 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q10PE

Expert-verified Found in: Page 860 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # 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?

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

## 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. ### Want to see more solutions like these? 