Question

Obtain an expression for e.m.f. induced in a coil rotating with uniform angular velocity in a uniform magnetic field. Show graphically the variation of e.m.f. with time $\left(t\right)$.
Resistance of a potentiometer wire is $0.1\Omega /cm$. A cell of e.m.f. $1.5V$ is balanced at $300cm$ on this potentiometer wire. Calculate the current and balancing length for another cell of e.m.f. $1.4V$ on the same potentiometer wire.

Answer 1

When a rectangular coil $EFGH$ is rotated about its axis in plane perpendicular to the plane of a field, then the magnetic flux $\overrightarrow{B}$ passing through the coil changes, as a result an induced e.m.f. gets developed.

Consider a coil which has area $A=lb$ and $n$ be the number of turns of the coil. It is being rolled in a uniform magnetic field $\left(B\right)$ with an uniform angular velocity $\omega$. At any instant of time, the angle between the normal of the coil and that of field of $B$, then the flux associated with the coil is

$\varphi =nBA\cos \theta$ ...(i)

The plane of the coil is perpendicular to the magnetic field then after time $t$, is rotated by angle $\theta$.

Hence, $\omega =\frac{\theta }{t}$ or $\theta =\omega t$

by equation (i) $\varphi =nBA\cos \omega t$ ...(ii)

By Faraday's law

$E=-\frac{d\varphi }{dt}=-\frac{d}{dt}$ $\left(nBA\cos \omega t\right)$

$E=nBA\omega \sin \omega t$

The graph showing the variation with time '$t$' of the e.m.f. induced in the coil.

$R=0.1\Omega /cm$, $V_2=1.4V$

$L_1=300cm$ $I_2=$?

$V_1=1.5V$, $L_2=$?

Current :

$I_1=\frac{V_2}{R}=\frac{1.4}{0.1}=14A$

Balancing length,

$K=\frac{V_1}{L_1}=\frac{1.5}{300}=5\times {10}^{-3}V/cm$

$L_2=\frac{V_2}{K}=\frac{1.4}{5\times {10}^{-3}}=280cm$.