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 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 B=B(t)k

(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.

Answers (1)

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



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


The force acting on the conductor is given by F=ilB\; \sin 90^{o}=ilB

Substituting the vlues, 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


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



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.


= decrease in kinetic energy

This proves that the decrease in kinetic energy of XY equals the heat lost in R.



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