Question

A conducting wire XY of mass m and neglibile 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 ˆ. magnetic field `B = B(t)hatk`.

1. Write down equation for the acceleration of the wire XY.
2. If B is independent of time, obtain v(t) , assuming v(0) = u₀.
3. For (b), show that the decrease in kinetic energy of XY equals the heat lost in R.

Answer 1

First we have to analyse the situation as shown in the figure. Let the parallel wires be at y = 0 and y = L and are placed along x-axis. Wire XY is along y-axis.

At t = 0, wire AB starts from x = 0 and moves with a velocity v.

Let at time t, wire is at x(t) = vt.   ....(Where x(t) is the displacement as a function of time).

Let us redraw the diagram as shown below.

(i) Let wire XY at t = 0 is at x = 0

And at t = t is at x = x(t)

Magnetic flux is a function of time `phi(t) = B(t) xx A`

∴ `phi(t) = B(t)l.x(t)`

`ε = - (dphi(t))/(dt) = - (dB(t))/(dt)l.x(t) - B(t)l. (dx(t))/(dt)`

`ε = (- dB(t))/(dt) l.x(t) - B(t)lv(t)`

The direction of induced current by Fleming's Right-Hand Rule or by Lenz's law is in clockwise direction in loop XY < AX.

`I = ε/R = (-l)/R [x(t) (dB(t))/(dt) + B(t)v(t)]`  ......(I)

The force acting on the conductor is F = B(t) I l sin 90°

F = B(t)I.l

F = `(B(t)lε)/R = (-B(t)l^2)/R [(-dB(t))/(dt).x(t) - B(t).v(t)]`

`(md^2x)/(dt^2) = (-B(t)l^2)/R [x(t) (dB(t))/(dt) + B(t)v(t)]`

Or `(d^2x)/(dt^2) = (-l^2)/(mR) B(t) [x(t) (dB(t))/(dt) + B(t).v(t)]` ......(II)

(ii) Now B is independent of time i.e. B does not change with time or it is constant

∴ `(dB)/(dt)` = 0, B(t) = B and v(t) = c  ......(III)

Put (III) in (II) we get

`(d^2x)/(dt^2) = (-l^2)/(mR) [0 + Bv]`

`(d^2x)/(dt^2) + (B^2l^2)/(mR) (dx)/(dt)` = 0

`(dv)/(dt) + (B^2l^2)/(mR) v` = 0

Integrating using variable separable from differential equation we have

`v = A exp((-l^2B^2t)/(mR))`

At t = 0, v = u

∴ `v(t) = u exp ((-l^2B^2t)/(mR))`  ......(IV)

(iii) Heat lost per second in (ii) where `(dB)/(dt)` = 0

H = I²R

Magnitude of current from equation I in (i) part

`I = (Blv)/R = (-l)/R [0 + Bv]`  ......`[because (dB)/(dt) = 0]`

Heat produced per second H = I²R

∴ `H = (B^2l^2v^2)/R^2 R`

`H = (B^2l^2)/R u^2 exp  (-2i^2B^2t)/(mR)` 

v from equation (IV) in (iii) part

Power lost = `int_0^t I^2 Rdt = (B^2l^2u^2)/R int_0^t e^((-2l^2B^2t)/(mR)) dt`  ....`[because v^2 = u^2 exp ^((-2l^2B^2t)/(mR))]`

Power lost = `(B^2l^2u^2)/R (mR)/(2l^2B^2) [1 - e^((-2l^2B^2t)/(mR))]`

= `(m u^2)/2 [1 - e^((-2l^2B^2t)/(mR))]`

= `(m u^2)/2 - m/2 u^2e^((-2l^2B^2t)/(mR))`

= `[(m u^2)/2 - (mv^2(t))/2]`

= initial K.E. – final K.E.

Power lost = decrease in kinetic energy

This proves that decrease in K.E of XY is equal to the heat lost in R.