A conducting wire XY of mass M and negligible resistance slides smothly 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)hatk`

Explanation

Let us assume that the parallel wiers at are `y=0`, i.e., along x-axis and y-l, At=0, XY has `x=0` i.e., along y-axis.
(i) Let the wire be at `x=x`(t) at time t.
The magnetic flux linked with the loop is given by
`theta=B.A =BAcos0=BA`
at any instant t
Magnetic flux = `B(t)(l xx x(t))`
Total emf in the circuit= emf due to change in field (along XYAC) + the motional emf across XY
`E=-(dphi)/(dt) = -(d(B(t))/(dt)lx(t) -B(t)lv(t)` [second term due to motional emf]
Electric current in clockwise direction is given by
`I=1/R E`
The force acting on the conductor is given by `F=ilBsin90^(@)= ilB`
Substituting the values, we have
Force `=(IB(t))/(R) [-(dB(t))/(dt) Ix(t) - B(t)Iv(t)]hati`
Applying Newton's second law of motion,
`m(d^(2)x)/(dt^(2))= -(I^(2)B(t))/(R) (dB)/(dt)x(t) - (I^(2)B^(2)(t))/(R) (dx)/(dt)`
Which is the required equation.
ii) If B is independent of time i.e., B=Constant
Or `(dB)/(dt)=0`
Substituting the above value in Eq(i), we have
`(d^(2)x)/(dt^(2))+(I^(2)B^(2))/(mR)(dx)/(dt)=0`
`(dv)/(dt)+(I^2B^2)/(mR)V=0`
Integrating using variable separable form of differential equation, we have
`V=Aexp((-I^(2)B^(2)t)/(mR))`
Applying given conditions
at `t=0v=u_0`
`v(t)=u_0exp(-I^2B^2t//mR)`
This is the required equation,
Since the power consumption is given by `P=I^2R`
Here,
`I^2R=(B^2I^2V^2(t))/(R^2)xxR`
`=(B^(2)I^(2))/(R) u_(0)^(2)exp(-2I^(2)B^(2)tI mR)`
Now, energy consumed in time interval dt is given by energy consumed `= Pdt=I^(2)Rdt`
Therefore, total energy consumed in time t
`=int_(0)^(t)I^(2)Rdt= (B^(2)I^(2))/(R) mu_(0)^(2) (mR)/(2I^(2)B^(2))[1-e^(-(l^(2)B^(2)t//mr)]`
`m/2u_(0)^(2)-m/2v^(2)(t)`
= decrease in kinetic energy.
This proves that the decrease in kinetic energy of XY equals the heat lost in R.