Electro Magnetic Induction And Alternating Currents Question 451
Question: is a circular loop of radius r and resistance R. A variable magnetic field of induction $ B=B_{0}{{e}^{-t}} $ is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to.
Options:
A) $ \frac{B_{0}^{2}\pi r^{2}}{R} $
B) $ \frac{B_{0}10r^{3}}{R} $
C) $ \frac{B_{0}^{2}{{\pi }^{2}}r^{4}R}{5} $
D) $ \frac{B_{0}^{2}{{\pi }^{2}}r^{4}}{R} $
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Answer:
Correct Answer: D
Solution:
- $ e=-\frac{d\phi }{dt}–A\frac{dB}{dt}=-\pi r^{2}\frac{d}{dt}(B_{0}{{e}^{-t}})=\pi r^{2}B_{0}{{e}^{-t}} $ At $ t=0 $ , $ e=\pi r^{2}B_{0} $ , $ P=\frac{e^{2}}{R}=\frac{{{(\pi r^{2}B_{0})}^{2}}}{R}=\frac{{{\pi }^{2}}r^{4}B_{0}^{2}}{R} $
BETA
