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## Biot-Savart

Assume a current going through a straight wire. We know that pieces of magnetite line up in a circle

From magnetic field experiments, we learned

For a circle has radius \(R\), the magnetic field strength \(B\) $$ B \propto \frac{I}{R} \nonumber $$

For electric fields, we know

For a straight wire with uniformly distributed with positive charge, the electric field \(E\) at distance \(R\) is proportional as $$ E \propto \frac{1}{R} \nonumber $$ This is because, according to Coulomb’s law the electric field for electric monopoles (individual charges) falls off as \(1/R^2\). As you integrate that over the length of the wire, you get the \(1/R\) electric field.

Even tough the direction is different, magnetic fields like electric fields fall off as \(1/R\). By analogy, it would be plausible that if magnetic monopoles existed, that the magnetic field would also fall off as \(1/R^2\).

The simple fact that the magnetic field around a current wire falls off as \(1/R\), suggest that if you cut this wire up in little elements \(dl\), that each one of those elements contributes to the magnetic fields in inverse R-squared law. By integrating out over the whole wire, would then get the \(1/R\) fields.

Biot and Savart formalized this

If you have a little current element \(d\vec l\), and the current \(I\) is in the same direction,

Then at distance \(r\), the contribution to the magnetic field \(d\vec B\) is $$ \shaded{ d\vec B = \frac{\mu_0 I}{4 \pi r^2}\,(d\vec l\times \hat r) } \tag{Biot-Savart} \label{eq:BiotSavart} $$

Where

- \(\hat r\) is to get the direction right
- \(\mu_0\) is the permeability of free space \(\approx 1.26\times 10^{-6}\,\rm{H/m}\)

### Examples

#### Straight wire

We can apply Biot-Savart’s law on a straight wire to find the magnetic field at distance \(R\)

Divide the wire in tiny segments \(d\vec l\), at distance \(r\). Then calculate \(dB\), and integrate it over the whole wire.

Using the right-hand rule, \(d\vec l\cdot\hat r\) points out of the page for any element along the wire. Therefore, \(d\vec B\) for all \(d\vec l\) have the same direction. This means we can calculate the net field at \(P\) by evaluating the scalar sum of \(dB\) $$ \left|d\vec l\times\hat r\right| = (dl)(1)\sin\theta = \sin\theta\,dl \nonumber $$

Distance \(r\) and the angle \(\theta\) using trig $$ \left\{ \begin{align*} r &= \sqrt{l^2 + R^2} \\ \sin\theta &= \frac{R}{\sqrt{l^2 + R^2}} \end{align*} \right. $$

Substituting these in \(\eqref{eq:BiotSavart}\) $$ \begin{align*} d\vec B &= \frac{\mu_0\,I}{4\pi r^2}\,\sin\theta\,dl \\ &= \frac{\mu_0\,I}{4\pi(l^2+R)^2}\,\frac{R}{\sqrt{l^2 + R^2}}\,dl \\ &= \frac{\mu_0 I R}{4\pi}\, \frac{1}{(l^2+R^2)^{3/2}}\,dl \\ \Rightarrow B &= \frac{\mu_0 I R}{4\pi} \int_{-\infty}^{\infty}\frac{1}{(l^2+R^2)^{3/2}}\,dl \\ \end{align*} $$

There is symmetry in \(O\) $$ \begin{align*} B &= \frac{\mu_0 I R}{4\pi} 2 \int_0^{\infty}\frac{1}{(l^2+R^2)^{3/2}}\,dl \\ &= \frac{\mu_0 I R}{2\pi} \int_0^{\infty}\frac{1}{(\underline{l}^2+R^2)^{3/2}}\,\underline{dl} \\ \end{align*} $$

Let \(u = \rm{atan}\frac{l}{R}\) $$ \begin{align*} u &= \rm{atan}\frac{l}{R} \\ \Rightarrow \underline{l} &= R\tan u \\ \Rightarrow \frac{dl}{du} &= \frac{d}{dl} R\tan u = R\,\rm{sec}^2 u \\ \Rightarrow \underline{dl} &= R\,\rm{sec}^2 u\,du \end{align*} $$

Substituting \(l = R\tan u\) and \(dl = R\,\rm{sec}^2 u\,du\) in the integral $$ \begin{align*} B &= \frac{\mu_0 I R}{2\pi} \int_0^{\infty}\frac{1}{\left[ (\underline{R\tan u})^2+R^2 \right]^{3/2}}\,\underline{R\,\rm{sec}^2 u\,du} \\ &= \frac{\mu_0 I R}{2\pi} \int_0^{\infty}\frac{R\,\rm{sec}^2\,u}{ R^3\,\left( \tan^2 u+1 \right)^{3/2} }\,du \\ &= \frac{\mu_0 I \bcancel{R^2}}{2\pi R^{\bcancel{3}}} \int_0^{\infty}\frac{\rm{sec}^2\,u}{\rm{sec}^3\,u}\,du = \frac{\mu_0 I}{2\pi R} \int_0^{\infty}\frac{1}{\rm{sec}\,u}\,du \\ &= \frac{\mu_0 I}{2\pi R} \int_0^{\infty}\cos u\,du = \frac{\mu_0 I}{2\pi R} \Big[ \sin u \Big]_0^{u=\infty} \\ \end{align*} $$

