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 30 - Electromagnetic Induction - Exercises and Problems - Page 874: 69


a.) $\Delta V_L = -I_0L\omega\cos(\omega t)$ b.) $I_0 = -1.27\cdot10^{-3}$ A

Work Step by Step

Assuming that "the potential difference across an inductor is measured along the direction of the current" flow, a.) $\displaystyle \Delta V_L = -L\frac{dI}{dt}\qquad\qquad$ given that $\qquad I(t) = I_0\sin(\omega t)$ $\Delta V_L = -L[I_0\cos(\omega t)(\omega)]$ $\Delta V_L = -I_0L\omega\cos(\omega t)$ b.) The angular frequency $\omega$ can be found by $\omega = 2\pi f$ Also, when we have the maximum voltage, we know that $\cos(\omega t)$ must be a maximum and its maximum value is always 1. $\qquad\qquad\rightarrow \cos_{max}(\omega t) = 1$ Isolating for $I_0$ we get $I_0 = \displaystyle -\frac{\Delta V_L}{L\omega\cos(\omega t)} = -\frac{\Delta V_{L, max}}{L(2\pi f)(1)}$ $I_0 = \displaystyle -\frac{20}{(50\cdot10^{-6})(2\pi\cdot500\cdot10^3)} = -1.27\cdot10^{-3}$ A
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