A long solenoid having 400 turns per cm carries a current 2A. A 100 turn coil of cross-sectional area 4 cm2 is placed co-axially inside the solenoid so that the coil is in the field produced by the solenoid. Find the emf induced in the coil if the current through the solenoid reverse its direction in 0.04 sec.

Explanation

Given : No of turns of long solenoid cm `(1)/(l)=400xx10^(2)m`
No. of turns inside (small) solenoid `N_(2)=100`
Area of inside (small) solenoid `A_(2)=4xx10^(-4)m`
Current through solenoid `I=2A`
Time taken to pass through solenoid t = 0.04 s
To find :
emf induced in the coil e = ?
`e=M.(dI)/(dt)`
`e=-M.(dI)/(dt)`
M - mutual inductance of the coil.
Formula :
`M=mu_(0)N_(1)N_(2)(A_(2))/(l)`
`=4xx3.14xx10^(-7)xx400xx10^(-2)xx100xx4xx10^(-4)`
`M=200.96xx10^(-5)H`
Solution :
`e=-200.9xx10-5xx(2)/(0.04)`
`=-100xx10^(4)xx10^(-5)`
e = -0.10 V
If the current through the solenoid reverse its direction e = -0.20V.