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: 64

Answer

See the detailed answer below.

Work Step by Step

First of all, we need to recall the induced potential difference through an inductor which is given by $$\Delta V_L=-L\dfrac{dI}{dt}$$ $$\Delta V_L=-L\dfrac{\Delta I}{\Delta t}$$ Plug the known; $$\Delta V_L=-(10\times 10^{-3})\dfrac{\Delta I}{\Delta t}\tag 1$$ We have, from the given graph, 4 stages. We will use (1) to find the potential difference in each stage. $\bullet$ For the first stage from $t=0$ s to $t=2$ ms. $$(\Delta V_L)_1=-(10\times 10^{-3})\dfrac{(2-0)}{(2-0)\times 10^{-3}}=\bf -10\;\rm V $$ $\bullet\bullet$ For the second stage from $t=2$ ms to $t=3$ ms. $$(\Delta V_L)_2=-(10\times 10^{-3})\dfrac{(2-2)}{(3-2)\times 10^{-3}}=\bf 0\;\rm V $$ $\bullet\bullet\bullet$ For the third stage from $t=3$ ms to $t=5$ ms. $$(\Delta V_L)_2=-(10\times 10^{-3})\dfrac{(-2-2)}{(5-3)\times 10^{-3}}=\bf 20\;\rm V $$ $\bullet\bullet\bullet\bullet$ For the fourth stage from $t=5$ ms to $t=6$ ms. $$(\Delta V_L)_3=-(10\times 10^{-3})\dfrac{(-2-[-2])}{(6-5)\times 10^{-3}}=\bf 0\;\rm V $$ Now we can draw the graph of the potential difference through these four stages. See the graph below.
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