A conducting wire XY of mass m and negligible resistance slides smoothly on two parallel conducting wires as shown in the 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 the 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), showing that the decrease in kinetic energy of XY equals the heat lost in.

Hint: The change in flux results in the induced emf.
Step 1: Letusassumethattheparallelwiresareaty=0,i.e.,alongx-axisandy=l.
Att=0,XYhasx=0i.e.,alongy-axis.
(i)Letthewirebeatx=x(t)attimet.
Themagneticfluxlinkedwiththeloopisgivenby;
ϕ=B.A=BAcos0°=BA
Atanyinstantt,Magneticflux=B(t)(l×x(t))
Totalemfinthecircuit=emfduetochangeinfield(alongXYAC)+themotionalemfacrossXY
ε=-dt=-dB(t)dt×lx(t)=B(t)lv(t)[secondtermduetomotionalemf]
Electriccurrentinclockwisedirectionisgivenby;
I=1Rε
Theforceactingontheconuctorisgivenby;F=ilBsin90°=ilB
Substitutingthevalues,wehave;
Force=IB(t)R-dB(t)dtIx(t)-B(t)Iv(t)i^
ApplyingNewton'ssecondlawofmotion,
md2xdt2=-I2B(t)R×dB(t)dtx(t)-I2B2(t)Rdxdt...(i)
whichistherequiredequation.
(ii)Step 2: IfBisindependentoftimei.e.,B=constant
OrdBdt=0
SubstitutingtheabovevalueinEq(i),wehave;
d2xdt2+I2B2mRdxdt=0
ordvdt+I2B2mRv=0
Integratingusingvariableseparableformofdifferentialequal,wehave;
v=A.exp-I2B2tmR
Applyinggivenconditions,att=0,v=u0
v(t)=u0×exp(-I2B2t/mR)
Thisistherequiredequation.
(iii)Step 3:SincethepowerconsumptionisgivenbyP=I2R
Here,I2R=B2I2v2(t)R2×R
=B2I2Ru02×exp(-2I2B2tmR)
Now,energyconsumedintimeintervaldtisgivenby;energyconsumed=Pdt=I2Rdt
Therefore,totalenergyconsumedintimet=0tI2Rdt=B2I2Ru02×0texp(-2I2B2tmR)
=B2I2Ru02×mR2I2B21-exp(-2I2B2tmR)=12muo2-12mvt2
=0tI2Rdt=B2I2Ru02mR2I2B21-e-l2B2t/mr
=m2uo2-m2v2t
=decreaseinkineticenergy.
ThisprovesthatthedecreaseinkineticenergyofXYequalstheheatlostinR.