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Magnetic Field of current carrying straight wire |
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| Nov10-06, 11:02 PM | #1 |
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Magnetic Field of current carrying straight wire
So I get a "cheat" sheet for my upcoming emag test. I would like to have a general expression for the magnetic field of a current carrying wire. Would someone let me know if I am on the right path here.
Lets say we have a section of a current carrying wire that has length [itex] L [/itex]. Lets say there is a point P that is located at [itex] P(\bar r, \bar \phi, \bar z ) [/itex] We will use cylindrical coordinates and denote the bottom of the wire as [itex]0[/itex], and the top of the wire as [itex] L [/itex]. Since we are in cylindrical coordinates, there will be no phi dependence, so the point can be expressed as: [itex] P(\bar r, 0, \bar z) [/itex] Thus, is my thought process correct here (I don't want to solve these integrals yet, if I am doing something wrong). Recall: [tex] \vec A = \frac{\mu_0 I}{4 \pi} \oint_{C'} \frac{\vec dl'}{R} [/tex] Thus, if we break the integral into two contours, [tex] \vec A = \frac{\mu_0 I}{4 \pi} \left( \int_{C'_1} \frac{\vec dl'}{R_1} + \int_{C'_2} \frac{\vec dl'}{R_2} \right) [/tex] [tex] \int_{C'_1} \frac{\vec dl'}{R_1} = \int_{0}^{\bar z} \frac{\hat z dz'}{\sqrt{z'^2+\bar r^2}} [/tex] [tex] \int_{C'_2} \frac{\vec dl'}{R_2} = \int_{\bar z}^{L} \frac{\hat z dz'}{\sqrt{[(L-\bar z)-z']^2+\bar r^2}} [/tex] Now if I solve these two integrals and plug into [itex] \vec A [/itex] and then get [itex] \vec B [/itex] by [itex] \vec B = \nabla \times \vec A [/itex] I should be all set right? (...I hope) |
| Nov11-06, 06:53 PM | #2 |
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Recognitions:
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Just use the law of Biot-Savart.
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| Nov11-06, 08:37 PM | #3 |
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| Nov12-06, 02:20 PM | #4 |
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Magnetic Field of current carrying straight wire
Ok, I used the Biot-Savart law, and it was surprisingly easier. The derivation was long, so I will not post it unless someone wants to see it. I came up with the following expression though:
[tex] \vec B = \hat \phi \frac{\mu_0 I}{4 \pi \bar r} (\alpha + \beta) [/tex] [tex] \alpha = \frac{L - \bar z}{\sqrt{\bar r^2 + (\bar z - L)^2}} [/tex] [tex] \alpha = \frac{\bar z}{\sqrt{\bar r^2 + \bar z^2}} [/tex] Where the line is from 0 to L, and the point is located at [itex] P(\bar r, \bar phi, \bar z [/tex]. I used the bar notation to represent constants. This expression could (and probably should) be cleaned up. Maybe taylor expand the sqrt expressions, or apply some type of simplification. I don't think it's beneficial for me to spend so much time on one problem, so I'm just going to drop it for the time being. Thanks for the suggestion Meir Achuz. |
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