A loop ABCD containing two resistors as shown in figure is placed in a uniform magnetic field B directed outward to the plane of page. Asliding conductor EF of length I and of negligible resistance moves to the right with a uniform velocity v as shown in Fig. Determine the current in each branch.

Explanation

The magnetic field inductionB, length l and the velocity v of the conductor EF are mutually perpendicular, hence the emf induced in it is e= B vl (from E to F)
`therefore` The effective electric circuit can be redrawn as shown in Fig

Let `i_(1), i_(2) and i` be the current through the branches DA, CB and EF respectively. Applying Kirchhoff.s law to the junction E or F, we have
`i_(1) + i_(2)=i` ....(i)
Applying Kirchhoff.s loop rule to the loops DAEFD and EBCFE respectively, we have
`i_(1)R_(1)- B vl= 0 or, i_(1)R_(1)= B vl` ...(ii)
and `-i_(2) R_(2) + Bv l= 0 or, i_(2)R_(2)= B vl` ...(iii)
From Eq. (ii) and (iii)
`i_(1) = (B v l)/(R_(1)) and i_(2)= (Bvl)/(R_(2))` Finally from Eqn (i), `i= i_(1)+ i_(2)= Bv l ((1 )/(R_(1)) + (1)/(R_(2)))`