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
`l_(1)=-l_(2)=(Blv)/(R),l=(2Blv)/(R)`
B
`l_(1)=l_(2)=(Blv)/(3R),l=(2Blv)/(3R)`
C
`l_(1)=l_(2)=l=(Blv)/(R)`
D
`l_(1)=l_(2)=(Blv)/(6R),l=(Blv)/(3R)`
Text Solution
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The correct Answer is:
B
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