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FrogPad

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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]. Let's 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)

Lets say we have a section of a current carrying wire that has length [itex] L [/itex]. Let's 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)

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