Question

A magnetic field B = B_o sin ( ωt )`hatk` wire AB slides smoothly over two parallel conductors separated by a distance d (Figure). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity?

Answer 1

In this problem, the emf induced across AB is motional emf due to its motion, and emf induced by change in magnetic flux linked with the loop change due to a change of magnetic field.

In figure CA and OB are long parallel conducting wires, connected by d and conductor CO. The resistance of ACOB is negligible.

Let wire AB at t = 0 is at x = 0 i.e., on Y-axis.

Now AB moves with velocity `vhati`

Let at any time t, position of conductor AB is x(t) = `vhati t`

Motional e.m.f across AB

`V_(AB) = (W_(AB))/q = (D.d)/q = (qvi xx B)/q d`

`V_(AB) = vhati xx B_0 sin ωt. hatk xx d`

⇒ `e_1 = (B_0 sin ωt)vd (-hatj)`

And emf due to change in field (along OBAC)

`e_2 = (d(phi_B))/(dt)`

`phi_B = (B_0  sin ωt)(x(t)d)`  ....(where, area A = xd)

`e_2 = - B_0 ω cos wtx(t)d`

Total emf in the circuit = emf due to change in field (along OBAC) + the emotional emf across AB

`e_1 + e_2 = - B_0d [ωx cos (t) + v sin (ωt)]`

The equivalent electrical diagram is shown in the diagram below.

Electric current in clockwise direction is given by

= `(B_0d)/R (ωx cos ωt + v sin ωt)`

The force acting on the conductor is given by F = `ilB` sin 90° = `ilB`

Substituting the values,

`vecF_m = (B_0d)/R (ωx cos ωt + v  sin ωt)(d)(B_0 sin ωt)(-hati)`

External force needed on wire is along positive x-axis to keep moving it with constant velocity is given by,

`vecF_(ext) = (B_0^2d^2)/R (ωx cos ωt + v  sin ωt)sin ωt(hati)`

This is the required expression for force.