Question

A metal bar AB can slide on two parallel thick metallic rails separated by a distance I. A resistance R and an inductance L are connected to the rails as shown in the figure. A long straight wire, carrying a constant current 70 is placed in the plane of the rails as shown. The bar AB is held at rest at a distance xQ from the long wire. At t = 0, it made to slide on the rails away from the wire. Answer the following questions.
(a) Find a relation among i,$\frac{di}{dt}$ and $\frac{d\varphi }{dt}$ where i is the current in the circuit and $\varphi$ is the flux of the magnetic field due to the long wire through the circuit.
(b) It is observed that at time $t=T$, the metal bar AB is at a distance of 2x${}_0$ from the long wire and the resistance $R$ carries a current Obtain an expression for the net charge that has flown through resistance $R$ from $t=0$ to t = T.
(c) The bar is suddenly stopped at time $T$. The current through resistance R is found to be $t_1$,/4 at time 2T.
Find the value of L/R in terms of the other given quantities.

Answer 1

(a) Applying Kirchhoff s second law, we get
$\frac{d\varphi }{dt}-iR-L\frac{di}{dt}=0$
or $\frac{d\varphi }{dt}=iR+L\frac{di}{dt}$ ...(i)
This is the desired relation between i, $\frac{di}{dt}and\frac{d\varphi }{dt}$
(b) Eq. (i) can be written as
$d\varphi =iRdt+Ldi$
Integrating we get
$\Delta \varphi =R\Delta q+L{i}_i$
$\Delta q=\frac{\Delta \varphi }{R}-\frac{L{i}_1}{R}$ ...(ii)
Here , $\Delta \varphi ={\varphi }_f-{\varphi }_i=\underset{x=2x_0}{\overset{x=x_0}{\int }}\frac{{\mu }_0{l}_0}{2\pi x}Idx$
$=\frac{{\mu }_0{i}_{0}l}{2\pi }ln(2)$
So, from Eq. (ii) charge flown through the resistance
upto time t = T, when current is $i_i$ is
$\Delta q=\frac{1}{R}[\frac{{\mu }_0{i}_{0}l}{2\pi }ln(2)-L{i}_1]$
(c) This is the case of current decay in an L-R circuit. Thus
$i=i_0{e}^{-t{\tau }_L}$ ...(iii)
Here ,$i=\frac{i_1}{4},i_0=i_1,t=(2T-T)=Tand{\tau }_L=\frac{L}{R}$
Substituting these values in Eq, (iii), we get
${\tau }_L=\frac{L}{R}=\frac{T}{ln4}$