Question

Prove theoretically, the relation between e.m.f. induced and rate of change of magnetic flux in a coil moving in a uniform magnetic field.

Answer 1

Let us consider a rectangular wire loop $PQRS$ of width $l$, and its plane perpendicular to a uniform magnetic field. The loop is being pulled out of magnetic field at a constant speed $v$. At any instant, let '$x$' be the length of loop in the magnetic field. As the loop moves towards right, the area of the loop inside the field changes by $dA=ldx=lvdt$. So the change in magnetic flux is given by
$d\varphi =BdA=Blvdt$
so $\frac{d\varphi }{dt}=\frac{Blvdt}{dt}=Bvl$
Now a straight current-carrying conductor of length $L$ experiences a magnetic force given by
$\overrightarrow{F}=\overrightarrow{IL}\times \overrightarrow{B}$
Its direction is given by Fleming's left and rule.
So, the forces $F_1$ and $F_2$ on wires $PR$ and $QS$ respectively are equal in magnitude and opposite in direction and have the same line of action. Hence they balance each other. There is no force on the wire $RS$ as it lies outside the field. The force $\overrightarrow{F_3}$ on the wire $PQ$ has magnitude $F_3=IBL$ directed towards left. To move the loop with constant velocity $\overrightarrow{v}$ an external force must be applied. So work done by the external agent is given by
$dw=Fdx=-IlBdx=-IBdA$
$=-Id{\varphi }_m$
So rate of doing work $P=\frac{dw}{dt}$
$=I(-\frac{d{\varphi }_m}{dt})$
but $P=EI$
so $E=-\frac{d{\varphi }_m}{dt}$