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 $
