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← JEE Main 2022 · 26 Jun · Shift 2
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JEE Main 2022 · 26 Jun · Shift 2, Physics question 42

A bar magnet having a magnetic moment of 2.0×1052.0\times10^5 J T−1^{-1}, is placed along the direction of uniform magnetic field of magnitude B=14×10−5B=14\times10^{-5} T. The work done in rotating the magnet slowly through 60° from the direction of field is :

Choose one option
Answer and worked solution

Official answer: A

Work. W=mB(1−cos⁡60°)=2×105×14×10−5×12=14W=mB(1-\cos60°)=2\times10^5\times14\times10^{-5}\times\frac12=14 J.

(A)

Topics:Magnetic dipole and bar magnet
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