Electro Magnetic Induction And Alternating Currents Question 337
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} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ P=\frac{e^{2}}{R};\ \ e=-\frac{d}{dt}(BA)=A\frac{d}{dt}(B_{o}{{e}^{-t}})=AB_{o}{{e}^{-t}} $
$ \Rightarrow P=\frac{1}{R}{{(AB_{o}{{e}^{-t}})}^{2}}=\frac{A^{2}B_{o}^{2}{{e}^{-2t}}}{R} $ At the time of starting t = 0 so $ P=\frac{A^{2}B_{o}^{2}}{R} $
$ \Rightarrow P=\frac{{{(\pi r^{2})}^{2}}B_{o}^{2}}{R}=\frac{B_{o}^{2}{{\pi }^{2}}r^{4}}{R} $
