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A magnetic field B=B_{0}\sin\; (\omega t)k covers a large region where a 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?

Answers (1)

Due to motion : e=Blv

Due to change in magnetic flux: e=-N\frac{d(\phi _{B})}{dt}

First we have to analyse the situation as shown in the figure let the parallel wires are at y=0 and y=d and are placed along x-axis. Wire AB is along y-axis.

Let us redraw the diagram as shown below.

At t=0,wire AB starts from x=0 and moves with a velocity v. Let at time t, wire is at x(t)=vt

(Where, x(t) is the displacement as a function of time).

Now, the motional emf across AB is


\Rightarrow e_{1}=(B_{0}\; \sin \omega t)vd (-\hat{j})

and emf due to change in field (along OBAC)

e_{2}=\frac{d(\phi _{B})}{dt}

\phi _{B}=(B_{0}\; \sin \omega t)(x(t)d)                                         (Where,area A=xd)

e_{2}=-B_{0}\omega \cos \; \omega tx(t)d

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

e_{1}+e_{2}=-B_{0}d\left [ \omega x\; \cos (\omega t)+v\sin (\omega t) \right ]

The equivalent electrical diagram is shown in the diagram below.

Electric current in clockwise direction is given by

=\frac{B_{0}d}{R}(\omega x\; \cos \omega t+v \sin \omega t)

The force acting on the conductor is given by F=ilB\; \sin 90^{o}=ilB

Substituting the values,

\overrightarrow{F}_{m}=\frac{B_{0}d}{R}(\omega x \cos \omega t+v \sin \omega t)(d)(B_{0}\; \sin \omega t)(-\hat{i})

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

\overrightarrow{F}_{m}=\frac{B_{0}^{2}d^{2}}{R}(\omega x \cos \omega t+v \sin \omega t)\sin \omega t(\hat{i})

This is the required expression for force.


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