Question

(a) State the principle of an ac generator and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A, rotating with a constant angular speed '$\omega$' in a magnetic field $\overrightarrow{B}$ , directed perpendicular to the axis of rotation.
(b) An aeroplane is flying horizontally from west to east with a velocity of 900 km/hour. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth's magnetic field is $5\times {10}^{-4}$ T and the angle of dip is ${30}^{\circ }$.

Answer 1

coil = N, cross-section = A, angular speed of rotation = $\omega$, magnetic field = $\overrightarrow{B}$ , then to find emf induced, Flux through the coil when its normal makes angle $\theta$ with the field,

$\varphi$ = BA cos $\theta$

When coil rotates with angular velocity $\omega$ and turns through $\theta$ in time t then $\theta =\omega t$

$\Rightarrow \varphi =BAcos\omega t$

When coil rotates, $\varphi$ changes to set an induced emf.

$\epsilon =\frac{-d\varphi }{dt}=-\frac{d}{dt}\left(BAcos\omega t\right)$

For N turns, total induced emf,

$\epsilon =NBA\omega sin\omega t$

(b) Given: Velocity (v) = 900 $\frac{km}{h}=900\times \frac{5}{18}\frac{m}{s}$

= $250m{s}^{-1}$

Wing span (l) = 20 m

Horizontal component of earth's field $\left(B_H\right)=5\times {10}^{-4}T$ Angle of dip $\left(\delta \right)={30}^{\circ }$

p.d. developed = ?

By $B_V=B_{H}tan\delta$

= $5\times {10}^{-4}\times tan{30}^{\circ }$

= $5\times {10}^{-4}\times \frac{1}{1.732}$

emf induced = $B_{V}lv$

= $\frac{5\times {10}^{-4}}{1.732}\times 20\times 250=1.44$ V