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 588: 79

Answer

- $r=5.25\times 10^{-5} \;\rm m$ - $p=3.3\times 10^{-4}\;\rm m$

Work Step by Step

Since the electron is moving on a circular path, so the force exerted on it is a centripetal force due to the magnetic field. This force is given by $$F_B=qB v\sin\theta =ma_r$$ Noting that $\theta=90^\circ$ since we need the perpendicular direction of the velocity of the magnetic field. Thus, $$ q B v_{\perp }=m\dfrac{v_{\perp}^2}{r}$$ Noting that the electron enters the magnetic field at an angle of $\theta=45^\circ$, so $v_{\perp}=v\sin45^\circ$ $$ q B =m\dfrac{v \sin45^\circ }{r}$$ where $q$ is the charge of the electron, and $m$ is its mass. Solving for $r$; $$ r= \dfrac{mv \sin45^\circ }{qB }$$ Plugging the known; $$ r= \dfrac{ 9.11\times10^{-31}\cdot 3\times 10^6 \cdot \sin45^\circ }{1.6\times10^{-19}\cdot 0.23} $$ $$r=\color{red}{\bf 5.25\times 10^{-5}}\;\rm m$$ Now we need to find the pitch $p$ and we know that the horizontal velocity component of the electron is constant. We also know that the pitch is the distance traveled during completing one circular path. So, $$p=v_xt=v\cos45^\circ \cdot \dfrac{2\pi r}{v\sin45^\circ }$$ $$p= \dfrac{2\pi r\cos45^\circ }{ \sin45^\circ }$$ Plugging the known; $$p= \dfrac{2\pi \cdot 5.25\times 10^{-5}\cdot \cos45^\circ }{ \sin45^\circ }$$ $$p=\color{red}{\bf 3.3\times 10^{-4}}\;\rm m$$
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