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 $$t = 0$$.
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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 $$\dfrac {dQ(t)}{dt} = \dfrac {BvdC}{RC} e^{-\dfrac {t}{RC}}$$.
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