Two parallel long smooth conducting rails separated by a distance `l` are connected by a movable conducting connector of mass `m`. Terminals of the rails are connected by the resistor `R` and the capacitor `C` as shown in figure. A uniform magnetic field `B` perpendicular to the plane of the rail is switched on. The connector is dragged by a constant force `F`. Find the speed of the connector as a function of time if the force `F` is applied at `t = 0`. Also find the terminal velocity of the connector.

Answer

B

Explanation

Let `v` be the veloicty of connector at some instant of time. Then,

`V_(ab)=Bvl, i_1=(Bvl)/R, q=C (Bvl)`
`:. i_2=(dq)/(dt)=CBI(dv)/(dt)`
Now, `i=i_1+i_2=(Bvl)/R+CBI (dv)/(dt)`
Magnetic force `F_m=ilB=(B^2l^2)/R.v+B^2l^2C.(dv)/(dt)`
Further `F_("net")=F-F_m`
or `m (dv)/(dt)=F-(B^2l^2)/Rv-B^2l^2C(dv)/(dt)`
`:. int_0^v(dv)/(F-(B^2l^2)/Rv)=int_0^1(dt)/(m+B^2l^2C)`
Integrating we get,
`V=(FR)/(B^2l^2)[1-1^(-((B^(2)l^(2))/(mR+RB^(2)l^(2)C))t)]`
Terminal velocity in this case is : `v_T=(FR)/(B^2l^2)`