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Q11P

Expert-verifiedFound in: Page 896

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A rectangular coil of N turns and of length a and width b is rotated at frequency f in a uniform magnetic field, as indicated in Figure. The coil is connected to co-rotating cylinders, against which metal brushes slide to make contact. (a) Show that the emf induced in the coil is given (as a function of time t) by **${\mathit{\epsilon}}{\mathbf{=}}{\mathbf{2}}{\mathit{\tau}}{\mathit{\tau}}{\mathit{f}}{\mathit{N}}{\mathit{a}}{\mathit{b}}{\mathit{s}}{\mathit{i}}{\mathit{n}}\mathbf{\left(}\mathbf{2}\mathbf{\tau}\mathbf{\tau}\mathbf{f}\mathbf{t}\mathbf{\right)}{\mathbf{=}}{{\mathit{\epsilon}}}_{{\mathbf{0}}}{\mathit{s}}{\mathit{i}}{\mathit{n}}\mathbf{\left(}\mathbf{2}\mathbf{\tau}\mathbf{\tau}\mathbf{f}\mathbf{t}\mathbf{\right)}$

The value of $Nab=0.7961{\mathrm{m}}^{2}$

- Turns of the coil N
- Length of the rectangular coil is a
- Width of the rectangular coil is b
- Frequency of rotation, $\mathrm{f}=60\frac{\mathrm{rev}}{\mathrm{s}}$
- Induced emf $\mathrm{\epsilon}=150\mathrm{V}$
- Magnetic field, B = 0.500 T

**Use the expression for magnetic flux in Faraday’s law and deriving it, prove the required equation. By using the same proved equation, calculate value of Nab.**

** **

**Faraday's*** law of electromagnetic induction *

Formulae are as follow:

Faraday's law,

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

Magnetic flux, $\phi =\varphi BdA$

Where** $\mathrm{\Phi}$** is magnetic flux, *B* is a magnetic field, *A* is an area, *𝜀* is emf.

It is given that, a coil of conducting wire is placed in a magnetic field B.

The number of turns in a coil are N. The coil is rotated at frequency f.

Let the area of each turn be A.

Angular velocity of the coil is $\omega =2\tau \tau f...................................................\left(1\right)$

The coil is rotated by an angle $\theta $, in time t then $\theta =\omega t..........................................................\left(2\right)$

Magnetic flux passing through the coil is given by,

$\phi =\varphi BdA=\phi BdA\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}\varphi dA=AreaofNturns=NA\phantom{\rule{0ex}{0ex}}\phi =NAB\mathrm{cos}\theta $

From equation 1,

$\phi =NAB\mathrm{cos}\omega t.........................................................\left(3\right)$

Using this in the emf induced in a coil,

$\epsilon =\frac{d\phi}{dt}\phantom{\rule{0ex}{0ex}}\epsilon =\frac{d}{dt}\left[NAB\mathrm{cos}\omega t\right]\phantom{\rule{0ex}{0ex}}\epsilon =-NAB[-\omega \mathrm{sin}\omega t]\phantom{\rule{0ex}{0ex}}\epsilon =NAB\omega \mathrm{sin}\omega t$

From equation 1,

$\epsilon =NAB2\tau \tau f\mathrm{sin}2\tau \tau ft$

Since, it is a rectangular coil, area $A=length\times width=a\times b$

$\epsilon =Nab2\tau \tau f\mathrm{sin}2\tau \tau ft$

This is an expression for the induced emf in the coil at any instant t.

When $\mathrm{sin}2\tau \tau ft$ = 1, the emf is maximum.

${\epsilon}_{max}={\epsilon}_{0}=NabB2\tau \tau f.........................................................\left(3\right)\phantom{\rule{0ex}{0ex}}\epsilon ={\epsilon}_{0}\mathrm{sin}2\tau \tau ft$

Hence, $\epsilon =2\tau \tau fNabB\mathrm{sin}2\tau \tau ft={\epsilon}_{0}\mathrm{sin}2\tau \tau ft$ is proved.

It is given that,

Frequency of rotation, $\mathrm{f}=60\frac{\mathrm{rev}}{\mathrm{s}}$

Induced emf $\mathrm{\epsilon}=150\mathrm{V}$

Magnetic field, B = 0.550 T

And now find Nab.

At maximum emf, $\mathrm{sin}2\tau \tau ft=1$ .

Using all this in equation 3,

$\mathrm{\epsilon}=\mathrm{NabB}2\mathrm{\tau \tau f}\phantom{\rule{0ex}{0ex}}\mathrm{Nab}=\frac{\mathrm{\epsilon}}{\mathrm{B}2\mathrm{\tau \tau f}}\phantom{\rule{0ex}{0ex}}\mathrm{Nab}=\frac{150}{0.5\times 2\times 3\times 3.14\times 60}\phantom{\rule{0ex}{0ex}}\mathrm{Nab}=0.7961{\mathrm{m}}^{2}$

Hence, the value of Nab is $0.7961{\mathrm{m}}^{2}$.

Therefore, use the expression for magnetic flux in Faraday’s law and by deriving it, prove the required equation. By using the same equation, calculate the value of Nab.

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