Magnetic Field of a Long Straight Wire: Difference between revisions

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Integrating this, we get the expression <math> B = \frac{\mu_0}{4\pi} \frac{LI}{z(\sqrt{z^2+(L/2)^2})} </math>
Integrating this, we get the expression <math> B = \frac{\mu_0}{4\pi} \frac{LI}{z(\sqrt{z^2+(L/2)^2})} </math>
==Direction of Magnetic Field==
If you are simply interested in finding the direction of the magnetic field, all you have to do is use the right hand rule. Point your right thumb in the direction of the current, and your hand will curl in the direction of the magnetic field. So, for this situation, we point our thumb in the y direction and, at a point on the +z axis, we can see that our fingers curl right, or towards the +x direction.
==References==
Matter and Interactions Vol. II

Revision as of 22:11, 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 [math]\displaystyle{ \vec{r} }[/math] 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{ \sqrt{(z^2+y^2)} }[/math]. As a result, your [math]\displaystyle{ \hat{r} = \frac{(0,-y,z)}{\sqrt{(z^2+y^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} = \Delta{y} (0,1,0) }[/math].

Now that we have everything we need, we can plug it into the equation and evaluate the cross product. As a result we get [math]\displaystyle{ \Delta \vec{B} =\frac{\mu_0}{4\pi} \frac{I \Delta {y}}{(z^2+y^2)^{3/2}} (z,0,0) }[/math]

The final step is to integrate this. Since it is centered at the origin, we have to integrate from -L/2 to L/2. So our equation looks like [math]\displaystyle{ \int\limits_{-L/2}^{L/2}\ \Delta \vec{B} =\frac{\mu_0}{4\pi} \frac{I}{(z^2+y^2)^{3/2}} (z,0,0) \delta {y} }[/math]

Integrating this, we get the expression [math]\displaystyle{ B = \frac{\mu_0}{4\pi} \frac{LI}{z(\sqrt{z^2+(L/2)^2})} }[/math]

Direction of Magnetic Field

If you are simply interested in finding the direction of the magnetic field, all you have to do is use the right hand rule. Point your right thumb in the direction of the current, and your hand will curl in the direction of the magnetic field. So, for this situation, we point our thumb in the y direction and, at a point on the +z axis, we can see that our fingers curl right, or towards the +x direction.

References

Matter and Interactions Vol. II