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Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553278

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Chapter
Section
BuyFindarrow_forward

Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553278
Chapter 31, Problem 10P
Textbook Problem
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A 510-turn solenoid has a radius of 8.00 mm and an overall length of 14.0 cm. (a) What is its inductance? (b) If the solenoid is connected in series with a 2.50-Ω resistor and a battery, what is the time constant of the circuit?

(a)

To determine
The inductance of the solenoid.

Explanation of Solution

Given info: The total number of turns in the solenoid is 510 , the radius of the solenoid 8.00mm and the complete length of the solenoid is 14.0cm .

Formula to calculate the inductance is,

L=μ0N2Al . (1)

Here,

L is the inductance of the solenoid.

μ0 is the free space permittivity.

N is the total number of turns.

A is the cross sectional area of the solenoid.

l is the length of the solenoid.

Formula to calculate cross sectional area is,

A=πr2

Here,

r is the radius of the solenoid.

Substitute 4π×107 for μ0 , 510 for N , 14cm for l and πr2 for A in equation (1),

L=μ0N2Al=4π×107×(510)2×πr214

(b)

To determine
The time constant of the circuit when the battery and a resistor is connected.

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Chapter 31 Solutions

Physics for Scientists and Engineers
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