Question:6.23
A conducting wire XY of mass m and negligible resistance slides smoothly on two parallel conducting wires as shown in figure. The closed circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field
(i) Write down an equation for the acceleration of the wire XY
(ii) If B is independent of time, obtain v(t), assuming v(0) = u0
(iii) For (ii), show that the decrease in kinetic energy of XY equals the heat lost in.
Answer:
Upon analyzing the situation:
The wire is parallel with y-axis at y=0 and y=L. It placed along the x-axis.
At t=0, the wire starts from x=0 and is moving with a velocity of v. At certain time wire is at $x(t)=vt. (x(t)$ is the displacement function). The diagram is redrawn as:
The magnetic flux linked with the loop is given by
$\phi _{m}=\vec{B}.\vec{A}=BA \cos \theta$
and as we know both $\vec{A}$ (area vector) and $\vec{B}$ (magnetic field vector) are directed along z-axis. So angle between them is 0.
So, $\cos \; 0^{o}=1\; \; \; \; \; \; (\because \theta =0^{o})$
$\Rightarrow \phi _{m}=BA$
At any instant of time t,
Magnetic flux $\phi _{m}=B(t)(1\times x(t))$
Emf induced due to change in magnetic field
$e_{1}=-\frac{d\phi }{dt}$
$\Rightarrow e_{1}=-\frac{dB(t)}{dt}lx(t)$
Emf induced due to motion
$e_{2}=Blv$
$e_{2}=B(t)lv(t)(-\hat{j})$
Total emf in the circuit = emf due to change in field (along XYAC) + the motional emf across XY
$E=-\frac{d\phi }{dt}=-\frac{dB(t)}{dt}lx(t)-B(t)lv(t)$
Electric current in clockwise direction (as shown in equivalent diagram) is given by
$I=\frac{E}{R}$
The force acting on the conductor is given by $F=ilB\; \sin 90^{o}=ilB$
Substituting the values, we have
$Force =\frac{lB(t)}{R}\left [- \frac{B(t)}{dt}l x (t)-B(t)lv(t)\right ]\hat{i}$
Applying Newton's second law of motion,
$m\frac{d^{2}x}{dt^{2}}=\frac{l^{2}B(t)}{R}\frac{dB}{dt}x(t)-\frac{l^{2}B^{2}(t)}{R}\frac{dx}{dt}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i)$
which is the required equation.
If B is independent of time, i.e., B = Constant
or $\frac{dB}{dt}=0$
substituting the above value in Eq (i), we have
$\frac{d^{2}x}{dt^{2}}+\frac{l^{2}B^{2}}{mR}\frac{dx}{dt}=0$
or $\frac{dv}{dt}+\frac{l^{2}B^{2}}{mR}v=0$
Intergrating using variable separable form of differential equation, we have
$v=A \; exp \left ( \frac{-l^{2}B^{2}t}{mR} \right )$
Applying given conditions, at $t=0, v=u_{0}$
$v(t)=u_{0}exp (-l^{2}B^{2}t/mR)$
This is the required equation.
Since the power consumption is given by
$p=I^{2}R$
Here,
$I^{2}R=\frac{B^{2}l^{2}v^{2}(t)}{R^{2}}\times R$
$=\frac{B^{2}I^{2}}{R}u_{0}^{2}exp (-2l^{2}B^{2}t/mR)$
Now, energy consumed in time interval dt is given by energy consumed $=Pdt=I^{2}Rdt.$ Therefore, toal energy consumed in time t.
$=\frac{m}{2}u_{0}^{2}-\frac{m}{2}v^{2}(t)$
= decrease in kinetic energy
This proves that the decrease in kinetic energy of XY equals the heat lost in R.
