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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.

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प्रश्न

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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उत्तर

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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पाठ 38: Electromagnetic Induction - Exercises [पृष्ठ ३१३]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 38 Electromagnetic Induction
Exercises | Q 98 | पृष्ठ ३१३

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