Ncert Exercises & Solutions

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) $\hat{k}$ .

(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)= $u_{0}$

(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 : Let us assume that the parallel wires are at y = 0, i . e ., along x - axis and y = l .

At t = 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;

ϕ = B . A = BAcos 0 ° = BA

At any instant t, Magnetic flux = B (t) (l \times x (t))

Total emf in the circuit = emf due to change in field (along XYAC) + the motional emf across XY

ε = - \frac{dϕ}{dt} = - \frac{dB (t)}{dt} \times lx (t) = B (t) lv (t) [second term due to motional emf]

Electric current in clockwise direction is given by;

I = \frac{1}{R} ε

The force acting on the conuctor is given by; F = ilBsin 90 ° = ilB

Substituting the values, we have;

Force = \frac{IB (t)}{R} [- \frac{dB (t)}{dt} Ix (t) - B (t) Iv (t)] \hat{i}

Applying Newton' s second law of motion,

m \frac{d^{2} x}{{dt}^{2}} = - \frac{I^{2} B (t)}{R} \times \frac{dB (t)}{dt} x (t) - \frac{I^{2} B^{2} (t)}{R} \frac{dx}{dt} . . . (i)

which is the required equation .

(ii) Step 2 : 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{I^{2} B^{2}}{mR} \frac{dx}{dt} = 0

or \frac{dv}{dt} + \frac{I^{2} B^{2}}{mR} v = 0

Integrating using variable separable form of differential equal, we have;

v = A . \exp (\frac{- I^{2} B^{2} t}{mR})

Applying given conditions, at t = 0, v = u_{0}

v (t) = u_{0} \times \exp (- I^{2} B^{2} t / mR)

This is the required equation .

(iii) Step 3 : Since the power consumption is given by P = I^{2} R

Here, I^{2} R = \frac{B^{2} I^{2} v^{2} (t)}{R^{2}} \times R

= \frac{B^{2} I^{2}}{R} {u_{0}}^{2} \times \exp (\frac{- 2 I^{2} B^{2} t}{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 = \frac{B^{2} I^{2}}{R} {u_{0}}^{2} \times \int_{0}^{t} \exp (\frac{- 2 I^{2} B^{2} t}{mR})

= \frac{B^{2} I^{2}}{R} {u_{0}}^{2} \times \frac{mR}{2 I^{2} B^{2}} [1 - \exp (\frac{- 2 I^{2} B^{2} t}{mR})] = \frac{1}{2} {mu}_{o}^{2} - \frac{1}{2} {mv (t)}^{2}

= \int_{0}^{t} I^{2} Rdt = \frac{B^{2} I^{2}}{R} {u_{0}}^{2} \frac{mR}{2 I^{2} B^{2}} [1 - e^{- (l^{2} B^{2} t / mr)}]

= \frac{m}{2} {u_{o}}^{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 .

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