Electro Magnetic Induction And Alternating Currents Question 469

Question: A 0.1 m long conductor carrying a current of 50 I A is perpendicular to a magnetic field of 1.21 mT. The mechanical power to move the conductor 1 with a speed of $ 1m{{s}^{-1}} $ is

Options:

A) 0.25 m W

B) 6.25 m W

C) 0.625 W

D) 1W

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ P=Fv=BIlV=1.25\times {{10}^{-3}}\times 50\times 0.1\times 1W $

    $ =6.25\times {{10}^{-3}}W=6.25mW $



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