Magnetic Field of a Long Straight Wire: Difference between revisions

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==Calculation of Magnetic Field==
==Calculation of Magnetic Field==


Imagine centering a wire on the y-axis and having a current run through the wire in the +y direction. We are interested in finding the magnetic field at some point along the z axis, say (0,0,z).
Imagine centering a wire on the y-axis and having a current run through the wire in the +y direction. We are interested in finding the magnetic field at some point along the z axis, say <math> (0,0,z) </math>.


From here, it is an integral problem where you take an arbitrary piece of the rod and plug it into the generic formula for change in magnetic field: <math> \vec{B}  =\frac{\mu_0}{4\pi} \frac{I(\vec{l} \times \hat{r})}{y^2+z^2}  </math>
From here, it is an integral problem where you take an arbitrary piece of the rod and plug it into the generic formula for change in magnetic field: <math> \vec{B}  =\frac{\mu_0}{4\pi} \frac{I(\vec{l} \times \hat{r})}{y^2+z^2}  </math>

Revision as of 16:39, 4 December 2015

In many cases, we are interested in calculating the electric field of a long, straight wire. -Claimed by Arjun Patra


Calculation of Magnetic Field

Imagine centering a wire on the y-axis and having a current run through the wire in the +y direction. We are interested in finding the magnetic field at some point along the z axis, say [math]\displaystyle{ (0,0,z) }[/math].

From here, it is an integral problem where you take an arbitrary piece of the rod and plug it into the generic formula for change in magnetic field: [math]\displaystyle{ \vec{B} =\frac{\mu_0}{4\pi} \frac{I(\vec{l} \times \hat{r})}{y^2+z^2} }[/math]

First, you can find the [math]\displaystyle{ \hat{r} }[/math]. The directional vector r is equal to [math]\displaystyle{ (0,0,z) - (0,y,0) = (0,-y,z) }[/math]. You get this by doing final position - initial position. Next, you can find the magnitude of r, and you will get [math]\displaystyle{ (x^2+y^2)^{3/2} }[/math]. As a result, your [math]\displaystyle{ \hat{r} = \frac{(0,-y,z)}{(x^2+y^2)^{3/2}} }[/math]

The last thing we need to calculate is the [math]\displaystyle{ \Delta \vec{L} }[/math]. This is nothing more than a unit vector that tells us what direction the current is flowing. Since we know that the current is flowing in the +y axis, our [math]\displaystyle{ \Delta \vec{L} = (0,1,0) }[/math].