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Question
The current in a long solenoid of radius R and having n turns per unit length is given by i= i0 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 = 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
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