Magnetic Field of a Long Straight Wire

Claimed by Katharine Bernart (Spring 2017)

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]

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]

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

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]

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.

References

[1] Matter and Interactions Vol. II

[2] OpenStax University Physics

This page was created by Arjun Patra