Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

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 $ε = 2 τ τ f N a b s i n (2 τ τ f t) = ε_{0} s i n (2 τ τ f t)$ . This is the principle of the commercial alternating-current generator. (b) What value of Nab gives an emf with $ε_{0} 150 V$ when the loop is rotated at $60.0 \frac{rev}{s}$ in a uniform magnetic field of 0.500 T ?

The value of $N a b = 0.7961 m^{2}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

Turns of the coil N
Length of the rectangular coil is a
Width of the rectangular coil is b
Frequency of rotation, $f = 60 \frac{rev}{s}$
Induced emf $ε = 150 V$
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:

Faraday's law,

$ε = \frac{d φ}{d t}$

Magnetic flux, $φ = ϕ B d A$

Where $Φ$ 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 $ω = 2 τ τ f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

The coil is rotated by an angle $θ$ , in time t then $θ = ω t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)$

Magnetic flux passing through the coil is given by,

$φ = ϕ B d A = φ B d A \cos θ ϕ d A = A r e a o f N t u r n s = N A φ = N A B \cos θ$

From equation 1,

$φ = N A B \cos ω t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)$

Using this in the emf induced in a coil,

$ε = \frac{d φ}{d t} ε = \frac{d}{d t} [N A B \cos ω t] ε = - N A B [- ω \sin ω t] ε = N A B ω \sin ω t$

From equation 1,

$ε = N A B 2 τ τ f \sin 2 τ τ f t$

Since, it is a rectangular coil, area $A = l e n g t h \times w i d t h = a \times b$

$ε = N a b 2 τ τ f \sin 2 τ τ f t$

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

When $\sin 2 τ τ f t$ = 1, the emf is maximum.

$ε_{m a x} = ε_{0} = N a b B 2 τ τ f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3) ε = ε_{0} \sin 2 τ τ f t$

Hence, $ε = 2 τ τ f N a b B \sin 2 τ τ f t = ε_{0} \sin 2 τ τ f t$ is proved.

Step 4: (b) Determining the value of Nab

It is given that,

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

Induced emf $ε = 150 V$

Magnetic field, B = 0.550 T

And now find Nab.

At maximum emf, $\sin 2 τ τ f t = 1$ .

Using all this in equation 3,

$ε = NabB 2 ττf Nab = \frac{ε}{B 2 ττf} Nab = \frac{150}{0.5 \times 2 \times 3 \times 3.14 \times 60} Nab = 0.7961 m^{2}$

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

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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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given

Step 2: Determining the concept

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

Step 4: (b) Determining the value of Nab

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