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 - General Problems - Page 624: 91

Answer

$\frac{L_2}{L_1}=2$. The inductance doubles.

Work Step by Step

According to the equation on page 609, the inductance of a solenoid is $$L=\frac{\mu_oN^2A}{\mathcal{l}}$$ Find the ratio of the two inductances. $$\frac{L_2}{L_1}=\frac{\frac{\mu_oN_2^2A_2}{\mathcal{l}_2}}{\frac{\mu_oN_1^2A_1}{\mathcal{l}_1}}=\frac{N^2_2 A_2 \mathcal{l}_1}{ N^2_1 A_1 \mathcal{l}_2}$$ The second solenoid has twice the diameter of the first, meaning the area has quadrupled: $A_2=4A_1$. The circumference of the second solenoid is twice as big, and each coil takes up exactly twice as much wire. The total length of wire is the same, so that means the number of turns is halved: $N_2=0.5N_1$. With only half as many turns, the length of the second solenoid is also half that of the first, $\mathcal{l}_2=0.5\mathcal{l}_1$ $$\frac{L_2}{L_1}=(\frac{N_2}{N_1})^2(\frac{A_2}{A_1})(\frac{\mathcal{l}_1}{\mathcal{l}_2})=(0.5)^2(4)(2)=2$$
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