A) \[\frac{{{a}^{2}}{{\tau }^{3}}}{4R}\]
B) \[\frac{{{a}^{2}}{{\tau }^{3}}}{3R}\]
C) \[\frac{{{a}^{2}}{{\tau }^{3}}}{6R}\]
D) \[\frac{{{a}^{2}}{{\tau }^{3}}}{2R}\]
Correct Answer: B
Solution :
[b]: The flux through the stationary loop in the problem is \[\phi =at(\tau -t)\]Induced emf, \[\varepsilon =-\frac{d\phi }{dt}=-\frac{d}{dt}[at(\tau -t)]=-[a\tau -2at]=(2at-a\tau )\]The amount of heat generated in the loop during a small time interval dt is \[dQ=\frac{{{\varepsilon }^{2}}}{R}dt=\frac{{{(2at-a\tau )}^{2}}}{R}dt\] Hence, the total heat generated is \[Q=\int_{0}^{\tau }{\frac{{{(2at-a\tau )}^{2}}}{R}}dt=\frac{1}{R}\int\limits_{0}^{\tau }{(4{{a}^{2}}{{t}^{2}}+{{a}^{2}}{{\tau }^{2}}-4{{a}^{2}}\tau t)dt}\]\[=\frac{1}{R}\left[ \frac{4}{3}{{a}^{2}}{{t}^{3}}+{{a}^{2}}{{\tau }^{2}}t-\frac{4}{2}{{a}^{2}}\tau {{t}^{2}} \right]_{0}^{\tau }=\frac{1}{3}\frac{{{a}^{2}}{{\tau }^{3}}}{R}\]
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