Substitute back \(u = \rm{atan}\frac{l}{R}\) $$ \begin{align*} B &= \frac{\mu_0 I}{2\pi R} \Big[ \sin\left(\rm{atan}\frac{l}{R}\right) \Big]_0^{l=\infty} \\ &= \frac{\mu_0 I}{2\pi R} \Big[ \sin\frac{\pi}{2} – \sin 0 \Big]_0^{l=\infty} \\ &= \frac{\mu_0 I}{2\pi R} \end{align*} $$

So, the magnetic field at distance \(R\) is from the wire $$ \shaded{ B = \frac{\mu_0 I}{2\pi R} } \tag{straight wire} \label{eq:straightwire} $$

If you take a radius of \(R=0.1\,\rm{m}\) and you have a current \(I=100\,\rm A\), then $$ \begin{align*} B &= \frac{\mu_0 I}{2\pi R} = \frac{(1.26\times10^{-6}) 100}{2\pi\,0.1} \\ &\approx 2\times10^{-4}\,\rm{T} \end{align*} $$

Compared to the Earth’s magnetic field of \(0.5\times10^{-4}\,\rm T\), it is not that significant. If you would go at a radius of \(R=1\,\rm m\), the field would be 10 times lower, \(2\times10^{-5}\,\rm T\). Then the magnetic field of the Earth already dominates substantially.

#### Circular wire loop

##### At center

We can apply Biot-Savart’s law on a current loop to find the magnetic field at the center

Divide the wire in tiny segments \(d\vec l\). Using the right-hand rule, \(d\vec l\cdot\hat r\) at the center points up for any element along the wire. Since \(d\vec l\perp \hat r\) $$ d\vec l\times\hat r = dl.1.\sin\frac{\pi}{2} = dl \nonumber $$

Substituting these in \(\eqref{eq:BiotSavart}\) $$ \begin{align*} d\vec B &= \frac{\mu_0\,I}{4\pi R^2}\,dl \\ \Rightarrow B &= \frac{\mu_0\,I}{4\pi R^2}\int_\rm{circle}\,dl \\ &= \frac{\mu_0\,I}{4\bcancel\pi R^{\bcancel 2}}\,2\bcancel\pi \bcancel{R} \\ &= \frac{\mu_oI}{2R} \end{align*} $$

So, the magnetic field at the center of a wire loop is $$ \shaded{ \vec B = \frac{\mu_oI}{2R}\,\hat z } \tag{wire loop center} \label{eq:wireloopcenter} $$

When you very far away, the electric fied configuration is very similar to that of a electric dipole.

##### Along axis

We may extend this approach to calculate the field anywhere along the center (\z\)-axis of the loop.

Based on the cylindrical symmetry, any outward component of \(d\vec B\) will be exactly balanced by an opposite-directed outward component from an opposing wire fragment. The magnetic field of the wire loop as a whole only has a \(z\)-component. In calculating the total field, we only need to integrate the z-components. $$ \begin{align*} d\vec l\times\hat r &= dl.1.\sin\theta, & \left( \sin\theta=\frac{R}{r} \right) \\ &= \frac{R}{r}\,dl & \end{align*} $$

Substituting this in \(\eqref{eq:BiotSavart}\) $$ \begin{align*} d\vec B &= \frac{\mu_0\,I}{4\pi (z^2 + R^2)}\, \frac{R}{r}\,dl, & \left(r=\sqrt{z^2+R^2}\right) \\ &= \frac{\mu_0\,I}{4\pi (z^2 + R^2)}\, \frac{R}{\sqrt{z^2 + R^2}}\,dl \\ &= \frac{\mu_0\,I\,R}{4\pi (z^2 + R^2)^{3/2}}\, dl \\ \Rightarrow B &= \frac{\mu_0\,I\,R}{4\pi (z^2 + R^2)^{3/2}}\, \int_\rm{circle} dl, & \left( \int_\rm{circle} dl = 2\pi R\right) \\ &= \frac{\mu_0\,I\,R}{\bcancel{4}_2\bcancel{\pi} (z^2 + R^2)^{3/2}}\, \bcancel{2}\bcancel{\pi} R \end{align*} $$

So, the magnetic field along the center axis of a wire loop is $$ \shaded{ \vec B = \frac{\mu_0\,I\,R^2}{2 (z^2 + R^2)^{3/2}} \hat z } \tag{wire loop axis} \label{eq:wireloopaxis} $$

##### Field lines

We would expect the magnetic field lines as

## Gauss’s law for magnetism

Looking from far away, ignoring what happens between the charges: the loops of electric field lines from a dipole, as similar to the magnetic field lines from the current loop

Gauss’ law tells us that the closed surface \(S\) integral of the electric flux is the charge inside the box divided by \(\varepsilon_0\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \iint_S \vec E \cdot d\vec A = \frac{\sum_i Q_i}{\varepsilon_0} \nonumber $$

For the electric field, there is clearly a charge inside the box.

No matter where in the magnetic field you make a closed surface \(S\), there is never any magnetic flux going through that surface. Because there (as far as we know) are no magnetic monopoles.

This brings us to the second of four Maxwell’s equations: **Gauss’s law for magnetism**
$$
\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint}
\shaded{
\oiint_S \vec B \cdot d\vec A = 0
}
\tag{Gauss’s law}
$$