A) \[0.8\times {{10}^{7}}\,\text{erg}\]
B) \[0.8\,erg\]
C) \[8\,J\]
D) \[0.4\,J\]
Correct Answer: A
Solution :
Work done, \[W=MB(cos{{\theta }_{1}}-cos{{\theta }_{2}}).\] When the magnet is rotated from \[{{0}^{o}}\] to\[{{60}^{o}},\] then work done is 0.8 J \[0.8=MB(cos{{0}^{o}}-cos{{60}^{o}})\] \[=\frac{MB}{2}\] \[MB=0.8\times 2=1.6\,N-m\] In order to rotate the magnet through an angle of \[{{30}^{o}},\] i.e., from \[{{60}^{o}}\] to\[{{90}^{o}},\] the work done is \[W=MB(\cos {{60}^{o}}-\cos {{90}^{o}})\] \[=MB\left( \frac{1}{2}-0 \right)=\frac{MB}{2}\] \[W=\frac{1.6}{2}=0.8\,J\] \[=0.8\times {{10}^{7}}\,erg\]
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