Answer 1

Given:-

Radius of the long solenoid = R

Number of turns per unit length of the long solenoid = n

Current in the long solenoid, i =  i₀ sin ωt

Number of turns in the small solenoid = N

Radius of the small solenoid = R

The magnetic field inside the long solenoid is given by

B = μ₀ni

Flux produced in the small solenoid because of the long solenoid, ϕ = (μ₀ni) × (NπR²)

(a) The emf developed in the small solenoid is given by

\[e =\frac{d\phi}{dt} = \frac{d}{dt}( \mu_0 niN\pi R^2 )\]

`e = μ_0nN πR^2(di)/(dt)`

Substituting i = i₀ sin ωt, we get

e = μ₀nNπR²i₀ω cos ωt

(b) Let the mutual inductance of the coils be m.

Flux ϕ linked with the second coil is given by

ϕ = (μ₀ ni) × (NπR²)

The flux can also be written as

ϕ = mi

∴ (μ₀ ni) × (NπR²) = mi

And,

m =  πμ₀nNR²