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A rectangular loop has a sliding connect...

A rectangular loop has a sliding connector PQ of length l and resistance R`Omega` and it is moving with a speed v as shown. The set up is placed in a uniform magnetic field going into the plane of the paper. The three currents `I_(1),I_(2)` and I are

A

`I_(1)=-I_(2)=(Blv)/R, I=(2Blv)/R`

B

`I_(1)=I_(2)=(Blv)/(3R),I+(2Blv)/(3R)`

C

`I_(1)=I_(2)=(Blv)/R`

D

`I_(1)=I_(2)=(Blv)/(6R),I=(2Blv)/(3R)`

Text Solution

Verified by Experts

The correct Answer is:
B
Current `I=(vBl)/(R//2+R)=(2vBl)/(3R)`
`I_(1)=I_(2)=(vBl)/(3R)`
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