Electro Magnetic Induction And Alternating Currents Question 342
Question: A wire cd of length l and mass m is sliding without friction on conducting rails ax and by as shown. The vertical rails are connected to each other with a resistance R between a and b. A uniform magnetic field B is applied perpendicular to the plane abcd such that cd moves with a constant velocity of
Options:
A) $ \frac{mgR}{Bl} $
B) $ \frac{mgR}{B^{2}l^{2}} $
C) $ \frac{mgR}{B^{3}l^{3}} $
D) $ \frac{mgR}{B^{2}l} $
Show Answer
Answer:
Correct Answer: B
Solution:
Due to magnetic field, wire will experience an upward force
$ F=Bil=B\ ( \frac{Bvl}{R} )\ l $
$ \Rightarrow F=\frac{B^{2}vl^{2}}{R} $
If wire slides down with constant velocity then
$ F=mg\Rightarrow \frac{B^{2}vl^{2}}{R}=mg\Rightarrow v=\frac{mgR}{B^{2}l^{2}} $
BETA
