Electro Magnetic Induction And Alternating Currents Question 416
Question: PQ is an infinite current carrying conductor. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity v as shown. The force needed to maintain constant speed of EF is.
Options:
A) $ \frac{1}{vR}{{[ \frac{{\mu_{0}}Iv}{2\pi }In( \frac{b}{a} ) ]}^{2}} $
B) $ {{[ \frac{{\mu_{0}}Iv}{2\pi }In( \frac{b}{a} ) ]}^{2}}\frac{1}{vR} $
C) $ {{[ \frac{{\mu_{0}}Iv}{2\pi }In( \frac{b}{a} ) ]}^{2}}\frac{v}{R} $
D) $ \frac{v}{R}{{[ \frac{{\mu_{0}}Iv}{2\pi }In( \frac{b}{a} ) ]}^{2}} $
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Answer:
Correct Answer: A
Solution:
- Induced emf $ \int\limits_{a}^{b}{Bvdx}=\int\limits_{a}^{b}{\frac{{\mu_{0}}I}{2\pi x}vdx} $
$ \Rightarrow $ Induced emf, $ E=\frac{{\mu_{0}}Iv}{2\pi }ln( \frac{b}{a} ) $
$ \Rightarrow $ Power dissipated $ =\frac{E^{2}}{R} $ Also, power $ =F.v\Rightarrow F=\frac{E^{2}}{vR} $ $ F={{[ \frac{{\mu_{0}}Iv}{2\pi }ln( \frac{b}{a} ) ]}^{2}}\frac{1}{vR} $
