Electro Magnetic Induction And Alternating Currents Question 450

Question: 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.

Options:

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}} $

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Answer:

Correct Answer: A

Solution:

  • 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}} $



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