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

Published by Pearson

# Chapter 29 - The Magnetic Field - Exercises and Problems - Page 830: 3

#### Answer

(a) $B = 0$ (b) $B = 0.16~T$ in the $+z$ direction (c) $B = 0.040~T$ in the $-z$ direction

#### Work Step by Step

We can use the Biot-Savart law to solve this question. (a) $B = \frac{\mu_0~q~v~sin~\theta}{4~\pi~r^2}$ $B = \frac{(4\pi\times 10^7~H/m)~(1.6\times 10^{-19}~C)~(1.0\times 10^7~m/s)~sin~0^{\circ}}{(4~\pi)~(0.01~m)^2}$ $B = 0$ (b) $B = \frac{\mu_0~q~v~sin~\theta}{4~\pi~r^2}$ $B = \frac{(4\pi\times 10^7~H/m)~(1.6\times 10^{-19}~C)~(1.0\times 10^7~m/s)~sin~90^{\circ}}{(4~\pi)~(0.01~m)^2}$ $B = 0.16~T$ By the right hand rule, the magnetic field is in the $+z$ direction. (c) $B = \frac{\mu_0~q~v~sin~\theta}{4~\pi~r^2}$ $B = \frac{(4\pi\times 10^7~H/m)~(1.6\times 10^{-19}~C)~(1.0\times 10^7~m/s)~sin~90^{\circ}}{(4~\pi)~(0.02~m)^2}$ $B = 0.040~T$ By the right hand rule, the magnetic field is in the $-z$ direction.

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