Question

Find the current in the sliding rod $$AB (resistance = R)$$ for the arrangement shown in Fig. $$\vec {R}$$ 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

Conductor $$AB$$ moves towards right with speed

$$v$$ and magnetic

$$\vec {B}$$

perpendicularly upward so angle between

$$B$$ and $$v$$ is

$$90^{\circ}$$ and an induced e.m.f.,? flows in loop $$AXYB$$.

$$\epsilon = \dfrac {d}{dt}\vec {B} \vec {A} $$

$$= \dfrac {d}{dt} BA\cos 0 = \dfrac {d}{dt} B(d\times vtimes t) = Bvd$$

$$\because$$ angle between $$\vec {B}$$ and $$\vec {A}$$ is

$$0^{\circ}$$.

$$\therefore \epsilon = B.v.d$$

at $$t = 0$$,

current starts increasing in loop along with inductor $$L$$ due to potential difference

By Kirchhoff's law in loop $$ABYX$$,

$$-L \dfrac {dI(t)}{dt} + Bvd = IR$$

$$\dfrac {LdI(t)}{dt} + RI(t) = Bvd$$

It is a differential equation.

$$I(t) = \dfrac {Bvd}{R} + Ae^{-\dfrac {Rt}{2}} .... (I)$$

at $$t = 0, I = 0$$

$$\therefore 0 = \dfrac {Bvd}{R} + Ae^{0} [\because e^{0} = 1]$$

$$A = \dfrac {-Bvd}{R}$$

$$\therefore I(t) = \dfrac {Bvd}{R} - \dfrac {Bvd}{R} e^{-\dfrac {Rt}{2}}$$ [from I]

$$I = \dfrac {Bvd}{R} \left [1 - e^{-\dfrac {Rt}{2}}\right ]$$

This is the required current expression.