A) \[\frac{\pi t\sigma {{a}^{4}}}{8}{{\alpha }^{2}}\]
B) \[\frac{\pi t\sigma {{a}^{4}}}{4}{{\alpha }^{2}}\]
C) \[\frac{\pi t\sigma {{a}^{4}}}{2}{{\alpha }^{2}}\]
D) \[\frac{2\pi t\sigma {{a}^{4}}}{3}{{\alpha }^{2}}\]
Correct Answer: A
Solution :
[a] Consider an elemental circle of thickness dr. The induced emf in the circular path of radius r is \[\varepsilon =\frac{d}{dt}(\pi {{r}^{2}}B)=\pi {{r}^{2}}\alpha \] The resistance of circular path is The length of the path being \[2\pi r\] and tdr is the cross sectional area of current flow. For the element the power dissipated inside the path is \[dP=\frac{{{\varepsilon }^{2}}}{R}=\frac{\pi t\sigma }{2}{{\alpha }^{2}}{{r}^{3}}dr\] The total dissipated power P is \[P=\frac{\pi t\sigma }{2}{{\alpha }^{2}}\int\limits_{0}^{a}{{{r}^{3}}}\] \[dr=\frac{\pi t\sigma {{a}^{4}}}{8}{{\alpha }^{2}}\]You need to login to perform this action.
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