A coil has 1000 turns and $500 \mathrm{~cm}^2$ as its area. The plane of the coil is placed at right angles…
A coil has 1000 turns and $500 \mathrm{~cm}^2$ as its area. The plane of the coil is placed at right angles to a magnetic induction field of $2 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$. The coil is rotated through $180^{\circ}$ in $0.2 \mathrm{~s}$. The average emf induced in the coil, in $\mathrm{mV}$, is
(a) 5
10
15
20
Solution
$N=1000, A=500 \mathrm{~cm}^2=500 \times 10^{-4}$
$=5 \times 10^{-2} \mathrm{~m}^2$
$B=2 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2, \theta_1=0^{\circ}$,
$\theta_2=180^{\circ}, \Delta t=0.2 \mathrm{~s}$
Initial flux linked with coil,
$\phi_1=N B A \cos \theta_1$
$=N B A \cos 0^{\circ}$
$=N B A$
Final flux, $\phi_2=N B A \cos 180^{\circ}$
$=N B A(-1)=-N B A$
Change in flux, $\phi=\phi_2-\phi_1$
$=-N B A-(N B A)$
$=-2 N B A$
$\therefore$ Induced emf, $\quad e=\frac{-\Delta \phi}{\Delta t}=-\frac{(-2 N B A)}{\Delta t}$
$=\frac{2 N B A}{\Delta t}$
$=\frac{2 \times 1000 \times 2 \times 10^{-5} \times 5 \times 10^{-2}}{0.2}$
$=10 \times 10^{-3} \mathrm{~V}$
$=10 \mathrm{mV}$