Q. 125.0( 2 Votes )

# <span lang="EN-US

Answer :

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 μ_{0}, I, N, n π r’ and R in above eqn.

**Therefore flowing through the galvanometer is**

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