Question

Consider the situation shown in figure. The wire PQ has mass m, resistance r and can slide on the smooth, horizontal parallel rails separated by a distance l. The resistance of the rails is negligible. A uniform magnetic field B exists in the rectangular region and a resistance R connects the rails outside the field region. At t = 0, the wire PQ is pushed towards right with a speed v0. Find (a) the current in the loop at an instant when the speed of the wire PQ is v, (b) the acceleration of the wire at this instant, (c) the velocity v as a functions of x and (d) the maximum distance the wire will move.
Figure

Answer 1

EMF is Product of Field, Velocity and Length

Explanation:

(a) When the speed is V Emf = Bℓv  

Resistance = r + r  

Current = $\frac {Blv}{r}$ + R  

(b) Force acting on the wire = iℓB

= $\frac {BlvlB}{R + r} = \frac {B^2l^2v}{R + r}$

Acceleration on the wire = $\frac {B^2l^2v}{m(R + r)}$

(c) v = $v_0 + at = v_0 - \frac {B^2l^2v}{m(R +r)t}$ (Force is opposite to velocity)

= $v_0 - \frac {B^2l^2x}{R + r}$

(d) a = $v \frac {dv}{dx} = \frac {B^2l^2v}{m(R + r)}$

⇒ $dx = \frac {dvm(R + r)}{B^2l^2}$

x = $\frac {m(R + r)v_0}{B^2l^2}$

Answer 2

Answer:

In the pictures above.

Explanation:

In part c) we are integrating and not using equations of motion because the equations are valid only when a is constant. But here the retardation depends on velocity, which is decreased by retardation. Hence we integrate. The rest are self-explanatory.