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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 .............. \(J\) 

  1. A \(14\)
  2. B \(8.4\)
  3. C \(4\)
  4. D \(1.4\)
Verified Solution

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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