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

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 29 - The Magnetic Field - Exercises and Problems - Page 830: 15

Answer

$B = 1.2\times 10^{-4}~T$

Work Step by Step

The equivalent resistance of the $20~\Omega$ resistor and the $30~\Omega$ resistor in series is $50~\Omega$ Since the potential difference across these resistors is $60~V$, the current in this section of the circuit is $I = \frac{60~V}{50~\Omega} = 1.2~A$ We can find the magnetic field strength at point A: $B = \frac{\mu_0~I}{2~\pi~r}$ $B = \frac{(4\pi\times 10^{-7}~T~m/A)~(1.2~A)}{(2~\pi)~(2.0\times 10^{-3}~m)}$ $B = 1.2\times 10^{-4}~T$
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