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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.

Answer»

10 H
8 H
3mH
30mH

Solution :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


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