Question

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area $A=10c{m}^2$ and length $=20cm$ . If one of the solenoids has $300$ turns and the other $400$ turns, their mutual inductance is $({\mu }_0=4\pi \times 10^{-7}Tm{A}^{-1})$

• 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

$M=\frac{\text{ fluxin(2) }}{\text{ currentin(1) }}$
$=\frac{{\varphi }_2}{i_1}$
$=\frac{B_1{A}_2{N}_2}{i_1}$
$=\frac{{\mu }_0{N}_1{i}_1}{l}\times \frac{A_2{N}_2}{i_1}$
$=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$
$A_1=A_2=A$
$A=\pi r_1^2=10c{m}^2,l=20cm,N_1$
$=300,N_2=400$.
$M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}=$
$\frac{4\pi \times 10^{-7}\times 300\times 400\times 10\times 10^{-4}}{0.20}$
$=2.4\pi \times 10^{-4}H$.