Question

A long solenoid having $200$ turns per centimeter carrying a current of $1.5A$. At the center of the solenoid 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 $0.05s$, the induced emf in the coil is

• A. $0.48V$
• B. $0.048V$
• C. $0.0048V$
• D. $48V$

Answer 1

The correct option is B $0.048V$

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