A) \[5/\sqrt{3}V\]
B) \[2\pi /\sqrt{3}V\]
C) \[(5\sqrt{3}+2\pi /3)\,V\]
D) \[(5\sqrt{3}+\pi )\,V\]
Correct Answer: C
Solution :
[c] The induced emf across the ends B and F due to motion of the loop, \[{{e}_{1}}=Bv(BF)=5\times 1\times 2\,\sin \,{{60}^{o}}=5\sqrt{3}\,V\] The induced emf across the loop due to change in magnetic field \[{{e}_{2}}=A\frac{dB}{dt}=\frac{\pi {{R}^{2}}}{3}\left( \frac{dB}{dt} \right)=\frac{\pi {{(1)}^{2}}}{3}\times 2=\frac{2\pi }{3}V\] So, \[e={{e}_{1}}+{{e}_{2}}=\left( 5\sqrt{3}+\frac{2\pi }{3} \right)V\]
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