Inductive reactance and length of the solenoid

AI Thread Summary
The discussion focuses on calculating the inductive reactance and length of a solenoid used as an inductor in a circuit. The solenoid has a radius of 8.0 mm and 170 turns per centimeter, connected to a 15 V rms source at 22 kHz, with an rms current of 0.037 A. The inductive reactance is expressed using the formula XL = 2πfL, while the self-inductance L can be calculated using L = μn²πr²l. The user initially struggled with finding L but ultimately resolved the problem independently. The thread highlights the application of relevant equations in solving inductance-related problems.
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Homework Statement


A solenoid with a radius of 8.0 10-3 m and 170 turns/cm is used as an inductor in a circuit. When the solenoid is connected to a source of 15 V rms at 22 kHz, an rms current of 3.7 10-2 A is measured. Assume the resistance of the solenoid is negligible.
(a) What is the inductive reactance?
(b) What is the length of the solenoid?

r = 8E-3 m
n = 170E2 turns/m
Vrms = 15 V
f = 22E3 Hz
Irms = 3.7E-2 A
Area = pir^2 = 2.01E-4 m^2

Homework Equations


XL = 2pifL

L = μn^2pir^2l

The Attempt at a Solution


I know how to utilize the equations, but my problem is finding L (self-inductance) in order to solve both part a and b. I try looking throughout my physics book, but nothing.

a) XL = 2*pi*22E3*L

b) L = 4piE-7*17000^2*3.14*8E-3*l
 
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Nevermind, I managed to figure the problem out myself.
 

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