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Found in: Page 896

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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# In Figure, a wire loop of lengths L = 40.0 cm and W = 25.0 cm lies in a magnetic field $\stackrel{\mathbf{\to }}{\mathbf{B}}$ .(a)What is the magnitude $\epsilon$ if $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}\mathbf{\left(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}}\mathbf{\right)}{\mathbf{y}}\stackrel{\mathbf{^}}{\mathbf{k}}$?(b)What is the direction (clockwise or counterclockwise—or “none” if 0) of the emf induced in the loop if $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}\mathbf{\left(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}}\mathbf{\right)}{\mathbf{y}}\stackrel{\mathbf{^}}{\mathbf{k}}$?(c)What is the ${\mathbf{\epsilon }}$ if $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}\mathbf{\left(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{s}}\mathbf{\right)}{\mathbf{t}}\stackrel{\mathbf{^}}{\mathbf{k}}$ (d)what is the direction if $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}\mathbf{\left(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{s}}\mathbf{\right)}{\mathbf{t}}\stackrel{\mathbf{^}}{\mathbf{k}}$ (e)What is the ${\mathbf{\epsilon }}$ if $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}\mathbf{\left(}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}\mathbf{.}\mathbf{s}}\mathbf{\right)}\mathbit{y}{\mathbf{t}}\stackrel{\mathbf{^}}{\mathbf{k}}$ (f)What is the direction if $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}\mathbf{\left(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{s}}\mathbf{\right)}{\mathbf{t}}\stackrel{\mathbf{^}}{\mathbf{k}}$(g)What is the ${\mathbf{\epsilon }}$ if $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}\mathbf{\left(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}\mathbf{.}\mathbf{s}}\mathbf{\right)}{\mathbit{x}}{\mathbf{t}}\stackrel{\mathbf{^}}{\mathbf{k}}$ (h)What is the direction if $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}\mathbf{\left(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}\mathbf{.}\mathbf{s}}\mathbf{\right)}{\mathbit{x}}{\mathbf{t}}\stackrel{\mathbf{^}}{\mathbf{k}}$(i)What is the if $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}\mathbf{\left(}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}\mathbf{.}\mathbf{s}}\mathbf{\right)}{\mathbit{y}}{\mathbf{t}}\stackrel{\mathbf{^}}{\mathbf{k}}$(j)What is the direction if $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}\mathbf{\left(}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}\mathbf{.}\mathbf{s}}\mathbf{\right)}{\mathbit{y}}{\mathbf{t}}\stackrel{\mathbf{^}}{\mathbf{k}}$

1. Magnitude of emf induced $\mathrm{\epsilon }=0$.
2. Direction of the emf is none.
3. Magnitude of emf induced $\mathrm{\epsilon }=6\mathrm{mV}$.
4. Direction of the emf is clockwise.
5. Magnitude of emf induced $\mathrm{\epsilon }=1\mathrm{mV}$.
6. Direction of the emf is clockwise.
7. Magnitude of emf induced $\mathrm{\epsilon }=0$.
8. Direction of the emf is none.
9. Magnitude of emf induced $\mathrm{\epsilon }=0$.
10. Direction of the emf is none.
See the step by step solution

## Step 1: Given

1. Length of loop, L = 40 cm = 0.4 m
2. Width of the loop, w = 25.0 cm=0.25 m

## Step 2: Determining the concept

By using the concept of the magnetic flux, Faraday’s law and Lenz’s law, find the magnitude and the direction of the induced emf in all the cases.

Faraday's law of electromagnetic induction states, Whenever a conductor is

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

Lenz's law states that the current induced in a circuit due to a change in a magnetic field is directed to oppose the change in flux and to exert a mechanical force which opposes the motion.

Formulae are as follow:

1. Faraday's law $\epsilon =\frac{d\phi }{dt}$
2. Magnetic flux $\phi =\varphi BdA$

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

## Step 3: (a) Determining the magnitude of emf induced ε

The magnetic flux passing through the coil is given by,

$\phi =\varphi B.dA=\varphi BdA\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}\phi =BA\mathrm{cos}\theta$

The emf induced in coil is given by,

$\epsilon =-\frac{d\phi }{dt}\phantom{\rule{0ex}{0ex}}\epsilon =-\frac{d}{dt}\left[BA\mathrm{cos}\theta \right].........................................................\left(1\right)\phantom{\rule{0ex}{0ex}}\epsilon =BA\mathrm{sin}\theta$

Now, for the rectangular coil,

$\mathrm{A}=0.4×0.25\phantom{\rule{0ex}{0ex}}\mathrm{A}=0.1{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{\epsilon }=0.1\mathrm{B}\mathrm{sin\theta }....................................................................\left(2\right)$

For $B=4×{10}^{-2}T/m\right)y\stackrel{^}{k}$

As the unit vector $\stackrel{^}{k}$ , it is along z axis and it is perpendicular to the coil. But it is not changing with time. Hence, from equation 1,

$\epsilon =-\frac{d}{dt}\left[BA\mathrm{cos}\theta \right]\phantom{\rule{0ex}{0ex}}\epsilon =0$

Hence, the emf induced in the coil is zero.

