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: 89

Answer

See the detailed answer below.

Work Step by Step

a) Let's start with wire M, we need to find the net magnetic force exerted on it. As we see in the figure below, the force exerted on wire M by the two other wires (N and P) is a repulsive force since the current in it is in opposite direction (the blue arrows). Thus, $$\sum F_{x,\rm on\;M}=F_{\rm N}\sin 30^\circ-F_{\rm P}\sin 30^\circ=B_{\rm N}I_{\rm M}L_{\rm M}\sin 30^\circ-B_{\rm P}I_{\rm M}L_{\rm M}\sin 30^\circ$$ Noting that the distances between the three wires are equal and are represented here by $d$ $$\sum F_{x,\rm on\;M}=L\left[\dfrac{\mu_0 \overbrace{I_NI_M}^{I^2}\sin 30^\circ}{2\pi d }-\dfrac{\mu_0 \overbrace{I_PI_M}^{I^2}\sin 30^\circ}{2\pi d }\right]=0$$ This means that the horizontal components of the two forces cancel each other since they have the same magnitude but opposite directions. Thus, the net force exerted on wire M is vertically upward since the components of the two forces are pointing up. $$\sum F_{\rm on\;M}=\sum F_{y,\rm on\;M}=F_{\rm N}\cos30^\circ+F_{\rm P}\cos 30^\circ=B_{\rm N}I_{\rm M}L_{\rm M}\cos30^\circ+B_{\rm P}I_{\rm M}L_{\rm M}\cos30^\circ$$ $$\sum F_{\rm on\;M}=L\left[\dfrac{\mu_0 \overbrace{I_NI_M}^{I^2}\cos 30^\circ}{2\pi d }+\dfrac{\mu_0 \overbrace{I_PI_M}^{I^2}\cos 30^\circ}{2\pi d }\right] $$ $$\dfrac{\sum F_{\rm on\;M}}{L}=\dfrac{2\mu_0I^2\cos 30^\circ}{2\pi d}$$ Plugging the known; $$\dfrac{\sum F_{\rm on\;M}}{L}=\dfrac{2\cdot 4\pi \cdot 10^{-7}\cdot 8^2 \cdot \cos 30^\circ}{2\pi \cdot 3.8\times10^{-2}}$$ $$\dfrac{\sum F_{\rm on\;M}}{L}=\color{red}{\bf 5.83\times10^{-4}}\;\rm N/m$$ ------------------------------------------------- Now we need to find the net force exerted on wire P (the green arrows). The force between P and N is an attractive force while the force between P and M is a repulsive force. $$\sum F_{x,\rm on\;P}=F_{\rm M}\sin30^\circ-F_{\rm N}$$ $$\sum F_{x,\rm on\;P}= B_MIL\sin30^\circ- B_NIL$$ $$\dfrac{\sum F_{x,\rm on\;P}}{L}=\dfrac{\mu_0I^2\sin30^\circ}{2\pi d}-\dfrac{\mu_0I^2}{2\pi d}$$ $$\dfrac{\sum F_{x,\rm on\;P}}{L}=\dfrac{\mu_0I^2}{2\pi d}\left[\sin30^\circ-1\right] $$ Plugging the known; $$\dfrac{\sum F_{x,\rm on\;P}}{L}=\dfrac{4\pi \cdot10^{-7}\cdot 8^2}{2\pi \cdot 3.8\times10^{-2}}\left[\sin30^\circ-1\right] $$ $$\dfrac{\sum F_{x,\rm on\;P}}{L}=\color{blue}{\bf -1.68\times10^{-4}}\;\rm N/m$$ $$\sum F_{y,\rm on\;P}=-F_{\rm M}\cos30^\circ= -B_MIL\cos30^\circ $$ $$\dfrac{\sum F_{y,\rm on\;P}}{L}=-\dfrac{\mu_0I^2\cos30^\circ}{2\pi d} $$ Plugging the known; $$\dfrac{\sum F_{y,\rm on\;P}}{L}=-\dfrac{4\pi \cdot10^{-7}\cdot 8^2\cdot \cos30^\circ}{2\pi \cdot 3.8\times10^{-2}} $$ $$\dfrac{\sum F_{y,\rm on\;P}}{L}=\color{blue}{\bf-2.92\times10^{-4}}\;\rm N/m$$ Therefore, $$\dfrac{\sum F_{\rm on\; P}}{L}=\sqrt{\left[\dfrac{\sum F_{y,\rm on\;P}}{L}\right]^2+\left[\dfrac{\sum F_{y,\rm on\;P}}{L}\right]^2} $$ $$\dfrac{\sum F_{\rm on\; P}}{L}=\sqrt{\left[-1.68\times10^{-4}\right]^2+\left[-2.92\times10^{-4}\right]^2}$$ $$\dfrac{\sum F_{\rm on\; P}}{L}=\color{red}{\bf 3.37\times10^{-4}}\;\rm N/m$$ And its direction is given by $$\theta_{\rm of\;P}=\tan^{-1}\left[\dfrac{-2.92\times10^{-4}}{-1.68\times10^{-4}}\right]=\bf 60^\circ$$ And since it is in the third quadrant, its direction is given by $$\theta_{\rm of\;P}=60^\circ+240^\circ=\color{red}{\bf 240^\circ}$$ ------------------------------------------------- Now we need to find the net force exerted on wire N (the red arrows). The force between N and P is an attractive force while the force between N and M is a