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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 Rate this question :

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