A) 0.2 J
B) 2.0 J
C) 4.18 J
D) 2 × 102 J
Correct Answer: A
Solution :
Magnetic moment of bar \[M={{10}^{4}}J/T\] \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\sqrt{2}M}{{{d}^{3}}}\] Hence work done \[W=\overrightarrow{M}.\overrightarrow{B}\] \[={{10}^{4}}\times 4\times {{10}^{-5}}\times \cos {{60}^{o}}=0.2\ J\]
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