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 μ0, I, N, n π r’ and R in above eqn.



Therefore flowing through the galvanometer is


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Find the current Physics - Exemplar

A metallic ring oPhysics - Exemplar

Find the current Physics - Exemplar

A rod of mass m aPhysics - Exemplar

A magnetic field Physics - Exemplar

A rectangular looPhysics - Exemplar

Consider an infinPhysics - Exemplar

ODBAC is a fixed Physics - Exemplar

A magnetic field Physics - Exemplar

A conducting wirePhysics - Exemplar