question_answer

A conducting disc of conductivity a has a radius 'a' and thickness 'f. If the magnetic field B is applied in a direction perpendicular to the plane of the disc changes with time at the rate of \[\frac{dB}{dt}=\alpha \]. Calculate the power dissipated in the disc due to the induced current.

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 \[P=\frac{\pi t\sigma }{2}{{\alpha }^{2}}\int\limits_{0}^{a}{{{r}^{3}}}\]   \[dr=\frac{\pi t\sigma {{a}^{4}}}{8}{{\alpha }^{2}}\]