College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 19 - Magnetism - Learning Path Questions and Exercises - Exercises - Page 695: 60


Magnetic force, $ \vec{F}=q (\vec{V} \times \vec{B}) $ Magnitude of centripetal force $\frac{mV^2}{r}$ Magnitude of Linear momentum $p= mv$ Kinetic energy $K=\frac{1}{2}mV^2$

Work Step by Step

Magnetic force $ \vec{F}=q (\vec{V} \times \vec{B}) $ Magnitude = F=qVB Now, centripetal force $\frac{mV^2}{r}= qVB$ Therefore $mV= qBr=P$ (This is magnitude of linear momentum, P) From here we can get the radius as well $r=\frac{mV}{qB}$ Kinetic energy, $K=\frac{1}{2}mV^2=\frac{m^2V^2}{2m}=\frac{P^2}{2m}$ We get the answers by putting the known values(please see the documents attached)
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.