Question

State Faraday's law of electromagnetic induction.
Figure shows a rectangular conductor PQRS in which the conductor PQ is free to move in a uniform magnetic field B perpendicular to the plane of paper.
The field extends from x = 0 to x b and is zero for x > b.
Assume that only the arm PQ possesses resistance r.
When the arm PQ is pulled outward from x = 0 to x = 2b and is then moved backward to x = 0 with constant speed v, obtain the expression for the flux and the induced emf.
Sketch the variation of these quantities with distance $0\le x\le 2b.$

Answer 1

Refer to Point $3$ Basic Concepts.
Let length of conductor $PQ=1$
As $x=0,$ magnetic flux $\phi =1$
When $PQ$ moves a small distance from $x$ to $x+dx$ then magnetic flux linked $=BdA=Bldx$
The magnetic field is from $x=0tox=b$, to so final magnetic flux
$\sum BIdx=BI\sum dx=BIb$ (increasing)
Mean magnetic flux from $x=0tox=b$ is $\frac{1}{2}BIb$
The magnetic flux from $x=btox=2b$ is zero
Induced emf, $\epsilon =-\frac{d\phi }{dt}=\frac{d}{dt}\left(BIdx\right)=-BI\frac{dx}{dt}=-BIv$
where $v=\frac{dx}{dt}$ velocity of arm $PQ$ from $x=0tox=b$.
During return from $x=2b$ to $x=b$ the induced emf is zero; but now area is decreasing so magnetic flux is decreasing and induced emf will be in opposite direction.
$\epsilon =BIv$