## Step 4: (b) Determining the direction of the emf, (clockwise or anticlockwise or none.)

From a),

Hence, the direction of the emf is none.

## Step 5: (c) Determining the magnitude of emf induced

For $\mathrm{B}=6×{10}^{-2}\mathrm{T}/\mathrm{s}\right)\mathrm{t}\stackrel{^}{\mathrm{k}}$

It is perpendicular to the coil and changes with time. From equation 2,

$\mathrm{\epsilon }=0.1\mathrm{Bsin\theta }\phantom{\rule{0ex}{0ex}}\mathrm{\epsilon }=0.1×6×{10}^{-2}\mathrm{sin}90\phantom{\rule{0ex}{0ex}}\mathrm{\epsilon }=0.1×6×{10}^{-2}×1\phantom{\rule{0ex}{0ex}}\mathrm{\epsilon }=0.1×6×{10}^{-2}\mathrm{V}\phantom{\rule{0ex}{0ex}}\mathrm{\epsilon }=6×{10}^{-3}\mathrm{V}\phantom{\rule{0ex}{0ex}}\mathrm{\epsilon }=6\mathrm{mV}$

Hence, emf induced in the loop will be 6 mV.

## Step 6: (d) Determining the direction of the emf, (clockwise or anticlockwise or none.)

Since, the magnetic field is directed into the page,

Hence, the direction of the induced emf is clockwise.

## Step 7: (e) Determining the magnitude of emf induced ε

For $\mathrm{B}=8×{10}^{-2}\mathrm{T}/\mathrm{m}.\mathrm{s}\right)\mathrm{yt}\stackrel{^}{\mathrm{k}}$,

which is in the direction of y axis, that is, along the width of the coil.

$\phi =\varphi B.dA\phantom{\rule{0ex}{0ex}}\phi =\varphi 8×{10}^{-2}yt×d\left(l×w\right)$

w is along x axis.

$\phi =l×8×{10}^{-2}t{\int }_{0}^{W}ydy\phantom{\rule{0ex}{0ex}}\phi =0.4×8×{10}^{-2}t×\frac{{w}^{2}}{2}$

$\phi =0.4×8×{10}^{-2}×t×\frac{{\left(0.25\right)}^{2}}{2}\phantom{\rule{0ex}{0ex}}\phi =0.1t×{10}^{-2}$

The emf induced can be calculated as,

$\mathrm{\epsilon }=\frac{\mathrm{d\phi }}{\mathrm{dt}}\phantom{\rule{0ex}{0ex}}\mathrm{\epsilon }=\frac{\mathrm{d}}{\mathrm{dt}}\left[0.1\mathrm{t}×{10}^{-2}\right]\phantom{\rule{0ex}{0ex}}\mathrm{\epsilon }=1×{10}^{-3}\mathrm{V}\phantom{\rule{0ex}{0ex}}\mathrm{\epsilon }=1\mathrm{mV}$

Hence, the emf induced in the coil is 1 mV.

## Step 8: (f) Determining the direction of the emf, (clockwise or anticlockwise or none.)

From e),

Hence, the direction of the induced emf will be clockwise.

## Step 9: (g) Determining the magnitude of emf induced ε

For $\mathrm{B}=3×{10}^{-2}\mathrm{T}/\mathrm{m}.\mathrm{s}\right)\mathrm{xt}\stackrel{^}{\mathrm{j}}$ , the magnetic field is directed along the y axis i.e. parallel to the coil.

The flux through the coil i.e. $\mathrm{\phi }=0$

$\epsilon =0$

Hence, the emf induced in the coil is zero.

## Step 10: (h) Determining the direction of the emf, (clockwise or anticlockwise or none.)

From g),

Hence, the direction of emf will be none.

## Step 11: (i) Determining the magnitude of emf induced ε

For$\mathrm{B}=5×{10}^{-2}\mathrm{T}/\mathrm{m}.\mathrm{s}\right)\mathrm{yt}\stackrel{^}{\mathrm{i}}$ , the magnetic field is directed along the x axis i.e. parallel to the coil.

The flux through the coil i.e. $\mathrm{\phi }=0$

$\epsilon =0$

Hence, the emf induced in the coil is zero.

## Step 12: (j) Determining the direction of the emf, (clockwise or anticlockwise or none.)

From i),

Hence, the direction of emf will be none.

Therefore, by using the concept of the magnetic flux, Faraday’s law and Lenz’s law, find the magnitude and the direction of the induced emf in all the cases.

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