A conducting rod shown in figure of mass `m` and length `l` moves on two frictionless horizontal parallel rails in the presence of a unifom magnetic field directed into the page., The rod is given an initial velocity `v_0` to the right and is released at `t=0`. Find s a function of time
a. the volocity of the rod
b. the induced current and
c. the magntitude of the induced emf.

Explanation

a. Let `v` be the velocity of the rod at time `t`
current in the ciruit at his moment is
`i=(Bvl)/R` …..i
From right had rule, we can see that this current is in counterclockwise direction.
The magnetic force is
`F_m=-ilB=-(B^2l^2)/Rv`
Hence, negative sign denotes that the force is to the left and retards the motion. This is the only horizontal force acting on the bar, and hence, Newton's second law applied to motion is horizontal direction gives
`m(dv)/(dt)=F_m=-(B^2l^2)/Rv`
`:. (dv)/v=-((B^2l^2)/(mR))dt`

Integrating this equation using the initial condition that
`v=v_0` at `t=0`, we find that
`int_(v_0)^v(dv)/v=(B^2l^2)/(mR) int_0^t dt`
Solving this equation we find that
`v =v_(0)e^(-t//tau)` ...........ii
where ,` tau=(mR)/(B^2l^2)`
This expression indicates that the velocity of the rod decreses expnentially with time under the action of the magnetic retarding force.
b. `i=(Bvl)/r`
substituting the value of `v` from Eqn (ii), we get
`i=(Blv_0)/R e^(-t//tau)`
c. `eiR=Blv_0e^(-t//tau)`
i. and e both decrease exonentially with time v-t, i=t and e-t graphs are as shown in figure.

Alternate solution: This problem can also be solved by energy conservation principle. Let at some instant velocity of the rod is `v`. As no external force is present. Energy is dissipated in the resistor at the cost of kinetic energy of the rod. Hence,
`(-(dK)/(dt))=` power dissipated in the resistor
or `-d/(dt)(1/2mv^2)=e^2/R`
or `-mv((dv)/(dt))=(B^2l^2v^2)/R` (as `e=Bvl`)
`:. (dv)/v=(B^2l^2)/(mR)dt`
`:. int_(v_0)^v (dv)/v=(B^2l^2)/(mR)dt`
or `v=v_0e^(-t//tau),` where `tau=(mR)/(B^2l^2)`