Given: 0 = 4 × 107 Tm/A. An air-core circular solenoid is shown in the figure be
ID: 2016396 • Letter: G
Question
Given: 0 = 4 × 107 Tm/A.
An air-core circular solenoid is shown in the
figure below. A current of 15 A is established
in the wire which makes up this solenoid.
The permeability of free space is 0 = 4 × 107 T·m/A.
I can't seem to get the diagram to upload. The length is 94cm, radius is 5cm. Current is 15A and it has a total of 517 turns. I calculate the magnetic field at the center to be .0103673 T(and it was correct) but now I can't figure out how to calculate the self inductance in mH of the energy stored (J). I am not sure if I am using the wrong formula or what. HELP!!
Explanation / Answer
Length of the wire is l = 94.0cm radius of the wire is r = 5.00cm current in the wire is i = 15.0A Number of turns is N = 517 self inductance coil is L = _0 N^2 A / l = 4*10^-7* ( 517)^2 * ( 5.00*10^-2 m)^2 / 94.0*10^-2 m = 2.80*10^-3 H = 2.80 mH b ) Energy stored in an inductor U = 1/2 L I^2 = 1/2 *2.80*10^-3 H * ( 15 A)^2 = 0.315 JRelated Questions
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