Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 34 - Electromagnetic Fields and Waves - Exercises and Problems - Page 1030: 31

Answer

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Work Step by Step

$$\color{blue}{\bf [a]}$$ From the given graph, we can see that: - $E=( -10^5\;\hat j)\;\rm V/m$ - $v= ( 10^6\;\hat i)\;\rm m/s$ - $\sum F=0\;\rm N$ Since the electron will move without deflecting, the net force exerted on it is zero, as mentioned above. Recalling that the net force exerted on a moving charge in an electromagnetic field is given by $$\sum F=q(\vec E+\vec v\times \vec B)$$ $$0=q(\vec E+\vec v\times \vec B)$$ Thus, $$\vec E=-\vec v\times \vec B$$ $$\vec B=\dfrac{\vec E}{-\vec v}$$ Thus, We can see that the electric force is downward, so the magnetic force is upward. So, according to the right-hand rule, the direction of the magnetic field is into the page. Plug the known; $$\vec B=\dfrac{10^5}{10^6 }=-(\color{red}{\bf 0.1}\;\hat k)\;\rm T$$ $$\color{blue}{\bf [b]}$$ We need to use Galilean fields transformation equations, $$\left.\begin{matrix} &\vec E_B=\vec E_A+\vec v_{BA}\times \vec B_A \\ & \vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}\\ \end{matrix}\right\}\tag 1$$ Assuming that B is the proton's frame. So, $$\vec E_B=\vec E_A+\vec v_{BA}\times \vec B_A$$ Plug the known; $$\vec E_B=( -10^5\;\hat j)+ ( 10^6\;\hat i)\times (-0.1\;\hat k)$$ $$\vec E_B=( -10^5\;\hat j)+ ( 10^5\;\hat j) $$ $$\vec E_B=\color{red}{\bf 0}\;\rm V/m$$ And, $$\vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}$$ Plug the known; $$\vec B_B=(-0.1\;\hat k)-\dfrac{( 10^6\;\hat i)\times ( -10^5\;\hat j)}{(3\times 10^8)^2}=-0.099\;\hat k$$ $$\vec B_B=(-\color{red}{\bf 0.1}\;\hat k) \;\rm T$$ $$\color{blue}{\bf [c]}$$ The experimenter on the proton's frame sees no electric force since $E_B=0$, and sees no magnetic force since the proton is at rest in his frame.
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