(= MB \left(\cos \theta_{1}-\cos \theta_{2}\right)\) \(\theta_{1}=0^{\circ}, \theta_{2}=60^{\circ}\) \(=2 \times 10^{5} \times 14 \times 10^{-5}(1-1 / 2)\) \(=14 \;J\)
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