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 - Problems - Page 586: 55

Answer

$E=1.87\times 10^6\frac{V}{m}$ It is perpendicular to velocity and magnetic field and in opposite direction to magnetic force.

Work Step by Step

In the given scenario: $F_E=F_B$ where $F_E$ and $F_B$ represent electric and magnetic force respectively. $qE=qvB$ This can be rearranged as: $E=vB$.......eq(1) We also know that in the given scenario, magnetic force is equal to the centripetal force $qvB=\frac{mv^2}{r}$ This simplifies to $v=\frac{qBr}{m}$ Thus equation(1) becomes $E=(\frac{qBr}{m})B$ $E=\frac{qB^2r}{m}$ We plug in the known values to obtain: $E=\frac{1.6\times 10^{-19}(0.566)^2\times 6.1\times 10^{-2}}{1.67\times 10^{-27}}=1.87\times 10^6\frac{V}{m}$ It is perpendicular to velocity and magnetic field and in opposite direction to magnetic force.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.