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}} $

Show Answer

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} $


जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक