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

Answer

B

Explanation

(b) Due to the movement of resistorR, an emf equal to Blv will be induced in it as shown in figure clearly,
`I=(I_1)+(I_2)`
Also, `(I_1)=(I_2)`
Solving the circuit, we get `(I_1)=(I_2)=(Blv)/(3R)`
and I=2(I_1)=(2 Blv)/(3R) `