Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 33 - Electromagnetic Induction - Exercises and Problems - Page 1000: 59

Answer

See the detailed answer below.

Work Step by Step

We know from, Faraday’s law of electric induction, that $$\oint \vec E\cdot d\vec s=A\left| \dfrac{dB}{dt}\right|$$ As we see below, the electric field is tangent to the curve at $r\gt R$ at any point on its surface. Hence, $$ E (2\pi r)=A\left| \dfrac{dB}{dt}\right|$$ $$ E (2\pi r)=\pi R^2\left| \dfrac{dB}{dt}\right|$$ Thus, $$\boxed{E=\dfrac{R^2}{2r}\left| \dfrac{dB}{dt}\right|}$$ To be sure, let's test at $r=R$, $$E=\dfrac{R }{2}\left| \dfrac{dB}{dt}\right|$$ which is the strength of the induced electric field inside the solenoid.
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.