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 - Conceptual Questions - Page 996: 11

Answer

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Work Step by Step

We know that the potential difference across an inductor is given by $$\Delta V_L=-L\dfrac{dI}{dt}$$ We are given $L$ and $\Delta V$ for both inductors, so we can't find which has a larger current than the other, but we can find $dI/dt$ which means we can find which one has a rapid change in current. $$\color{blue}{\bf [a]}$$ No. $$\color{blue}{\bf [b]}$$ Yes, and to find which is which, we need to use the formula above for the two conductors. Solving this formula for $dI/dt$, $$ \left|\left(\dfrac{dI}{dt}\right)_1\right|=\left|\dfrac{(\Delta V_L)_1}{-L_1}\right| =\left|\dfrac{2}{2}\right|=\bf 1\;\rm A/s$$ $$ \left|\left(\dfrac{dI}{dt}\right)_2\right|=\left|\dfrac{(\Delta V_L)_2}{-L_2}\right| =\left|\dfrac{4}{1}\right|=\bf 4\;\rm A/s$$ Thus, Inductor 2 is the one at which the current is changing more rapidly. $$\color{blue}{\bf [c]}$$ In both cases, the current is decreasing since $\dfrac{dI}{dt}\lt 0$ and $\Delta V_L\gt 0$
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