Question

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 $\left(\mu_{0}=4 \pi \times 10^{-7} \right. T \left. m A ^{-1}\right)$

• A. $2.4\pi \times 10^{- 4}H$
• B. $2.4\pi \times 10^{- 5}H$
• C. $4.8\pi \times 10^{- 4}H$
• D. $4.8\pi \times 10^{- 5}H$

Answer 1

Answer: (A)

Mutual inductance

$M =\frac{\text { fluxin(2) }}{\text { currentin(1) }}$
$=\frac{\phi_2}{i_1}$
$=\frac{B_1 A_2 N_2}{i_1}$
$=\frac{\mu_0 N_1 i_1}{l} \times \frac{A_2 N_2}{i_1}$
$=\frac{\mu_0 N_1 N_2 A}{l}$
$A_1=A_2=A$
$A=\pi r_1^2=10\, cm ^2, l=20 \,cm , N_1$
$=300, N_2=400$.
$M=\frac{\mu_0 N_1 N_2 A}{l}=$
$\frac{4 \pi \times 10^{-7} \times 300 \times 400 \times 10 \times 10^{-4}}{0.20}$
$=2.4 \pi \times 10^{-4} \,H$.