Saved Bookmarks
| 1. |
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» 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. |
|