Question

A long solenoid having $200$ turns per cm carries a current of $1.5$ amp. At the centre of it is placed a coil of$100$ turns of cross-sectional area $3.14 \times 10^{-4}m^{2}$ having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within $0.05\, sec$, the induced e.m.f (in V) in the coil is

Answer 1

$B =\mu_{0} ni = \left(4\pi \times10^{-7}\right)\left(200 \times10^{-2}\right) \times1.5$
$= 3.8 \times10^{-2}Wbm^{2}$
Magnetic flux through each turn of the coil
$\phi =BA= \left(3.8 \times10^{-2}\right)\left(3.14 \times10^{-4}\right) = 1.2 \times10^{-5}$ weber
When the current in the solenoid is reversed, the change in magnetic flux
$=2 \times\left(1.2 \times10^{-5}\right)=2.4 \times10^{-5}$weber
Induced e.m.f.$= N \frac{d\phi}{dt} =100 \times\frac{2.4 \times10^{-5}}{0.05} = 0.048V$