repulsive force. $$\sum F_{x,\rm on\;N}=F_{\rm P}-F_{\rm M}\sin30^\circ$$ $$\sum F_{x,\rm on\;N}= B_PIL- B_MIL\sin30^\circ$$ $$\dfrac{\sum F_{x,\rm on\;N}}{L}=\dfrac{\mu_0I^2}{2\pi d}-\dfrac{\mu_0I^2\sin30^\circ}{2\pi d}$$ $$\dfrac{\sum F_{x,\rm on\;N}}{L}=\dfrac{\mu_0I^2}{2\pi d}\left[1-\sin30^\circ \right] $$ Plugging the known; $$\dfrac{\sum F_{x,\rm on\;N}}{L}=\dfrac{4\pi \cdot10^{-7}\cdot 8^2}{2\pi \cdot 3.8\times10^{-2}}\left[1-\sin30^\circ \right] $$ $$\dfrac{\sum F_{x,\rm on\;N}}{L}=\color{blue}{\bf 1.68\times10^{-4}}\;\rm N/m$$ $$\sum F_{y,\rm on\;N}=-F_{\rm M}\cos30^\circ= -B_MIL\cos30^\circ $$ $$\dfrac{\sum F_{y,\rm on\;N}}{L}=-\dfrac{\mu_0I^2\cos30^\circ}{2\pi d} $$ Plugging the known; $$\dfrac{\sum F_{y,\rm on\;N}}{L}=-\dfrac{4\pi \cdot10^{-7}\cdot 8^2\cdot \cos30^\circ}{2\pi \cdot 3.8\times10^{-2}} $$ $$\dfrac{\sum F_{y,\rm on\;N}}{L}=\color{blue}{\bf-2.92\times10^{-4}}\;\rm N/m$$ Therefore, $$\dfrac{\sum F_{\rm on\; N}}{L}=\sqrt{\left[\dfrac{\sum F_{y,\rm on\;P}}{L}\right]^2+\left[\dfrac{\sum F_{y,\rm on\;P}}{L}\right]^2} $$ $$\dfrac{\sum F_{\rm on\; N}}{L}=\sqrt{\left[ 1.68\times10^{-4}\right]^2+\left[-2.92\times10^{-4}\right]^2}$$ $$\dfrac{\sum F_{\rm on\; N}}{L}=\color{red}{\bf 3.37\times10^{-4}}\;\rm N/m$$ And its direction is given by $$\theta_{\rm of\;N}=\tan^{-1}\left[\dfrac{-2.92\times10^{-4}}{ 1.68\times10^{-4}}\right]=\bf 60^\circ$$ And since it is in the fourth quadrant, its direction is given by $$\theta_{\rm of\;N}=360^\circ-60^\circ=\color{red}{\bf 300^\circ}$$ ------------------------------------------------- b) In the second figure below, we can find the dot at which we need to find the net magnetic field. We used the right-hand rule, to find the magnetic field direction exerted by each wire at the given point. Thus, from the figure below, $$\sum B_x=B_{\rm by\;M}\cos30^\circ+B_{\rm by\;N}\cos30^\circ+B_{\rm by\;P}\cos60^\circ$$ $$\sum B_x=\dfrac{\mu_0 I\cos30^\circ}{2\pi (\frac{d}{2})}+\dfrac{\mu_0 I\cos30^\circ}{2\pi (\frac{d}{2})}+\dfrac{\mu_0 I\cos60^\circ}{2\pi (\frac{\sqrt{3}\;d}{2})}$$ The distances are found from the geometry of the second figure below. $$\sum B_x=\dfrac{\mu_0 I}{ \pi d}\left[\cos30^\circ+\cos30^\circ +\dfrac{ \cos60^\circ}{ \sqrt{3} }\right]$$ Plugging the known; $$\sum B_x=\dfrac{4\pi \cdot 10^{-7}\cdot 8}{ \pi \cdot 3.8\times10^{-2}}\left[\cos30^\circ+\cos30^\circ +\dfrac{ \cos60^\circ}{ \sqrt{3} }\right]$$ $$\sum B_x=\color{Green}{\bf 1.7\times10^{-4}}\;\rm T$$ ---- $$\sum B_y=-B_{\rm by\;M}\sin 30^\circ-B_{\rm by\;N}\sin 30^\circ+B_{\rm by\;P}\sin 60^\circ$$ $$\sum B_y=-\dfrac{\mu_0 I\sin 30^\circ}{2\pi (\frac{d}{2})}-\dfrac{\mu_0 I\sin 30^\circ}{2\pi (\frac{d}{2})}+\dfrac{\mu_0 I\sin 60^\circ}{2\pi (\frac{\sqrt{3}\;d}{2})}$$ $$\sum B_y=-\dfrac{\mu_0 I}{ \pi d}\left[\sin 30^\circ+\sin 30^\circ -\dfrac{ \sin 60^\circ}{ \sqrt{3} }\right]$$ Plugging the known; $$\sum B_y=-\dfrac{4\pi \cdot 10^{-7}\cdot 8}{ \pi \cdot 3.8\times10^{-2}}\left[\sin 30^\circ+\sin 30^\circ -\dfrac{ \sin 60^\circ}{ \sqrt{3} }\right]$$ $$\sum B_y=\color{Green}{\bf -4.2\times10^{-5}}\;\rm T$$ The net magnetic field is given by $$\sum B=\sqrt{\left(\sum B_x\right)^2+\left(\sum B_y\right)^2}$$ $$\sum B=\sqrt{\left(1.7\times10^{-4}\right)^2+\left(-4.2\times10^{-5}\right)^2}$$ $$\sum B=\color{red}{\bf 1.75\times10^{-4}}\;\rm T$$ And its direction is given by $$\alpha_B=\tan^{-1}\left[\dfrac{-4.2\times10^{-5}}{1.7\times10^{-4}}\right]=\bf -13.9^\circ$$ And since it is in the fourth quadrant, $$\alpha_B=360^\circ -13.9^\circ\approx\color{red}{\bf 346^\circ}$$
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