Question

A magnetic field `B = B_(0) sin (omega t) hat k` covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d, Fig. The wires are in the x-y plane. The wire AB (of length d) has resistance R and parallel wires have negligible resistance. If AB is moving with velocity `upsilon`, what is the current in the circuit ? What is the force needed to keep the wire moving at constant velocity ?

Answer 1

Let us assume that the parallel wires at are y=0 i.e., along x-axis and `y=d`. At `t=0`, AB has `x=0`, i.e, along y-axis and moves with a velocity v. Let at time t, wire is at `x(t)=vt`.
Now, the motional emf across AB is `=(B_0sinomegat)vd(-hatj)`
emf due to change in field (along) OBAC)
`=-B_0omegacosomegatx(t)d`
Total emf in the circuit =emf due to change in field (along OBAC)+the motional emf across AB
`=-B_0d[omegaxcos(omegat)+vsin(omegat)]`
Electric current in clockwise direction is given by
`=(B_0d)/(R)(omegaxcosomegat+vsinomegat)`
The force acting on the conductor is given by `F=ilBsin90^@=ilB`
Substituting the values, we have
Force needed along `i=(B_0d)/(R)(omegaxcosomegat+vsinomegat)xxdxxB_0sinomegat`
`=(B_0^2d^2)/(R)(omegaxcosomegat+vsinomegat)sinomegat`
This is the required expression for force.