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}} $
BETA
