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JEE Mains · Physics · STD 12 - Magnetism and matter
  • JEE Mains 2022
  • Medium
  • Numerical

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 .............. \(J\) 

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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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