Magnetic Field of a Long Straight Wire

In many cases, we are interested in calculating the electric field of a long, straight wire. Wires can create magnetic fields if they have a current flowing through them. If no current is flowing, then there will be no magnetic fields created. Below are steps that explain the derivation of the formula for calculating the magnetic field as well as how to calculate the direction of magnetic field. [1]

Calculation of Magnetic Field

Derivation

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] [1]

Approximation

If you know that L>>r, then you know that [math]\displaystyle{ \sqrt{r^2+(L/2)^2} = L/2 }[/math]. Therefore, if you have a really long wire and you are trying to find the magnetic field of a point relatively close to the rod, you can use the approximation [math]\displaystyle{ B = \frac{\mu_0}{4\pi} \frac{2I}{r} }[/math] [1]

In most situations, at least in the scope of this course, it is stated in the question whether or not you can use the approximation or not. This can come in a few different forms: "assume r<<L," "assume the length of the wire is much longer than than the distance from the wire to the observation location, etc.". However, if it is not explicitly stated, a good rule of thumb is that if the length of the wire is 100 times+ the r, we can use the approximation formula. Otherwise, it is smart to just play it safe and use the full formula.

Application with a Compass

As you have already seen in lab (or you will see soon, depending on how far along the semester you are), if you run a current through a wire and place a compass near it, the needle of the compass deflects due to the magnetic field from this wire. In fact, the amount of deflection that there is from the wire can actually help you calculate the approximate magnetic field of the wire. This is because there is a particular relationship between the needle of the compass, the magnetic field of the wire, and the magnetic field of the Earth itself that looks like this:

[math]\displaystyle{ |B_w| = |B_E|*tan(theta) }[/math]

The Earth itself has a magnetic field anywhere on the Earth (obviously based off location it varies slightly but it is easier to assume so) of approximately 2e^-5 T. We ourselves normally do not notice this by looking at an ordinary compass, but it is actually the reason that a compass' needle points north! In the formula above, theta is the amount of degrees that, when placed directly above or underneath, the compass needle deflects from North. This is a very useful formula to have in lab or in the real world if it is unknown how much current is running through a wire.

Examples

Easy

The magnitude of the magnetic field 50 cm from a long, thin, straight wire is 8.0μT. What is the current through the long wire? [2]

Use the approximation formula.

[math]\displaystyle{ B = \frac{\mu_0}{4\pi} \frac{2I}{0.5m} }[/math]

[math]\displaystyle{ I = B \frac{4\pi}{\mu_0} }[/math]

[math]\displaystyle{ I = 20A }[/math]

An ordinary compass that points north when not in the vicinity of a magnetic field is pointing at 325 degrees. A wire with a current is then placed directly above the compass and the needle now points at 240 degrees. What is the magnitude of the magnetic field of this wire?

[math]\displaystyle{ |B_w| = |B_E| * tan(theta) }[/math]

[math]\displaystyle{ theta = 240 - 325 = -85 degrees }[/math]

[math]\displaystyle{ |B_w| = (2e-5) * tan(-85) }[/math]

[math]\displaystyle{ |B_w| = 2.29e-4 T }[/math]

Medium

The current of a thin, straight wire 2m long is 74A. What is the magnitude of the magnetic field at a location 0.35m away and perpendicular to the center of the wire?

Use the full formula.

[math]\displaystyle{ B = \frac{\mu_0}{4\pi} \frac{LI}{z(\sqrt{z^2+(L/2)^2})} }[/math]

[math]\displaystyle{ B = \frac{\mu_0}{4\pi} \frac{2 * 74}{0.35(\sqrt{0.35^2+(2/2)^2})} }[/math]

[math]\displaystyle{ B = 39.912 }[/math]μT

Hard

There are two wires, separated by a distance of 80 meters on the x-axis. The left wire has a current running through it of 5 A, while the right wire has a current running through it of 12 A. The length of the left wire is 2 meters, while the length of the right wire is 3 meters. Find the total magnetic field at a point on the x-axis directly in between the two wires.

Use the full formula.

[math]\displaystyle{ B_l = \frac{\mu_0}{4\pi} \frac{LI}{z(\sqrt{z^2+(L/2)^2})} }[/math]

[math]\displaystyle{ B_l = \frac{\mu_0}{4\pi} \frac{2*5}{40(\sqrt{40^2+(2/2)^2})} }[/math]

[math]\displaystyle{ B_l = 6.25e-10 }[/math] T

[math]\displaystyle{ B_r = \frac{\mu_0}{4\pi} \frac{LI}{z(\sqrt{z^2+(L/2)^2})} }[/math]

[math]\displaystyle{ B_r = \frac{\mu_0}{4\pi} \frac{3*12}{-40(\sqrt{(-40)^2+(2/2)^2})} }[/math]

[math]\displaystyle{ B_r = -2.25e-9 }[/math] T

[math]\displaystyle{ B_n = B_left + B_right }[/math]

[math]\displaystyle{ B_n = 6.25e-10 + (-2.25e-9) }[/math]

[math]\displaystyle{ B_n = -1.62e-9 }[/math] T

There are two wires, pictured below, both with current of 10 A. Find the magnitude and direction of the net magnetic field at the point P.

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. [1]

Example 1:

A wire sits directly in the z-axis with current flowing out of the page (positive z direction). In what direction is the magnetic field at a point to the right of the wire?

Solution:

If you point your thumb in the direction of the current, you can see that on the right side of your curled fingers, your fingers curl directly up, so the direction of the magnetic field is the positive y direction.

Connectedness

I am very interested in clean energy storage and production, which typically involves long wires at some point between where the energy is generated and where electricity is used. It is important to understand all the forces involved with an electrical current, so that if something goes wrong, you can determine where the problem is and why it might be occurring so that you can fix it. [1]

I am an electrical engineer major, so all of the material within this class is vastly important, not only for the following courses required for a EE major, but also for the field once we graduate and find a job. However, this concept specifically interests me because for a while I had a very difficult time finding the direction of the magnetic field in situations such as these, so this is my way of giving back in an effort to make sure future students don't run into the same problem. [2]

History

The magnetic field of a wire was first discovered during an experiment by Hans Christian Oersted (1777-1851) of Denmark in 1820. There are many rumors of this story that this discovery was actually an accident, but Oersted claims that it was, indeed, based off predictions that he had made beforehand. This experiment consisted of running a current through a wire and placing a compass underneath it to see if there was any effect. And the effect he found changed the world forever: he had discovered the important relationship between electricity and magnetism. Before this, the world had taken note of the similarities between electricity and magnetism but nobody had truly "proved" this relationship up until this point. Oersted then went on to write his groundbreaking scientific paper "Experiments on the effect of a current of electricity on the magnetic needle," which shocked and awed the rest of the scientific world. This was the birth of physics 2. If it had not been for Oersted, we might not be taking this very class at Georgia Tech! While this finding falls directly into the category of "Magnetic Field of a Long Straight Wire," it also may very well be the most important discovery by any physicist in history (this is up for debate but this is just my opinion on the topic; nonetheless, it is extremely crucial). To wrap up the story, the content of Oersted's paper was marvelous . . . and, thus, Emag was born!

References

[1] Matter and Interactions Vol. II

[2] OpenStax University Physics

[3] Skulls in the Stars

This page was created by Arjun Patra

This page was edited by Brandon Baker