A rectangular loop with a sliding conductor of length `l` is located in a uniform magnetic field perpendicular to the plane of loop. The magnetic induction perpendicular to the plane of loop Is equal to `B`. The part ad and be has electric resistance `R_1` and `R_2`, respectively. The conductor starts moving with constant acceleration `a_0`, at time `t = 0`. Neglecting the self-inductance of the loop and resistance of conductor. Find

(a) the current through the conductor during its motion.
(b) the polarity of `abcd` terminal.
(c) external force required to move the conductor with the given acceleration.

Answer

A, B

Explanation

a. At time `t, v=a_0t`
Motional emf, `V=Bvl, Ba_0l t`
Total resistance `=(R_1R_2)/(R_1+R_2)`
`:. i=((Ba_0l t)(R_1+R_2))/(R_1R_2)`
b. From right hand rule, we can see that points `a` and `b` will be at higher potential and `c` and `d` at lower potentials.
`F_m=(ilB)=(B^2l^2a_0t)/(R_1R_2)(R_1+R_2)`
Let `F` be the external force applied the
`F-F_m=ma_0`
`:. F=F_m+ma_0=(B^2l^2a_0t)/(R_1R_2)+ma_0`