Question

The mean diameter of a silicon iron ring is 100 mm and its cross-sectional area is 500 mm². The ring has a coil of 600 turns. When a current of magnitude 0.8 A is passed through the windings, the relative permeability of 4000 is observed. What will be the induced emf (approximately) when the current is reduced to zero in 60 ms?
1. -27 V
2. 27 V
3. -38 V
4. 38 V

Answer 1

Correct Answer - Option 3 : -38 V

Faraday’s Laws of Electromagnetic Induction:

First Law: It tells us about the condition under which an EMF is induced in a conductor or coil and may be stated as under:

When the magnetic flux linking a conductor or coil changes, an EMF is induced in it.

Second Law: It gives the magnitude of the induced EMF in a conductor or coil and may be stated as under:

The magnitude of the EMF induced in a conductor or coil is directly proportional to the rate of change of flux linkages i.e.

Induced EMF (E) ∝  \(\frac{Nϕ_1-Nϕ_2}{t}\)

In differential form, we have,

E = \(N\frac{dϕ}{dt}\)

The direction of induced EMF is given by Lenz’s law.

The magnitude and direction of induced EMF should be written as:

E = - \(N\frac{dϕ}{dt}\)  .... (1)

Consider a magnetic circuit having 'N; numbers of turns, 'A' cross-sectional area, 'L' is the length and current 'I'

We know that,

MMF = NI = ϕS

or, ϕ = \(\frac{NI}{S}\)

Where S is reluctance and given by,

S = \(\frac{L}{μ_oμ_rA}\)

From equation (1),

E = - \(N\frac{d(NI/S)}{dt}\)

or, E = - \(\frac{N^2}{S}\frac{dI}{dt}\) = - \(L\frac{dI}{dt}\) 

Application:

Given,

d = 100 mm ⇒ L = πd = 100π mm

N = 600

μ_r = 4000

A = 500 mm²

Hence,

S = \(\frac{L}{μ_oμ_rA}\) = \(\frac{100\pi\ mm}{4\pi \times 10^{-7}\times 4000\times 500\ mm^2}\)

or, S = \(\frac{L}{μ_oμ_rA}\) = \(\frac{100}{4 \times 10^{-7}\times 4000\times 500\ mm}\)

or, S = \(\frac{L}{μ_oμ_rA}\) = \(\frac{10^{12}}{8 \times10^6}=\frac{10^6}{8}\ H^{-1}\)

Hence, L = \(\frac{N^2}{S}= \frac{600\times 600\times 8}{10^6}=\frac{72}{25}\ H\)

Since, the current is reduced to zero in 60 ms.

Hence, \(\frac{dI}{dt}=\frac{0.8}{60}\times 1000=\frac{80}{6}\)

Hence, E = - \(L\frac{dI}{dt}\) = \(-\frac{72}{25}\times \frac{80}{6}=-\frac{960}{25}=-38.4\ V\)