Q. A short solenoid of length 4 cm, radius 2 cm and 100 turns is placed inside and on the axis of a long solenoid of length 80 cm and 1500 turns. A current of 3 A flows through the short solenoid. The mutual inductance of two solenoids is
A) $2.96 \times 10^{- 4}H$
B) $5.3 \times 10^{- 5}H$
C) $3.52 \times 10^{- 3}H$
D) $8.3 \times 10^{- 5}H$
Explanation
(a) $M_{21} = \frac{\mu_{0}N_{1}N_{2}A_{2}}{l_{2}}$
$\therefore M_{21} = \frac{\begin{array}{r}
\left( 4 \times 3.14 \times 10^{- 7} \right) \times 1500 \times 100 \times \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\{ 3.14\left( 2 \times 10^{- 2} \right)^{2} \right\}
\end{array}}{80 \times 10^{- 2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ M_{21} = 2.96 \times 10^{- 4}H$
$\Rightarrow \ \ \ \ \ M_{12} = M_{21} = 2.96 \times 10^{- 4}H$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.