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 830: 5


$B = (0,0,0.28\times 10^{-5})~T$

Work Step by Step

We can find the magnetic field at the point: $B = \frac{\mu_0~q~v~sin~\theta}{4~\pi~r^2}$ $B = \frac{(4\pi\times 10^7~H/m)~(1.6\times 10^{-19}~C)~(2.0\times 10^7~m/s)~sin~45^{\circ}}{(4~\pi)~(2~\sqrt{2}~m)^2}$ $B = 0.28\times 10^{-5}~T$ By the right hand rule, the magnetic field is in the $+z$ direction. We can write the magnetic field as a vector: $B = (0,0,0.28\times 10^{-5})~T$
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