Explanation

(a) The induced emf is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = - M\frac{di}{dt}$ …(i)
Where $M\ $is coefficient of mutual inductance, $\frac{di}{dt}$ is rate of change of current.
Also, mutual inductance of two coaxial solenoids is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M = \frac{\mu_{0}N_{1}N_{2}A}{l}$ …(ii)
[image:6_image73.emf]
From Eqs. (i) and (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = \frac{\mu_{0}N_{1}N_{2}A}{l} \times \frac{di}{dt}$
Given, $N_{1} = 2000,N_{2} = 300,A = 1.2 \times 10^{- 3}m^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{di}{dt} = \frac{2 - ( - 2)}{0.25} = \frac{4}{0.25}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = \frac{4\pi \times 10^{- 7} \times 2000 \times 300 \times 1.2 \times 10^{- 3} \times 4}{0.3 \times 0.25}\ $
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = \frac{4 \times 3.14 \times 2 \times 3 \times 1.2 \times 4 \times 10^{- 5}}{0.3 \times 0.25}$
$\ \Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ |e| = 4.8 \times 10^{- 2}V$