Magnetic Field of current carrying straight wire

[FrogPad]

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)

 

[Meir Achuz]

Just use the law of Biot-Savart.

 

[FrogPad]

Just use the law of Biot-Savart.

Does that make the math any easier? I'll see what I can do with that, but it seems like it would be easier to do this way. Those integrals are lengthy though (I let Maple solve the more complex one).

 

[FrogPad]

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:

$$\vec B = \hat \phi \frac{\mu_0 I}{4 \pi \bar r} (\alpha + \beta)$$
$$\alpha = \frac{L - \bar z}{\sqrt{\bar r^2 + (\bar z - L)^2}}$$
$$\alpha = \frac{\bar z}{\sqrt{\bar r^2 + \bar z^2}}$$

Where the line is from 0 to L, and the point is located at [itex]P(\bar r, \bar phi, \bar z [/tex].<br /> <br /> 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.[/itex]