Question

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 (38-E24). 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 ; 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 nit (e) Find the velocity of the wire as a function of time ft) Find the displacement of the wire as a function ai 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.

Answer 1

(a)When the speed of wire is v emf developed =Blv

(b)Induced current is the wire $=\frac{Blv}{R}$

(from b to a)

(c)Downward acceleration of the wire

$=\frac{mg-F}{m}$ due to the current

$=\frac{ilB}{m}=g-\frac{B^2{l}^{2}v}{Rm}$

(d) Let the wire start moving with constant velocity then acceleration =0

$\frac{B^2{l}^2{v}_m}{Rm}=g$

$\Rightarrow v_m=\frac{gRm}{B^2{l}^2}$

(e)Since $\frac{dv}{dt}=a$

$\Rightarrow \frac{dv}{dt}=\left(\frac{\frac{mg-B^2{l}^{2}v}{R}}{m}\right)$

$\Rightarrow \frac{dv}{\frac{mg-\frac{B^2{l}^{2}v}{R}}{m}}=dt$

${\int }_0^v\frac{mdv}{mg-\frac{B^2{l}^{2}v}{R}}={\int }_0^{t}dt$

$\Rightarrow \frac{m}{\frac{B^2{l}^2}{R}}{\left[log\right(mg-\frac{B^2{l}^{2}v}{R}\left)\right]}_0^v=t$

$\Rightarrow \frac{-mR}{B^2{l}^2}\left[log\right(mg-\frac{B^2{l}^{2}v}{R})-log(mg\left)\right]=t$

$\Rightarrow log\left[\frac{mg-\frac{B^2{l}^{2}v}{R}}{mg}\right]=\frac{-t{B}^2{l}^2}{mR}$

$\Rightarrow log[1-\frac{B^2{l}^{2}v}{Rmg}]=\frac{-t{B}^2{l}^2}{mR}$

$\Rightarrow 1-\frac{B^2{l}^{2}v}{Rmg}=e^{-\frac{t{B}^2{l}^2}{mR}}$

$\Rightarrow (1-e^{-\frac{t{B}^2{l}^2}{mR}})=\frac{B^2{l}^{2}v}{Rmg}$

$v=\frac{Rmg}{B^2{l}^2}(1-e^{-\frac{B^2{l}^{2}t}{mR}})$

$\Rightarrow v=v_m(1-e^{-\frac{gt}{v_m}})[\because v_m=\frac{Rmg}{B^2{l}^2}]$

(f) Since $\frac{ds}{dt}=v$

$\int ds=\int v.dt$

$s=v_m{\int }_0^t(1-e^{-\frac{gt}{v_m}}).dt$

$=v_m.(t+\frac{v_m}{g}.e^{-\frac{dt}{v_m}})$

$=(v_{m}t+\frac{v_m^2}{g}{e}^{-\frac{dt}{v_m}})-\frac{v_m^2}{g}$

$=v_{m}t-\frac{v_m^2}{g}(1-e^{-\frac{dt}{v_m}})$

$\left(g\right)\frac{d}{dt}\left(mgs\right)=mg.\frac{ds}{dt}$

$=mg.v_m(1-e^{-\frac{gt}{v_m}})$

$\frac{d_H}{dt}=l^{2}R=R.{\left(\frac{lBv}{R}\right)}^2$

$=\frac{l^2{B}^2{v}^2}{R}$

$=\frac{l^2{B}^2}{R}.v_m^2{(1-e^{-\frac{gt}{v_m}})}^2$

After steady state; i.e. $t\to \infty$

$\frac{d}{dt}\left(mgs\right)=mg{v}_m$

$\frac{d_H}{dt}=\frac{l^2{B}^2}{R}{v}_m^2$

$=\frac{l^2{B}^2}{R}.v_m.\frac{mgR}{l^2{B}^2}$

$=mg{v}_m$

Hence after steady state

$\frac{d_H}{dt}=\frac{d}{dt}mgs$