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### Fundamentals Of Physics

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 ${\mathbit{\epsilon }}{\mathbf{=}}{\mathbf{2}}{\mathbit{\tau }}{\mathbit{\tau }}{\mathbit{f}}{\mathbit{N}}{\mathbit{a}}{\mathbit{b}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}\mathbf{\left(}\mathbf{2}\mathbf{\tau }\mathbf{\tau }\mathbf{f}\mathbf{t}\mathbf{\right)}{\mathbf{=}}{{\mathbit{\epsilon }}}_{{\mathbf{0}}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}\mathbf{\left(}\mathbf{2}\mathbf{\tau }\mathbf{\tau }\mathbf{f}\mathbf{t}\mathbf{\right)}$. This is the principle of the commercial alternating-current generator. (b) What value of Nab gives an emf with ${{\mathbit{\epsilon }}}_{{\mathbf{0}}}{\mathbf{150}}{\mathbf{}}{\mathbf{V}}$ when the loop is rotated at ${\mathbf{60}}{\mathbf{.}}{\mathbf{0}}\frac{\mathbf{rev}}{\mathbf{s}}$ in a uniform magnetic field of 0.500 T ?

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

See the step by step solution

## Step 1: Given

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

## Step 2: Determining the concept

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 states, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follow:

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

Magnetic flux, $\phi =\varphi BdA$

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

## Step 3: (a) To prove ε=2ττfNabBsin2ττft=ε0sin2ττft

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,

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

From equation 1,

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

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

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

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.

## Step 4: (b) Determining the value of Nab

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,

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

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.