Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Integrated Concepts Referring to the situation in the previous problem: (a) What current is induced in the ring if its resistance is 0.0100 Ω? (b) What average power is dissipated? (c) What magnetic field is induced at the center of the ring? (d) What is the direction of the induced magnetic field relative to the MRI’s field?

The value of current (I) is \({\rm{0}}{\rm{.304\;A}}\).
The potential difference in current will be dissipated power is \(9.24 \times {\rm{1}}{{\rm{0}}^{ - 4}}\;{\rm{W}}\).
The magnetic field induced at the center of rings is \({\rm{17}}{\rm{.4}}\;\mu {\rm{T}}\).
The direction of the induced magnetic field is opposite

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Concept Introduction

The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, which is a vector field. In a magnetic field, a moving charge experiences a force that is perpendicular to both its velocity and the magnetic field.

Step 2: Calculate the value of current

(a)

The resistance is given, and we've just determined the electromotive force or potential difference \({\rm{\varepsilon }} = {\rm{\Delta V}}\).

In this case, we can simply determine the current using Ohm's law as follows:

\(\begin{aligned}{c}{\rm{I}} &= \frac{{\Delta {\rm{V}}}}{{\rm{R}}}\\ &= \frac{{3.04 \times {{10}^{ - 3}}\;{\rm{V}}}}{{{\rm{0}}{\rm{.0100}}\;\Omega }}\\ &= {\rm{0}}{\rm{.304\;A}}\end{aligned}\)

Therefore, the value of the current is \({\rm{0}}{\rm{.304\;A}}\).

Step 3: Calculate the potential difference

(b)

The product of the potential difference with the current will be the dissipated power. This results in

\(\begin{aligned}{c}{\rm{P}} &= {\rm{VI}}\\ &= \left( {3.04 \times {\rm{1}}{{\rm{0}}^{ - {\rm{3}}}}\;{\rm{V}}} \right) \times {\rm{0}}{\rm{.304}}\;{\rm{A}}\\ &= 9.24 \times {\rm{1}}{{\rm{0}}^{ - 4}}\;{\rm{W}}\end{aligned}\)

Therefore, the potential difference in current will be dissipated power is \({\rm{0}}{\rm{.9\;mW}}\).

Step 4: Calculate the magnitude of the ring

(c)

The magnetic field created at the ring's center is known to have a magnitude of

\({\rm{B = }}\frac{{{{\rm{\mu }}_{\rm{0}}}{\rm{I}}}}{{{\rm{2R}}}}\)

As a result of which we may calculate it as

\(\begin{aligned}{c}{\rm{B}} &= \frac{{\left( {{\rm{4\pi }} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 7}}}}\;{{T \cdot m} \mathord{\left/

{\vphantom {{T \cdot m} A}} \right.

\kern-\nulldelimiterspace} A}} \right)\; \times \left( {{\rm{0}}{\rm{.304}}\;{\rm{A}}} \right)}}{{{\rm{2}} \times \left( {{\rm{0}}{\rm{.011}}\;{\rm{m}}} \right)}}\\ &= 17.4 \times {\rm{1}}{{\rm{0}}^{ - 6}}\;T\left( {\frac{{1\;\mu T}}{{{\rm{1}}{{\rm{0}}^{ - 6}}\;T}}} \right)\\ &= {\rm{17}}{\rm{.4}}\;\mu {\rm{T}}\end{aligned}\)

Therefore, the magnitude of rings at the center is \({\rm{17}}\;{\rm{\mu T}}\).

Step 5: Direction of magnetic field

(d)

The direction of the induced field will be opposite to the field whose flux increase created it, according to Lenz's law.

Therefore, the direction of the induced magnetic field is the opposite.

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Concept Introduction

Step 2: Calculate the value of current

Step 3: Calculate the potential difference

Step 4: Calculate the magnitude of the ring

Step 5: Direction of magnetic field

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Concept Introduction

Step 2: Calculate the value of current

Step 3: Calculate the potential difference

Step 4: Calculate the magnitude of the ring

Step 5: Direction of magnetic field

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