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Electromagnetic Induction — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction5 Marks
Question
A magnetic field $\text{B}=\text{B}_0\sin(\omega\text{t})\hat{\text{k}}$ 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 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?
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Answer

Key concept: 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 change of magnetic field.
Due to motion: e = Blv
Due to change in magnetic flux: $\text{e}=-\text{N}\frac{\text{d}(\phi_\text{B})}{\text{dt}}$
First we have to analyse the situation as shown in the figure, Let the parallel wires are at y = 0 and y = d and 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 dispacement as a function of time).
Now, the motional emf across AB is
$\text{e}_1=\frac{\text{B}}{\text{v}}$
$\Rightarrow\ \text{e}_1=\big(\text{B}_0\sin\omega\text{t}\big)\text{vd}(-\hat{\text{j}})$
and emf due to change in field (along OBAC)
$\text{e}_2=-\frac{\text{d}(\phi_\text{B})}{\text{dt}}$
$\phi_\text{B}=(\text{B}_0\sin\omega\text{t})(\text{x}(\text{t})\text{d}) \ \ (\text{where, area A}=\text{xd})$
$\text{e}_2=-\text{B}_0\omega\cos\omega\text{tx}(\text{t})\text{d}$
Total emf in the circuit = emf due to change in field (along OBAC) + the motional emf across AB
$\text{e}_1+\text{e}_2=-\text{B}_0\text{d}\big[\omega\text{x}\cos(\omega\text{t})+\text{v}\sin(\omega\text{t})\big]$
The equivalent electrical diagram is shown in the diagram below.

Electric current in clockwise direction is given by
$=\frac{\text{B}_0\text{d}}{\text{R}}(\omega\text{x}\cos\omega\text{t}+\text{v}\sin\omega\text{t})$
The force acting on the conductor is given by $\text{F}=\text{ilB}\sin90^\circ=\text{ilB}$.
Substituting the values,
$\vec{\text{F}_\text{M}}=\frac{\text{B}_0\text{d}}{\text{R}}(\omega\text{t}\cos\omega\text{t}+\text{v}\sin\omega\text{t})(\text{d})(\text{B}_0\sin\omega\text{t})(-\hat{\text{j}})$
External force needed on wire is along positive x-axis to keep moving it with constant velocity is given by,
$\vec{\text{F}}_\text{ext}=\frac{\text{B}_0\text{d}}{\text{R}}(\omega\text{t}\cos\omega\text{t}+\text{v}\sin\omega\text{t})\sin\omega\text{t}$
This is the required expression for force.

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