A) \[\frac{a{{\tau }^{3}}}{3R}\]
B) \[\frac{{{a}^{2}}{{\tau }^{3}}}{3R}\]
C) \[\frac{{{a}^{3}}{{\tau }^{3}}}{3R}\]
D) \[\frac{a\tau }{3R}\]
Correct Answer: B
Solution :
[b]: Induced emf, \[E=-\frac{d\phi }{dt}=2at-a\tau ,\] Current through the loop,\[I=\frac{E}{R}=\frac{2at-a\tau }{R}\]?(i) dQ = amount of heat generated in a time\[dt={{I}^{2}}Rdt\] \[\therefore \]Total amount of heat generated in that time is \[Q=\int_{{}}^{{}}{dQ}=\int\limits_{0}^{\tau }{{{I}^{2}}Rdt}=\int\limits_{0}^{\tau }{\frac{{{(2at-a\tau )}^{2}}Rdt}{{{R}^{2}}}}\](Using (i)) By solving integration, we get\[Q\approx \frac{{{a}^{2}}{{\tau }^{3}}}{3R}\]
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