Biot-Savart Law for Currents

Vaishnavi Ramanan - Fall 2018

Biot-Savart Law

The Biot-Savart Law can be used for more than just single moving charges; it can also be used to calculate the magnetic field for a large number of charges. One notable reason to do so is to find the magnetic field of a portion of a wire where there can be many moving charges. When we use Biot-Savart Law to find the magnetic field of a short wire, we can apply it to a variety of shapes.

A Mathematical Model

First We start off with the original version of the Biot-Savart Law. [math]\displaystyle{ \vec B=\frac{\mu_0}{4 \pi } \frac{q\vec v\times\hat r}{r^2} }[/math], where [math]\displaystyle{ \frac{\mu_0}{4 \pi } = 1 \times 10^{-7}\frac{Tm^2}{Cm/s}, }[/math]

Because we are dealing with a portion of wire [math]\displaystyle{ \mathrm{d}\boldsymbol{\ell} }[/math] long with an Area A containing n moving particles with charge q, we find that the total number of moving charges is equal to |q|(nAv) which is also equal to I, the current in the wire. [math]\displaystyle{ B = \frac{\mu_0I}{4\pi}\frac{\mathrm{d}\boldsymbol{\ell} \times \mathbf{\hat r}}{r^2}, }[/math]

Because the shape of the current carrying wire can vary from a straight wire to a loop, we must integrate over the region of the wire.

[math]\displaystyle{ \Delta B = \frac{\mu_0I}{4\pi}\int_{\mathrm{wire}}\frac{\mathrm{d}\boldsymbol{\ell} \times \mathbf{\hat r}}{r^2}, }[/math]

The key point is that there are [math]\displaystyle{ nA\Delta l }[/math] electrons in a short length of wire, each moving with average speed [math]\displaystyle{ \vec v }[/math], so that the sum of all the [math]\displaystyle{ q\vec v }[/math] contributions is [math]\displaystyle{ nA \Delta l|q|\vec v = I\Delta l. }[/math]

When applying the Biot-Savart Law to a Long Straight Wire, we follow a set of steps,

Step 1: Cut Up the Distribution into Pieces and Draw [math]\displaystyle{ \Delta B }[/math].

Step 2: Write an Expression for the Magnetic Field Due to One Piece.

Step 3: Add Up the Contributions of All the Pieces.

Step 4: Check the Result.

Where the Magnetic Field of a Straight Wire is shown by,

[math]\displaystyle{ B = \frac{\mu_0}{4\pi}\frac{LI}{r(r^2 + (L/2)^2)^{1/2}} }[/math] for length [math]\displaystyle{ L }[/math], conventional current [math]\displaystyle{ I }[/math], a perpendicular distance [math]\displaystyle{ r }[/math] from the center of the wire, or, [math]\displaystyle{ B = \frac{\mu_0}{4\pi}\frac{2I}{r} }[/math] if [math]\displaystyle{ L\;\gt \;\gt r. }[/math]

When using the Biot-Savart Law for Currents, it is crucial to understand the direction of the magnetic field created by a current. For this, we use the right hand rule. If we curl our fingers and extend our thumb, similar to a thumbs up position, and point our thumb in the direction of the current, our fingers curl in the direction of the magnetic field.

Example

For a long wire of length L positioned along the x axis with current flowing in the positive x direction

First, we start off with our adjusted Biot-Savart Formula for a slice of wire.

[math]\displaystyle{ \Delta B = \frac{\mu_0I}{4\pi}\int_{\mathrm{wire}}\frac{\mathrm{d}\boldsymbol{\ell} \times \mathbf{\hat r}}{r^2}, }[/math]

Second, we must find [math]\displaystyle{ r }[/math], the vector pointing from the source to the observation location. In this case, we will choose an observation location y above the rod.

[math]\displaystyle{ r = obs - source = \lt 0,y,0\gt - \lt x,0,0\gt = \lt -x,y,0\gt }[/math]. which has a magnitude of [math]\displaystyle{ \sqrt(x^2+y^2) }[/math]

We see that [math]\displaystyle{ \hat r = \frac{r}{|r|} }[/math] .

[math]\displaystyle{ \hat r = \frac{\lt -x,y,0\gt }{\sqrt(x^2+y^2))} }[/math]

We then have to express [math]\displaystyle{ \Delta \boldsymbol{\ell} }[/math] in terms of our variable of integration, x. [math]\displaystyle{ \Delta \boldsymbol{\ell} }[/math] = [math]\displaystyle{ \Delta x\lt 1,0,0,\gt }[/math]

Our new equation after substituting our new variables is [math]\displaystyle{ \Delta B = \frac{\mu_0I\Delta x\lt 1,0,0,\gt }{4\pi(x^2+y^2)} \times \frac{\lt -x,y,0\gt }{\sqrt(x^2+y^2))} }[/math]

Finding the cross product of the above vectors gives us a product in the +z direction. [math]\displaystyle{ \Delta B = \frac{\mu_0I\Delta x}{4\pi(x^2+y^2)^(\frac{3}{2})} }[/math]

We are finally ready to integrate. Because we are integrating the entire rod our limits are [math]\displaystyle{ \int\limits_{-L/2}^{L/2}\ }[/math]

[math]\displaystyle{ \int\limits_{-L/2}^{L/2}\ = \frac{\mu_0I\Delta x}{4\pi(x^2+y^2)^(\frac{3}{2})} }[/math]

We find that our final answer is [math]\displaystyle{ B= \frac{\mu_0}{4\pi}\frac{LI}{y\sqrt(y^2+(L/2)^2)}\hat z }[/math]

A Computational Model

The following link shows the magnetic field produced by small segments of wire in a loop individually. For a long straight wire, we see that there is a circular magnetic field surrounding the wire with current. The following link does a stepwise visual of the contributions of each part of the wire at an observation location a distance r from the wire.

http://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/17-Bwire-with-r

We see that along the axis of the wire, each contribution not on the axis is negated due to symmetry and the resulting magnetic field is all along the wire.

http://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/17-B-loop-with-r-dB.

See also

[Right Hand Rule[1]]

[Direction of magnetic fields[2]] Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?