Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 29 - The Magnetic Field - Exercises and Problems - Page 834: 66


$0.046$ T

Work Step by Step

If the proton is moving in a circular arc then the magnetic force on the proton acts as a centripetal force pulling it closer to the center of the circle. Thus, $\Sigma \vec{F}_c = N\vec{F}_B = m\vec{a}_c$ Where $N$ is the number of magnets that have a magnetic force on the proton. $N(q\vec{v} \times \vec{B}) = \displaystyle \frac{m\vec{v}^2}{R}$ The magnetic field is always perpendicular to the velocity (since the magnetic field is either into or out of the page and the velocity is either left or right) so $q\vec{v} \times \vec{B} = q\vec{v}\vec{B}\sin(90) = q\vec{v}\vec{B}$ And the question does not give us the radius, however, and instead gives us the arc length of the circle. We can calculate the radius with this information since (let $\vec{s}$ be arc length) $\vec{s} = \theta R\qquad\qquad\qquad\qquad$ For an entire circle, $\theta = 2\pi$ $1.0 = 2\pi R $ $R = \displaystyle \frac{1.0}{2\pi}$ Using all of this information we can now calculate $\vec{B}$ $Nq\vec{v}\vec{B} = \displaystyle \frac{m\vec{v}^2}{R}$ $\displaystyle \vec{B} = \frac{m\vec{v}}{NqR} = \frac{(1.67\cdot10^{-27})(2.5\cdot10^7)}{36(1.6\cdot10^{-19})(\frac{1.0}{2\pi})} = 0.046$ T
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