Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 21 - Magnetic Forces and Magnetic Fields - Problems - Page 608: 7

Answer

$1.1\times 10^{-2}$ $N$

Work Step by Step

The magnetic fore acting on a particle having charge $q$ and moving with a velocity $(\vec v)$ in a uniform magnetic field $(\vec B)$ is given by $\vec F_{m}=q(\vec v\times \vec B)$ The electric force experienced by a charge $q$ due to external electric field $(\vec E)$ is given by $\vec F_{e}=q\vec E$ Let, the directions of both fields ($\vec B$ and $\vec E$) are along positive $x$ axis. The velocity $(\vec v)$ of the particle is perpendicular to both fields. Therefore we are assuming that the direction of $\vec v$ is along posive y axis. Thus, the magnetic force acting on the particle is $|\vec F_{m}|=qvB\sin90^{\circ}$, along negative $z$ axis or, $|\vec F_{m}|=(1.8\times 10^{-6})\times (3.1\times 10^{6})\times (1.2\times 10^{-3})\times 1$ $N$, along negative $z$ axis or, $|\vec F_{m}|=6.696\times 10^{-3}$ $N$, along negative $z$ axis And the electric force acting on the particle is $|\vec F_{e}|=qE$, along positive $x$ axis or, $|\vec F_{e}|=(1.8\times 10^{-6})\times (4.6\times10^{3})$ $N$, along positive $x$ axis or, $|\vec F_{e}|=8.28\times10^{-3}$ $N$, along positive $x$ Now, the magnitude of the net force that acts on the charge is given by $|\vec F|=\sqrt{|\vec F_{m}|^{2}+|\vec F_{e}|^{2}}$ or, $|\vec F|=\sqrt{(6.696\times 10^{-3})^{2}+(8.28\times 10^{-3})^{2}}$ $N$ $|\vec F|\approx1.1\times 10^{-2}$ $N$ Therefore, the magnitude of the net force that acts on the charge is $1.1\times 10^{-2}$ $N$.
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