A conducting wire ab of length l, resistance r and mass m starts sliding at t = 0 down a smooth, vertical, thick pair of connected rails as shown in figure. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the rails. (a) Write the induced emf in the loop at an instant t when the speed of the wire is v. (b) What would be the magnitude and direction of the induced current in the wire? (c) Find the downward acceleration of the wire at this instant. (d) After sufficient time, the wire starts moving with a constant velocity. Find this velocity v_m . (e) Find the velocity of the wire as a function of time. (f) Find the displacement of the wire as a function of time. (g) Show that the rate of heat developed in the wire is equal to the rate at which the gravitational potential energy is decreased after steady state is reached.
Figure
  

(a) When the speed of the wire is v, the emf developed in the loop is, e = Blv.

(b)
Magnitude of the induced current in the wire, I = $\frac{\mathit{B}\mathit{l}\mathit{v}}{\mathit{R}}$
As wire is moving the magnetic flux passing through the loop is increasing with time. Therefore, the direction of the current should be as such to oppose the change in magnetic flux. Therefore in order to induce the current in anticlockwise direction the current flows from b to a
(c)
Due to motion of the wire in the magnetic field there is a force in upward direction (perpendicular to the wire).
The magnitude of the force on the wire carrying current i is given by $F=ilB$

The net force on the wire = $mg-F=mg-ilB$
Downward acceleration of the wire due to current, a$=\frac{mg-F}{m}$
$$\begin{gathered} a=\frac{mg-ilB}{m} \\ a=g-\frac{B^2{l}^{2}v}{\mathrm{R}m} \left[\because i=\frac{Blv}{R}\right] \end{gathered}$$
(d) Let the wire start moving with a constant velocity.
Now,
Let the speed of the wire be v_m
As speed is constant, acceleration, a = 0
$a=g-\frac{B^2{l}^2{v}_{\mathrm{m}}}{Rm} =0$
$$\begin{gathered} \frac{B^2{l}^2{v}_{\mathrm{m}}}{Rm}=g \\ \Rightarrow v_{\mathrm{m}}=\frac{gRm}{B^2{l}^2} \end{gathered}$$
(e)
The acceleration of the wire can be expressed as time rate of change of velocity
 $\frac{dv}{dt}=a$
$$\begin{gathered} \therefore \frac{dv}{dt}=\left(\frac{mg-B^2{l}^{2}v/R}{m}\right) \\ \Rightarrow \frac{dv}{{\displaystyle \frac{mg-B^2{l}^{2}v/R}{m}}}=dt \\ \underset{0}{\overset{v}{\int }}\frac{m dv}{mg-{\displaystyle \frac{B^2{l}^{2}v}{R}}}=\underset{0}{\overset{t}{\int dt}} \\ \Rightarrow \frac{m}{{\displaystyle -\frac{B^2{l}^2}{R}}}{\left[\log \left(mg-\frac{B^2{l}^{2}v}{R}\right)\right]}_0^v=t \\ \Rightarrow \frac{-mR}{B^2{l}^2}\left[\log \left(mg-\frac{{\mathrm{B}}^2{l}^{2}v}{\mathrm{R}}\right)-\log \left(mg\right)\right]=t \\ \Rightarrow \log \left[\frac{mg-{\displaystyle \frac{B^2{l}^{2}v}{R}}}{mg}\right]=\frac{-t{B}^2{l}^2}{mR} \\ \Rightarrow \log \left[1-\frac{{\mathrm{B}}^2{l}^{2}v}{\mathrm{R}mg}\right]=\frac{-t{B}^2{l}^2}{m\mathrm{R}} \\ \Rightarrow 1-\frac{B^2{l}^{2}v}{Rmg}=e^{-\frac{t{B}^2{l}^2}{mR}} \\ \Rightarrow \left(1-e^{-\frac{t{B}^2{l}^2}{mR}}\right)=\frac{B^2{l}^{2}v}{Rmg} \\ v=\frac{Rmg}{B^2{l}^2}\left(1-e^{-\frac{{\mathrm{B}}^2{l}^{2}t}{m\mathrm{R}}}\right) \\ \Rightarrow v=v_{\mathrm{m}}\left(1-e^{-\frac{gt}{v_{\mathrm{m}}}}\right) \left[ \because v_m=\frac{Rmg}{B^2{l}^2}\right] \end{gathered}$$
(f)
The velocity can be expressed as time rate of change of position
x is the position of the wire at instant t.
∵ $\frac{dx}{dt}=v$
Thus, the displacement of the wire can be expressed as:
$s={\int }_0^{t}dx={\int }_0^{t}v\mathrm{d}t$
$$\begin{gathered} \therefore s=x_t-x_0=v_m\underset{0}{\overset{t}{\int }}\left(1-e^{-\frac{gt}{v_m}}\right).dt \\ s=v_m.{\left(t+\frac{v_m}{g}.e^{-\frac{gt}{v_m}}\right)}_0^t \\ s=\left(v_{m}t+\frac{{v_m}^2}{g}{e}^{-\frac{gt}{v_m}}\right)-\frac{{v_m}^2}{g} \\ s=v_{m}t-\frac{{v_m}^2}{g}\left(1-e^{-\frac{gt}{v_m}}\right) \end{gathered}$$

(g)
Rate of development of heat in the wire is given by P = V × i
 $$\begin{gathered} V=Blv \\ i=\frac{Blv}{R} \end{gathered}$$
Therefore, the rate of development of heat in the wire is given by
$$\begin{gathered} P=Blv\times \frac{Blv}{R}=\frac{B^2{l}^2{v}^2}{R} \\ P=\frac{B^2{l}^2{v_m}^2{\left(1-e^{-\frac{gt}{v_m}}\right)}^2}{R} \left[\because v=v_m\left(1-e^{-\frac{gt}{v_m}}\right)\right] \end{gathered}$$

Rate of decrease in potential energy is given by
$$\begin{gathered} \frac{\mathrm{d}U}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(mgx\right)=mg.\frac{dx}{dt} \\ \frac{\mathrm{d}U}{\mathrm{d}t}=mgv \\ \frac{\mathrm{d}U}{\mathrm{d}t}=mg.v_m\left(1-e^{-\frac{gt}{v_m}}\right) \left[\because v=v_m\left(1-e^{-\frac{gt}{v_m}}\right) \right] \end{gathered}$$
After the steady state, i.e., t → ∞,
$$\begin{gathered} \frac{dU}{dt}=mg{v}_{\mathrm{m}} \\ P=\frac{l^2{B}^2}{R}{v_m}^2 \\ P=\frac{l^2{B}^2}{R}\times v_m\times \frac{mg\mathrm{R}}{l^2{B}^2} \left[\because v_m=\frac{mg\mathrm{R}}{l^2{B}^2}\right] \\ P=mg{v}_m \end{gathered}$$
Thus, after the steady state, $P=\frac{\mathrm{d}U}{\mathrm{d}t}$