A) \[0.8\times {{10}^{7}}\,erg\]
B) 0.8 erg
C) 8J
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 \[0.8=MB\,\,(\cos {{0}^{o}}-\cos {{60}^{o}})\] \[=\frac{MB}{2}\] \[M=0.8\times 2=1.6\,N-m\] In order to rotate the magnet through an angle of \[{{30}^{o}}\], i.e., from \[{{60}^{o}}\] to \[{{90}^{o}}\], the work done is \[W=MV(\cos {{60}^{o}}-\cos {{90}^{o}})\] \[=MB\left( \frac{1}{2}-0 \right)=\frac{MB}{2}\] \[W=\frac{1.6}{2}=0.8\,\,J\] \[=0.8\times {{10}^{7}}erg\]
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