Question

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A conducting ring of radius b is placed coaxially in a long solenoid of radius a (b < a) having n turns pe
unit length. A current i = i, cos ot flows through the solenoid. The induced emf in the ring is
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Answer 1

Answer:

Induced current in the coil is given as

$EMF = \mu_o n \pi b^2 i_o\omega sin\omega t$

Explanation:

As we know that the magnetic field along the axis of the solenoid is given as

$B = \mu_o ni$

now we have

magnetic flux linked with the ring is given as

$\phi = B.A$

$\phi = (\mu_o n i)(\pi b^2)$

now we have

$EMF = \frac{d\phi}{dt}$

so we have

$EMF = \mu_o n \pi b^2\frac{di}{dt}$

given that

$i = i_o cos\omega t$

so we have

$\frac{di}{dt} = i_o \omega sin\omega t$

so we have

$EMF = \mu_o n \pi b^2 i_o\omega sin\omega t$

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Topic : Faraday's law

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Answer 2

The induced emf in the ring is μ₀ni₀πb²ω sin ωt

Given:

Current flowing through the solenoid = i = i₀ cos ωt

Explanation:

The magnetic field inside the solenoid is given by the formula:

B = μ₀ni

On substituting 'i', we get,

B = μ₀n(i₀ cos ωt)

The induced emf is given as:

e = dΦ/dt

e = d(πb²μ₀ni₀ cos ωt)/dt

e = (πb²μ₀ni₀) d(cos ωt)/dt

On differentiating 'cos θ' we get 'sin θ'

e = (πb²μ₀ni₀)ω(sin ωt)

∴ e = μ₀ni₀πb²ω sin ωt