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 solenoid has $200$ turns and the other $400$ turns, their mutual inductance is $\left({\mu }_0=4\pi \times {10}^{-7}Tm{A}^{-1}\right)$.

• A. $1.6\pi \times {10}^{-4}H$
• B. $4.8\pi \times {10}^{-5}H$
• C. $2.4\pi \times {10}^{-4}H$
• D. $2.4\pi \times {10}^{-5}H$

Answer 1

The correct option is A $1.6\pi \times {10}^{-4}H$

Given :

$A_1=A_2={10}^{-3}{m}^2$

$l_1=l_2=0.2m$

$N_1=200\text{and}{N}_2=400$

To find : $M$

Solution :

We know that $M=\frac{{\mu }_{0}N1N2A}{l}$

Hence, $M=\frac{4\pi \times {10}^{-7}\cdot 200\cdot 400\cdot {10}^{-3}}{0.2}$

$M=1.6\pi \times {10}^{-4}H$

The correct opt: {A}