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hiigaranace
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Homework Statement
A long, insulated wire with a resistance of 18.6 ohms/m is to be used to construct a resistor. First, the wire is bent in half, and then the doubled wire is wound in a cylindrical form as shown in the above figure. The diameter of the cylindrical form is 2.6 cm, its length is 23 cm, and the total length of wire is 8.3 m. Find the inductance of this wire-wound resistor.
Homework Equations
L = [tex]\frac{phi}{I}[/tex]
L = [tex]\mu_{0}[/tex]n2(Area)(length)
The Attempt at a Solution
First, find the number of loops per unit length:
circumference = pi * .026cm = .0816814
Total loops = total length / circumference = 8.3 / .0817 = 101.614 turns
Loops/m = Loops / length = 101.614 / .23 = 441.80
Now, plug that into the equation:
[tex]\mu[/tex]n2 * A * l
4[tex]\pi[/tex]E-7 * 441.8012 * ([tex]\pi[/tex]* .0132) * .23
= 2.995E-5
Not entirely sure what I'm doing wrong here...
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