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 587: 75

Answer

$51.8^\circ$

Work Step by Step

The protons will move on a circular path (actually part of it as shown in the given figure) when they enter the small area of a magnetic field. This will bend the proton beam and we need to find the angle of this curve. First, we need to find the radius of curvature which is given by $$qvB=\dfrac{mv^2}{r}$$ Hence, $$r=\dfrac{mv}{qB}\tag 1$$ From the geometry of the figure below, we can see that $$\sin\theta=\dfrac{0.1}{r}$$ Plugging from (1); $$\sin\theta=\dfrac{x}{\dfrac{mv}{qB}}=\dfrac{xqB}{mv}$$ Thus, $$\theta=\sin^{-1}\left(\dfrac{0.1qB}{mv}\right)=\sin^{-1}\left(\dfrac{0.05\cdot 1.6\times10^{-19}\cdot 0.41}{1.67\times10^{-27}\cdot 2.5\times10^6}\right)$$ $$\theta=\color{red}{\bf51.8}^\circ$$
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