Magnetic Field from a Current Segment

You may have read about how to find the magnetic field from a very long wire of current. Now, what is the magnetic field from a single segment? Suppose we have the configuration shown below. Your observation point is at the origin, and the segment of current $I$ runs in a straight line from $\langle - L,0,0\rangle$ to $\langle 0, - L,0\rangle$.

Facts

The current in the segment is $I$.
The observation point is at the origin.
The segment stretches from from $\langle - L,0,0\rangle$ to $\langle 0, - L,0\rangle$.

Lacking

$\overset{\rightarrow}{B}$

Approximations & Assumptions

The current is steady, and the wire segment is uniform.

Representations

We represent the Biot-Savart Law for magnetic field from a current as

$$ \overset{\rightarrow}{B} = \int\frac{\mu_{0}}{4\pi}\frac{I \cdot d\overset{\rightarrow}{l} \times \overset{\rightarrow}{r}}{r^{3}} $$

We represent the situation with diagram given above.

Solution

Below, we show a diagram with a lot of pieces of the Biot-Savart Law unpacked. We show an example $\text{d}\overset{\rightarrow}{l}$, and a separation vector $\overset{\rightarrow}{r}$. Notice that $\text{d}\overset{\rightarrow}{l}$ is directed along the segment, in the same direction as the current. The separation vector $\overset{\rightarrow}{r}$ points as always from source to observation.

For now, we write

$$ \text{d}\overset{\rightarrow}{l} = \langle\text{d}x,\text{d}y,0\rangle $$

and

$$ \overset{\rightarrow}{r} = {\overset{\rightarrow}{r}}_{obs} - {\overset{\rightarrow}{r}}_{source} = 0 - \langle x,y,0\rangle = \langle - x, - y,0\rangle $$

Notice that we can rewrite $y$ as $y = - L - x$. This is a little tricky to arrive at, but is necessary to figure out unless you rotate your coordinate axes, which would be an alternative solution to this example. If finding $y$ is troublesome, it may be helpful to rotate. We can take the derivative of both sides to find $\text{d}y = - \text{d}x$. We can now plug in to express $\text{d}\overset{\rightarrow}{l}$ and $\overset{\rightarrow}{r}$ in terms of $x$ and $\text{d}x$:

$$ \text{d}\overset{\rightarrow}{l} = \langle\text{d}x, - \text{d}x,0\rangle $$$$ \overset{\rightarrow}{r} = \langle - x,L + x,0\rangle $$

Now, a couple other quantities that we see will be useful:

$$ \text{d}\overset{\rightarrow}{l} \times \overset{\rightarrow}{r} = \langle 0,0,\text{d}x(L + x) - ( - \text{d}x)( - x)\rangle = \langle 0,0,L\text{d}x\rangle = L\text{d}x\widehat{z} $$$$ r^{3} = (x^{2} + (L + x)^{2})^{3/2} $$

The last thing we need is the bounds on our integral. Our variable of integration is $x$, since we chose to express everything in terms of $x$ and $\text{d}x$. Our segment begins at $x = - L$, and ends at $x = 0$, so these will be the limits on our integral. Below, we write the integral all set up, and then we evaluate using some assistance some Wolfram Alpha.

$$ \begin{matrix} \overset{\rightarrow}{B} & = \int\frac{\mu_{0}}{4\pi}\frac{I \cdot d\overset{\rightarrow}{l} \times \overset{\rightarrow}{r}}{r^{3}} \\ & = \int_{- L}^{0}\frac{\mu_{0}}{4\pi}\frac{IL\text{d}x}{(x^{2} + (L + x)^{2})^{3/2}}\widehat{z} \\ & = \frac{\mu_{0}}{2\pi}\frac{I}{L}\widehat{z} \end{matrix} $$