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 - General Problems - Page 589: 92

Answer

$5.82\times10^{-5}\;\rm T$, $185.8^\circ$

Work Step by Step

First of all, we need to find the angles of the given triangle, as you see in the figure below. As the author mentioned, we can use the law of cosines. $$a^2=b^2+c^2-2ab\cos A$$ Thus, from the figure below $$13^2=12^2+ 8.2^2-[2\cdot 12\cdot 8.2]\cos \angle 1$$ Hence, $$\measuredangle{1} =\cos^{-1}\left[\dfrac{12^2+ 8.2^2-13^2}{2\cdot 12\cdot 8.2}\right]=\bf 77.6^\circ$$ And $\measuredangle 2$ is given by $$12^2=13^2+ 8.2^2-[2\cdot 13\cdot 8.2]\cos \angle 2$$ $$\measuredangle{2} =\cos^{-1}\left[\dfrac{13^2+ 8.2^2-12^2}{2\cdot 13\cdot 8.2}\right]=\bf 64.4^\circ$$ Therefore, the third angle is given by $$\measuredangle 3=180^\circ-64.4^\circ -77.6^\circ=\bf 38^\circ$$ We know that the magnetic field must be perpendicular to the distance between the source of the field and the point at which we need to measure the strength of the field. As we see in the second figure below. In the second figure below, we found all the angles we need. You can figure these angles out by noticing that the segment $\overline{ ML}$ is perpendicular to the magnetic field direction $B_{by\rm \;M}$, and $\overline{ KL}$ is perpendicular to the magnetic field direction $B_{by\rm \;K}$. The net magnetic field at point L is given by; $$\sum B_x=-B_{by\;\rm M}\cos 12.4^\circ-B_{by\;\rm K}\cos25.6^\circ$$ $$\sum B_x=-\left[\dfrac{\mu_0I\cos 12.4^\circ}{2\pi {\overline{ML}}}+\dfrac{\mu_0I\cos25.6^\circ}{2\pi {\overline{KL}}}\right]$$ $$\sum B_x=-\dfrac{\mu_0I}{2\pi }\left[\dfrac{\cos 12.4^\circ}{ {\overline{ML}}}+\dfrac{\cos25.6^\circ}{ {\overline{KL}}}\right]$$ Plugging the known; $$\sum B_x=-\dfrac{4\pi\cdot 10^{-7} \cdot 19.2}{2\pi }\left[\dfrac{\cos 12.4^\circ}{ {0.12 }}+\dfrac{\cos25.6^\circ}{ {0.13}}\right]$$ $$\sum B_x=\color{blue}{\bf -5.79\times10^{-5}}\;\rm T$$ $$\sum B_y= B_{by\;\rm M}\sin 12.4^\circ-B_{by\;\rm K}\sin 25.6^\circ$$ $$\sum B_y= \dfrac{\mu_0I\sin 12.4^\circ}{2\pi {\overline{ML}}}-\dfrac{\mu_0I\sin 25.6^\circ}{2\pi {\overline{KL}}} $$ $$\sum B_y= \dfrac{\mu_0I}{2\pi }\left[\dfrac{\sin 12.4^\circ}{ {\overline{ML}}}-\dfrac{\sin 25.6^\circ}{ {\overline{KL}}}\right]$$ Plugging the known; $$\sum B_y= \dfrac{4\pi\cdot 10^{-7} \cdot 19.2}{2\pi }\left[\dfrac{\sin 12.4^\circ}{ {0.12 }}-\dfrac{\sin 25.6^\circ}{ {0.13}}\right]$$ $$\sum B_y=\color{blue}{\bf -5.89\times10^{-6}}\;\rm T$$ Therefore, the magnitude of the net magnetic field at point L is given by; $$B=\sqrt{\left(\sum B_x\right)^2+\left(\sum B_y\right)^2}$$ $$B=\sqrt{\left(-5.79\times10^{-5} \right)^2+\left(-5.89\times10^{-6}\right)^2}$$ $$B=\color{red}{\bf5.82\times10^{-5}}\;\rm T$$ And its direction relative to the positive $x$-direction is given by $$\theta_B=\tan^{-1}\left[\dfrac{\sum B_y}{\sum B_x}\right]=\tan^{-1}\left[\dfrac{ -5.89\times10^{-6} }{-5.79\times10^{-5}}\right]=\bf 5.81^\circ$$ And since it is in the third quadranet, so $$\theta_B=180^\circ+5.81^\circ=\color{red}{\bf185.8^\circ}$$
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