A long solenoid of radius 2 cm has 100 turns/cm and carries a current of 5 A. A coil of radius 1 cm having 100 turns and a total resistance of 20 Ω is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer.
Given:
Radius of solenoid ![]()
No. of turns in the solenoid ![]()
Current in the solenoid ![]()
Radius of second coil ![]()
No. of turns in the coil ![]()
Resistance of the coil ![]()
We know that,
Magnetic field inside solenoid (B) is given by formula
![]()
Where,
n=no. of turns per unit length
i=current through solenoid
Magnetic flux(ϕ) through the coil is given by the formula
![]()
![]()
Where B=magnetic field intensity
A=area of cross section of the coil
θ =angle between area vector and magnetic field
magnetic field inside solenoid is perpendicular to the coil
initially flux through the coil is given by
![]()
When the current in the solenoid is reversed in direction of magnetic field gets reversed and flux through the coil now m=becomes
![]()
![]()
Now,
Average induced emf in time interval Δt is given by
…(i)
Where
are flux across the cross section at time intervals
respectively
Putting these values in eqn.(i) we get
![]()
Current (i) through the coil of resistance R can be calculated as
![]()
Hence the charge (Q) passing through the coil in time Δt is
![]()
Putting the values of μ0, I, N, n π r’ and R in above eqn.
![]()
Therefore flowing through the galvanometer is ![]()
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