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

Options

A. `I_(1) = I_(2) = (B l upsilon)/(3 R), I = (2 B l upsilon)/(3R)`
B. `I_(1) = I_(2) = I = ( B l upsilon)/(R)`
C. `I_(1) = I_(2) = (B l upsilon)/(6 R), I = (B l upsilon)/(3R)`
D. `I_(1) = I_(2) = (B l upsilon)/(R), I = (2 B l upsilon)/(R)`

Answer

A

Explanation

Induced emf produced in the arm PQ is e = Blv. It means there is a source of emf e ( = Blv) in arm QP with resistance R in tha arm and other two resistance each of value R in parallel to it. Therefore. Effective resistance
`R' = (R xx R)/(R + R) + R = (3R)/(2)`
Current, `I = (e)/(R') = (Blv)/(3 R // 2) = (2 B l v)/( 3R)` In Fig. `I_(1) = I_(2) = (I)/(2) = (Blv)/(3R)`