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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 L. Let's say there is a point P that is located at P(\bar r, \bar \phi, \bar z )
We will use cylindrical coordinates and denote the bottom of the wire as 0, and the top of the wire as L. Since we are in cylindrical coordinates, there will be no phi dependence, so the point can be expressed as: P(\bar r, 0, \bar z)
Thus, is my thought process correct here (I don't want to solve these integrals yet, if I am doing something wrong).
Recall:
\vec A = \frac{\mu_0 I}{4 \pi} \oint_{C'} \frac{\vec dl'}{R}
Thus, if we break the integral into two contours,
\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)
\int_{C'_1} \frac{\vec dl'}{R_1} = \int_{0}^{\bar z} \frac{\hat z dz'}{\sqrt{z'^2+\bar r^2}}
\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}}
Now if I solve these two integrals and plug into \vec A and then get \vec B by \vec B = \nabla \times \vec A I should be all set right? (...I hope)
Lets say we have a section of a current carrying wire that has length L. Let's say there is a point P that is located at P(\bar r, \bar \phi, \bar z )
We will use cylindrical coordinates and denote the bottom of the wire as 0, and the top of the wire as L. Since we are in cylindrical coordinates, there will be no phi dependence, so the point can be expressed as: P(\bar r, 0, \bar z)
Thus, is my thought process correct here (I don't want to solve these integrals yet, if I am doing something wrong).
Recall:
\vec A = \frac{\mu_0 I}{4 \pi} \oint_{C'} \frac{\vec dl'}{R}
Thus, if we break the integral into two contours,
\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)
\int_{C'_1} \frac{\vec dl'}{R_1} = \int_{0}^{\bar z} \frac{\hat z dz'}{\sqrt{z'^2+\bar r^2}}
\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}}
Now if I solve these two integrals and plug into \vec A and then get \vec B by \vec B = \nabla \times \vec A I should be all set right? (...I hope)
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