A bar magnet having a magnetic moment of 4×104 JT-1 is free to rotate in a horizontal plane. A horizontal magnetic field B=3×10−4 T exits in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 600 from the field is
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a
12 J
b
6 J
c
2 J
d
0.6 J
answer is B.
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Detailed Solution
Work done in turning the magnet of moment M=4×104 JT-1 in a field B=3×10-4 T through an angle 600 is W = Uf-Ui = [-MBcos60o] - [-MBcos0o] = MB2 ⇒W=4×104×3×10−42 ⇒W=12×12⇒W=6 J .