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The current in a long solenoid of radius R and having n turns per unit length is given by i= i_{0} sin ωt. A coil having N turns is wound around it near the centre. Find (a) the induced emf in the coil and (b) the mutual inductance between the solenoid ant the coil.

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#### Solution

Given:-

Radius of the long solenoid = R

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

Current in the long solenoid, i = i_{0} 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 = μ_{0}ni

Flux produced in the small solenoid because of the long solenoid, ϕ = (μ_{0}ni) × (NπR^{2})

(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 = i_{0} sin ωt, we get

e = μ_{0}nNπR^{2}i_{0}ω cos ωt

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

Flux ϕ linked with the second coil is given by

ϕ = (μ_{0} ni) × (NπR^{2})

The flux can also be written as

ϕ = mi

∴ (μ_{0} ni) × (NπR^{2}) = mi

And,

m = πμ_{0}nNR^{2}

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