Electro Magnetic Induction And Alternating Currents Question 317
Question: A wire of length 1 m is moving at a speed of 2ms?1 perpendicular to its length and a homogeneous magnetic field of 0.5 T. The ends of the wire are joined to a circuit of resistance 6 W. The rate at which work is being done to keep the wire moving at constant speed is [Roorkee 1999]
Options:
A) $ \frac{1}{12}W $
B) $ \frac{1}{6}W $
C) $ \frac{1}{3}W $
D) 1W
Show Answer
Answer:
Correct Answer: B
Solution:
Rate of work $ =\frac{W}{t}=P=Fv; $
also $ F=Bil=B,( \frac{Bvl}{R} )\ l $
Þ $ P=\frac{B^{2}v^{2}l^{2}}{R}=\frac{{{(0.5)}^{2}}\times {{(2)}^{2}}\times {{(1)}^{2}}}{6}=\frac{1}{6}W $
BETA
