U-frame `ABCD` and a sliding rod `PQ` of resistance `R`, start moving with velocities v and `2v` respectively, parallel to a long wire carrying current `i_0`. When the distance `AP =1` at `t = 0`, determine the current through the inductor of inductance `L` just before connecting rod `PQ` loses contact with the U-frame.

Answer

B

Explanation

Since `PQ` and `DC` both cut the lines of field.
`:. ` motional emf will be indued across both of them.
Integrating, potential difference across
`dximplies intde=int_a^(2v) v (mu_0i_0)/(2pix) dx`
`e_(Dc)=(vmu_0i_0)/(2pi)` ln 2 with `D` at higher potential
`e_(PQ)=(2vmu_0i_0)/(2pi)` ln 2 with `P` at higher potential
The relative velocity of the rod `PQ` w.r.t `U` frame
Now relative velocity of the rod `PQ` w.r.t `U` frame
`v_("net")=2v-v=v`
Now, time taken by its to loose the contact `t=l/v`

From equivalent electrical network
Net emf in the closed loop `QPDC`.
`e=e_(PQ)=e_(DC)=(vmu_0i_0)/(2pi) ln 2`
Growth of current in the `L-R` circuit is given by
`i=i_0(1-e^((-t)/L))=(e/R)(1-e^((-tR)/L))`
At time `t=l/v`
`implies i=(e/R)(1-e^(-(Rl)/(vL)))`