Q. Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area $A = 10{cm}^{2}$ and length=20 cm. If one of the solenoids has 300 turns and the other 400 turns , their mutual inductance is $(\mu_{0} = 4\pi \times 10^{- 7}{TmA}^{- 1})$
A) $2.4\pi \times 10^{- 5}H$
B) $4.8\pi \times 10^{- 4}H$
C) $4.8\pi \times 10^{- 5}H$
D) $2.4\pi \times 10^{- 4}H$
Explanation
(d) $M = \frac{\mu_{0}N_{1} \times N_{2} \times A}{l}$
Where, $N_{1}$=300 turns, $N_{2}$=400 turns, $A = 10{cm}^{2}$ and $l$=20cm.
Substituting the values in the given formula, we get
$M = 2.4\pi \times 10^{- 4}H$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.