Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 20 - Magnetism - Problems - Page 586: 63

Answer

\begin{array}{|c|c|c|} \hline B_0\;{(\rm T)}& B\;{(\rm T)} & \color{red}\mu\;{(\rm T.m/A)}\\ \hline 0 & 0 &{\rm \color{red}{undefined}} \\\\ \hline 0.13\times10^{-4} & 0.0042&\color{red}{ 4.06\times10^{-4}}\\\\ \hline 0.25\times10^{-4} & 0.010 & \color{red}{ 5.03\times10^{-4}} \\\\ \hline 0.5\times10^{-4}& 0.028 & \color{red}{ 7.04\times10^{-4}} \\\\ \hline 0.63\times10^{-4}& 0.043& \color{red}{ 8.85 \times10^{-4} }\\\\ \hline 0.78\times10^{-4}& 0.095& \color{red}{ 15.3 \times10^{-4} }\\\\ \hline 1.0\times10^{-4}& 0.45& \color{red}{ 56.5 \times10^{-4} }\\\\ \hline 1.3\times10^{-4}& 0.67& \color{red}{ 64.8 \times10^{-4} }\\\\ \hline 1.9\times10^{-4} & 1.01& \color{red}{ 66.8 \times10^{-4} }\\\\ \hline 2.5\times10^{-4} & 1.18& \color{red}{ 59.3 \times10^{-4} }\\\\ \hline 6.3\times10^{-4} & 1.44& \color{red}{ 28.7 \times10^{-4} }\\\\ \hline 13\times10^{-4} & 1.58& \color{red}{ 15.3 \times10^{-4} }\\\\ \hline 130\times10^{-4} & 1.72 & \color{red}{ 1.66 \times10^{-4} }\\\\ \hline 1300\times10^{-4} & 2.26 & \color{red}{ 0.218 \times10^{-4} }\\\\ \hline 10000\times10^{-4} & 3.15 & \color{red}{ 0.040 \times10^{-4} }\\\\ \hline \end{array}

Work Step by Step

We know that the magnetic permeability is given by $$B_0=\mu_0 n I$$ and hence, $$B=\mu n I$$ Divide $B$ by $B_0$ and solve for $\mu$; $$\dfrac{B}{B_0}=\dfrac{\mu n I}{\mu_0 n I}$$ Thus, $$ \mu=\dfrac{\mu_0 B}{B_0} $$ $$\boxed{\mu=4\pi \times10^{-7}\left(\dfrac{B}{B_0 }\right)} $$ We need to use this formula to find the values of $\mu$s, as you see in the table below. \begin{array}{|c|c|c|} \hline B_0\;{(\rm T)}& B\;{(\rm T)} & \color{red}\mu\;{(\rm T.m/A)}\\ \hline 0 & 0 &{\rm \color{red}{undefined}} \\\\ \hline 0.13\times10^{-4} & 0.0042&\color{red}{ 4.06\times10^{-4}}\\\\ \hline 0.25\times10^{-4} & 0.010 & \color{red}{ 5.03\times10^{-4}} \\\\ \hline 0.5\times10^{-4}& 0.028 & \color{red}{ 7.04\times10^{-4}} \\\\ \hline 0.63\times10^{-4}& 0.043& \color{red}{ 8.85 \times10^{-4} }\\\\ \hline 0.78\times10^{-4}& 0.095& \color{red}{ 15.3 \times10^{-4} }\\\\ \hline 1.0\times10^{-4}& 0.45& \color{red}{ 56.5 \times10^{-4} }\\\\ \hline 1.3\times10^{-4}& 0.67& \color{red}{ 64.8 \times10^{-4} }\\\\ \hline 1.9\times10^{-4} & 1.01& \color{red}{ 66.8 \times10^{-4} }\\\\ \hline 2.5\times10^{-4} & 1.18& \color{red}{ 59.3 \times10^{-4} }\\\\ \hline 6.3\times10^{-4} & 1.44& \color{red}{ 28.7 \times10^{-4} }\\\\ \hline 13\times10^{-4} & 1.58& \color{red}{ 15.3 \times10^{-4} }\\\\ \hline 130\times10^{-4} & 1.72 & \color{red}{ 1.66 \times10^{-4} }\\\\ \hline 1300\times10^{-4} & 2.26 & \color{red}{ 0.218 \times10^{-4} }\\\\ \hline 10000\times10^{-4} & 3.15 & \color{red}{ 0.040 \times10^{-4} }\\\\ \hline \end{array} Now we can easily plot the dots and draw the graph, as you see below.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.