Question

Find the current in the sliding rod $$AB (resistance = R)$$ for the arrangement shown in Fig. $$\vec {B}$$ is constant and is out of the paper. Parallel wires have no resistance. $$\vec {V}$$ is constant. Switch $$S$$ is closed at time $$t = 0$$.

Answer 1

Induced current $$I$$ in loop $$ABYX = I_{i} = \dfrac {\epsilon}{R} = \dfrac {-1}{R} \dfrac {dQ}{dt}$$
$$I_{i} = \dfrac {1}{R}\cdot \dfrac {d}{dt} \vec {B} . \vec {A}$$
$$I_{t} = \dfrac {vBd}{R}$$
Angle between $$\vec {B}$$ and $$\vec {A}$$ is zero. Directin of $$I$$ from $$A$$ to $$B$$ is given in Fleming's right hand rule.
As switch $$S$$ is closed at $$t = 0$$.
Current of first question will charge the capacitor. Let $$Q(t)$$ be the charge on capacitor.
$$\therefore Q (t) = C.V$$
$$Q(t) = C.I_{C}.R$$
$$\therefore I_{c} = \dfrac {Q(t)}{RC}$$
The capacitor opposes the flow of charge, so net current in circuit
$$I = I_{i} - I_{c}$$
$$I = \dfrac {Bvd}{R} - \dfrac {Q(t)}{RC}$$
$$\dfrac {dQ(t)}{dt} = \dfrac {Bvd}{R} - \dfrac {Q(t)}{RC}$$
$$\dfrac {Bvd}{R} = \dfrac {dQ(t)}{dt} + \dfrac {Q(t)}{RC}$$
Or $$\dfrac {dQ(t)}{dt} + \dfrac {Q(t)}{RC} =\dfrac {Bvd}{R}$$
So the solution of linear differential equation is
$$Q(t) = \dfrac {\dfrac {Bvd}{R}}{\dfrac {1}{RC}} + Ae^{\dfrac {-t}{RC}}$$
$$Q(t) = Bvd C + Ae^{-t/RC}$$
Initially at $$t = 0, Q = 0$$
So $$A = -Bvd\ C$$
So $$Q(t) = BvdC - Bvd\ C e^{\dfrac {t}{RC}} [\because e^{0} = 1]$$
$$Q(t) = BvdC \left [1 - e^{-\dfrac {t}{RC}}\right ]$$
Current in circuit $$I = \dfrac {dQ(t)}{dt} = \dfrac {Bvd}{R} e^{-\dfrac {1}{RC}}$$

or $$\dfrac {dQ(t)}{dt} = \dfrac {BvdC}{RC} e^{-\dfrac {t}{RC}}$$.