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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 30 / Problem 3E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Solved: A 10.0-cm-long solenoid of diameter 0.400 cm is

Chapter 30, Problem 3E

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QUESTION:

Problem 3E

A 10.0-cm-long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

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QUESTION:

Problem 3E

A 10.0-cm-long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

ANSWER:

Solution 3E

A 10.0-cm-long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center.

Step 1 of 2:

A coil is wound around a solenoid, so magnetic flux from the solenoid passes through the coil.

The magnetic field inside the solenoid can be expressed in terms of number of turns per unit length ${n}_{1}=\frac{{{N}_{1}}_{\ }^{\ }}{l}$

${B}_{1}$ = ${\mu{}}_{o}{n}_{1}{i}_{1}$ = $\frac{{\mu{}}_{o}{N}_{1}{i}_{1}}{l}$

The flux through a cross section of the solenoid equals ${B}_{1}A$ . As we mentioned above, this also equals the flux ${\Phi{}}_{B2}$ through each turn of the outer second coil, independent of its cross-sectional area.

The mutual induction $M$ is then given by

$M$ = $\frac{{N}_{2\ }{\Phi{}}_{B2}}{{i}_{1}}$ = $\frac{{N}_{2\ }\ {B}_{1}A}{{i}_{1}}$ = $\frac{{N}_{2\ }}{{i}_{1}}$ $\frac{{\mu{}}_{o}{N}_{1}{i}_{1}}{l}$ A = $\frac{{\mu{}}_{o}A{N}_{1}{N}_{2}}{l}$

Where ${N}_{1}$ and ${N}_{2}$ are number of turns in solenoid and second coil respectively.

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