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 $\frac{d\phi}{dt}$ where i is the current in the circuit and $\phi$ 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\phi}{dt} -iR -L\frac{di}{dt}=0$
or $ \frac{d\phi}{dt}=iR + L \frac{di}{dt} $ ...(i)
This is the desired relation between i, $\frac{di}{dt} \, and \, \frac{d\phi}{dt}$
(b) Eq. (i) can be written as
$d\phi =iRdt +Ldi$
Integrating we get
$\Delta \phi =R \Delta q +Li_i$
$ \Delta q =\frac{\Delta\phi}{R}-\frac{Li_1}{R} $ ...(ii)
Here , $\Delta \phi =\phi _f -\phi _i =\int\limits_{x=2x_0}^{x=x_0} \frac{\mu _0l_0 }{2\pi x}I dx$
$ =\frac{\mu_0i_0l}{2\pi} ln(2)$
So, from Eq. (ii) charge flown through the resistance
upto time t = T, when current is $i_i$ is
$\Delta q =\frac{1}{R}\bigg[\frac{\mu_0i_0l}{2\pi}ln(2)-Li_1\bigg]$
(c) This is the case of current decay in an L-R circuit. Thus
$ i=i_0e^{-t \tau_L} $ ...(iii)
Here ,$i=\frac{i_1}{4}, i_0 =i_1 , t=(2T-T)=T \, and \, \tau_L =\frac{L}{R}$
Substituting these values in Eq, (iii), we get
$ \tau_L =\frac{L}{R}=\frac{T}{ln 4}$