A) 0.2 C
B) 0.4 C
C) 2 C
D) 4 C
Correct Answer: B
Solution :
According to Faraday's second law of electromagnetic induction emf \[e=-\frac{d\phi }{dt}\] or \[IR=\frac{-d\phi }{dt}\] or \[Idt=-\frac{d\phi }{R}\] \[\because \] \[Idt=q\] \[\therefore \] \[q=-\frac{d\phi }{R}\] \[\because \] \[d\phi =\frac{NBA(\cos \,{{\theta }_{2}}-\cos {{\theta }_{1}})}{R}\] Given, \[N=500,\] \[R=50\Omega ,\] \[A=0.1{{m}^{2}}\] \[\therefore \] \[q=\frac{500\times (0.2)\times (0.1)\times (\cos \,{{180}^{o}}-\cos {{0}^{o}})}{50}\] \[\Rightarrow \] \[q=0.4C\]You need to login to perform this action.
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