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.

Answer 1

(a) When wire PQ is moving with a speed v, the emf induced across it is given by
e = Blv
Total resistance of the circuit = r + R
∴ Current in the circuit, i = $\frac{Blv}{r+R}$

(b) Force acting on the wire at the given instant, F = ilB

On substituting the value of i from above, we get
$F=\frac{\left(Blv\right)\left(lB\right)}{(R+r)}=\frac{B^2{l}^{2}v}{R+r}$
Acceleration of the wire is given by
a$=\frac{B^2{l}^{2}v}{m(R+r)}$

(c) Velocity can be expressed as:
v = v₀ + at = $v_0-\frac{B^2{l}^{2}v}{m(R+r)}t$ (As force is opposite to velocity)
Velocity as the function of x is given by
$v=v_0-\frac{B^2{l}^{2}x}{m(R+r)}$

$$\begin{gathered} \left(d\right)a=v\frac{dv}{dx}=\frac{B^2{l}^{2}v}{m(R+r)} \\ dx=\frac{m(R+r)}{B^2{l}^2}dv \end{gathered}$$
On integrating both sides, we get
$x=\frac{m(R+r)v_0}{B^2{l}^2}$