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Physics Question on Inductance

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A=10 cm2A=10\, cm ^{2} and length =20 cm=20\, cm. If one of the solenoid has 300300 turns and the other 400400 turns, their mutual inductance is (μ0=4π×−7TmA−1)\left(\mu_{0}=4 \pi \times^{-7} Tm A^{-1}\right)

A

2.4π×10−5H2.4 \pi \times10^{-5} H

B

4.8π×10−4H4.8 \pi \times10^{-4} H

C

4.8π×10−5H4.8 \pi \times10^{-5} H

D

2.4π×10−4H2.4 \pi \times10^{-4} H

Answer

2.4π×10−4H2.4 \pi \times10^{-4} H

Explanation

Solution

M=μ0N1N2A1M =\frac{\mu_{0} N _{1} N _{2} A }{1} =4π×10−7×300×400×10×10−40.2=\frac{4 \pi \times 10^{-7} \times 300 \times 400 \times 10 \times 10^{-4}}{0.2} =2.4π×10−4H=2.4 \pi \times 10^{-4} H