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A conducting wire $$XY$$ of mass $$m$$ and negligible resistance slides smoothly on two parallel conducting wires as shown in Fig. The closed circuit has a resistance $$R$$ due to $$AC. AB$$ and $$C$$ are perfect conductors. There is a magnetic field $$B = B(t)\hat {k}$$.
Write down equation for the acceleration of the wire $$XY$$.
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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 $$\phi (t) = B(t) \times A$$
$$\therefore \phi (t) = B(t)[l.x(t)]$$
$$\epsilon = -\dfrac {d\phi (t)}{dt} = -\dfrac {dB(t)}{dt} l.x(t) - B(t) l.\dfrac {dx(t)}{dt}$$
$$\epsilon = \dfrac {-dB(t)}{dt} l.x(t) - B(t) lv(t)$$
The direction of induced current by Fleming's Right Hand Rule or by Lenz's law is in clockwise direction in loop $$XY < AX$$.
$$I = \dfrac {\epsilon}{R} = \dfrac {-l}{R} \left [x(t) \dfrac {dB(t)}{dt} + B(t)v(t)\right ] .... (I)$$
The force acting on the conductor is $$F = B(t) I1\sin 90^{\circ}$$
$$F = B(t)I . 1$$
$$F = \dfrac {B(t)l\epsilon}{R} = \dfrac {-B(t)l^{2}}{R} \left [\dfrac {-dB(t)}{dt} x(t) - B(t) . v(t)\right ]$$
$$\dfrac {md^{2}x}{dt^{2}} = \dfrac {-B(t)l^{2}}{R} \left [x(t) \dfrac {dB(t)}{dt} + B(t)v(t)\right ]$$
Or $$\dfrac {d^{2}x}{dt^{2}} = \dfrac {-l^{2}}{mR}B(t) \left [x(t) \dfrac {dB(t)}{dt} + B(t) . v(t)\right ] ....(II)$$.
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