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 $0.05sec$, the induced e.m.f (in V) in the coil is

Answer 1

Answer: 0.048

$B={\mu }_{0}ni=\left(4\pi \times 10^{-7}\right)\left(200\times 10^{-2}\right)\times 1.5$
$=3.8\times 10^{-2}Wb{m}^2$
Magnetic flux through each turn of the coil
$\varphi =BA=\left(3.8\times 10^{-2}\right)\left(3.14\times 10^{-4}\right)=1.2\times 10^{-5}$ weber
When the current in the solenoid is reversed, the change in magnetic flux
$=2\times \left(1.2\times 10^{-5}\right)=2.4\times 10^{-5}$weber
Induced e.m.f.$=N\frac{d\varphi }{dt}=100\times \frac{2.4\times 10^{-5}}{0.05}=0.048V$