Find mutual inductance of two coaxial solenoids of equal length 30 cm with inner one surrounded by bigger one. Their area of cross section are `20 "cm"^2` and `40 "cm"^2`. They have windings of 40 turn/cm and 10 turn/cm.

Options

A. 10 H
B. 8 H
C. 3mH
D. 30mH

Answer

C

Explanation

Total no. of turns in smaller solenoid is `N_1`=(40)(30)=1200 and no. of turns in bigger solenoid is `N_2` =(10)(30)=300
Now, mutual inductance of given system is given by :
`M=(mu_0N_1N_2a)/l`
where a=area of cross -section of smaller solenoid.
`therefore M=((4pixx10^(-7))(1200)(300)(20xx10^(-4)))/(30xx10^(-2))`
`therefore M=3.0144xx10^(-3)` H
`therefore` M=3.0144 mH `approx` 3mH