A magnetic field of $2 \times 10^{-2} \mathrm{~T}$ acts at right angles to a coil of area $100…
A magnetic field of $2 \times 10^{-2} \mathrm{~T}$ acts at right angles to a coil of area $100 \mathrm{~cm}^2$ with 50 turns. The average e.m.f. induced in the coil is 0.1 V , when it is removed from the field in time $t$. The value of ' $t$ ' is (in second)
0.1 s
$\quad 0.01 \mathrm{~s}$
1 s
20 s
Solution
Using Lenz law,
Induced e.m.f. is given by, $e=-N \frac{d \phi}{d t}$
Change in magnetic flux $=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{dB}}{\mathrm{dt}} \mathrm{A} \cos \theta$
From (i),
$\begin{aligned} e & =-N \frac{d B}{d t} A \cos \theta \\ \therefore \quad e & =-\frac{N\left(B_2-B_1\right) A \cos \theta}{t}\end{aligned}$
$\begin{array}{ll}\therefore & t=\frac{-(50) \times\left(0-2 \times 10^{-2}\right) \times\left(100 \times 10^{-4}\right) \times \cos 0^{\circ}}{0.1} \\ \therefore & t=0.1 \mathrm{~s}\end{array}$