Q. A solenoid has 2000 turns wound over a length of 0.30 m. The area of its cross section is $1.2 \times 10^{- 3}\ m^{2}$. Around its central section, a coil of 300 turns is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 s, then the emf induced in the coil is equal to
A) $6\ \times 10^{- 4}\ $V
B) $4.8 \times 10^{- 2}\ $V
C) $6\ \times 10^{- 2}\ $V
D) 48 kV
Explanation
(b) $e = \frac{Mdi}{dt} = \ \left( \frac{\mu_{0}N_{1}N_{2}A}{l} \right)\frac{di}{dt}$
$= \ \frac{4\pi \times 10^{- 7} \times 2000 \times 300 \times 1.2\ \times 10^{- 3}(4)}{0.3 \times 0.25}$
$= 4.8\ \times 10^{- 2}$ V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.