Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 21 - Electromagnetic Induction and Faraday's Law - Problems - Page 622: 46

Answer

$M= \mu_o \frac{ N_1 N_2 A}{\mathcal{l}}$.

Work Step by Step

Use equation 21–2b to find the emf induced in the short coil. The relevant area is the common cross-sectional area, and the magnetic field B is the field created by the solenoid. The current changing in the solenoid causes the flux in the turns of wire to change, leading to an emf and their mutual inductance. $$\epsilon=-N_2\frac{A\Delta B}{\Delta t}=- N_2\frac{A\Delta (\mu_oN_1I_1)/\mathcal{l}|}{\Delta t}$$ $$= -N_2\frac{ N_1\mu_o A}{\mathcal{l}}\frac{\Delta I_1}{\Delta t}$$ Compare this to equation 21–8a. $$\epsilon =-M\frac{\Delta I_1}{\Delta t}$$ The mutual inductance M is the proportionality constant in the equation. $$M= \mu_o \frac{ N_1 N_2 A}{\mathcal{l}}$$
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