A) 3.6 A
B) 1.6 A
C) 6.2 A
D) 4.2 A
Correct Answer: C
Solution :
Given that r = 0.20 m, t = 10s, \[R=2\,\Omega \] \[B=(4.0\,T/{{S}^{2}}){{t}^{2}}+(2.0\,T/S)\,t+3.0\,T\] \[\frac{dB}{dt}=8\,t+2\] From \[E=-\frac{d\phi }{dt}=-A\frac{d\phi }{dt}\] \[E=-\pi {{r}^{2}}[8\,t+2]\] \[=3.14\times 0.2\times 0.2\times 82\] = 10.30 Total \[E=10.30+2\] \[=12.30\,V\] \[l=\frac{E}{R}\] \[=\frac{12.30}{2.0}=6.15\] \[\approx 6.2\,A\]You need to login to perform this action.
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