## Biot-Savart law; Gauss’s law for magnetism



My notes of the excellent lecture 14 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

## 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 $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \oiint_S \vec B \cdot d\vec A = 0 } \tag{Gauss’s law}$$

## Gauss’ law



My notes of the excellent lectures 3 and 12 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

Gauss’ law relates the electric field through a closed surface with the charge contained within that surface.

## Electric Flux

Electric flux is the amount of electric field passing through a surface.

### Open surface

Bring an open surface in an electric field. The normal vector $$\hat n$$ is perpendicular to the tiny surface $$dA$$ and chosen to be upwards. The electric field at $$dA$$ is $$\vec E$$

The electric flux $$d\phi$$ that goes through $$dA$$ is defined as the scalar \shaded{ \begin{align*} d\phi = \vec E \cdot \hat n\, dA = \vec E \cdot d\vec A = E\,dA\,\cos\theta \end{align*} } \nonumber

The flux for the whole surface $$A$$ follows from the $$\int dA$$.

### Closed surface

By convention, the normal $$\hat n$$ of a closed surface is from the inside to the outside of the surface.

The total flux through region $$R$$ follows from the integral $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \phi = \iint_R \vec E \cdot d\vec A } \quad \left[\frac{Nm^2}{C}\right] \tag{flux}$$

Notes

• The circle in the integral sign reminds us that it is a closed surface.
• If the flux is $$0$$, then what flows in also flows out.
• If more flows out than in, the flux is positive.

#### Example

A point charge $$+Q$$ in a sphere of radius $$R$$

On the sphere, the flux is the same everywhere. Both $$d\vec A$$ and $$\vec E$$ are radial ($$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} d\vec A\parallelsum\vec E$$), so the dot-product becomes a multiplication of scalars $$|\vec E|\,|d\vec dA|\,\cos 0=E.dA.1$$ $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \phi = \oiint \vec E \cdot dA = E \underbrace{\oiint\,dA}_\text{surface} = 4\pi R^2 \, E \nonumber$$

The electric field $$E_R$$ at distance $$R$$ $$\vec E = \frac{Q}{4\pi\varepsilon_0 R^2}\,\hat r \nonumber$$

Substituting $$E_R$$ in the expression for $$\phi$$ $$\phi = \bcancel{4\pi R^2} \, \frac{Q}{\bcancel{4\pi R^2}\,\varepsilon_0} = \frac{Q}{\varepsilon_0} \nonumber$$

Note:

• This flux $$\phi$$ is independent of the distance $$R$$. This is not surprise, because if you think of it as air flowing out, then all the air has to come out somehow no matter how big the sphere is. This implies that we could have taken any kind of surface around the point charge, and have found the same result.
• The same relation holds for any number of charges inside the surface.

## Gauss’ law

Gauss’ law relates the electric field through a closed surface with the charge contained within that surface.

### Definition

The law was first formulated by Joseph-Louis Lagrange, an Italian mathematician and astronomer (1773), followed by Carl Friedrich Gauss, a German mathematician and physicist (1835).

This brings us to Gauss’ law, also known as Maxwell’s first equation $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \phi = \oiint_S \vec E \cdot d\vec A = \frac{\sum_i Q_i}{\varepsilon_0} } \tag{Gauss’ law} \label{eq:gauss}$$

Where

• $$Q_i$$ is any charge inside any closed surface $$S$$
• $$\varepsilon_0$$ is the permittivity of free space.

Note

• If the flux is $$0$$, it means there is no net charge inside the surface.
• Gauss’ law always holds, no matter how weird the charge distribution inside the surface is, or how weird the shape of the surface.
• Both Coulomb’s and Gauss’ law make the electric field with a charge $$Q$$.

Gauss’ law holds, but will not help you calculate $$\vec E$$ when the charges are not distributed symmetrically. Here we will use spherical symmetry, cylindrical symmetry and flat planes with uniform charge distributions.

### Example: using spherical symmetry

Consider a sphere with radius $$R$$ and charge $$Q$$ uniformly distributed over the surface. We want to know the electric field at distance $$r$$ inside and outside the sphere.

The key is choosing the Gaussian surface right. We choose a sphere with radius $$r$$.

Symmetry arguments

1. Because of the symmetry the electric field is the same anywhere on the Gaussian surface.
2. The electric field is either pointing radially outwards or radially inwards $$\Longrightarrow$$ $$d\vec A$$ and $$d \vec E$$ are parallel or anti-parallel
3. $$\Longrightarrow$$ so the dot-product becomes a multiplication of scalars.

The surface area of the Gaussian surface is $$4\pi r^2$$.

#### Inside the charged sphere, for $$r\lt R$$

The enclosed charge is $$0$$. Applying equation $$\eqref{eq:gauss}$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} \\ \Rightarrow E \oiint_S dA &= \frac{0}{\varepsilon_0} \\ \Rightarrow 4\pi r^2 E &= 0 \\ \Rightarrow E &= 0 \end{align*}

That means that anywhere inside the sphere, there is no electric field! This is non-trivial. All distributed charges cancel out on the inside.

#### Outside the charged sphere, for $$r\gt R$$

The enclosed charge is $$Q$$. Applying equation $$\eqref{eq:gauss}$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} \\ \Rightarrow E \oiint_S dA &= \frac{Q}{\varepsilon_0} \\ \Rightarrow E &= \frac{Q}{4\pi r^2\varepsilon_0} \end{align*}

Note:

• The direction is radially outwards is $$Q$$ is positive, and inwards otherwise.
• This is non-trivial as well, because it behaves as a point charge at the center of the sphere as we have seen in the first example.

#### Conclusion

The electric field

• inside the sphere is $$0$$,
• outside the sphere, it is the same as if there was a point charge at the center of the sphere.

Plot of the electric field, $$E(r)$$

Key to Gauss’ law that the electric field falls of at $$\frac{1}{r^2}$$. A gravitational field also falls of at $$\frac{1}{r^2}$$ $$\Longrightarrow$$ if there was an hollow spherical planet, there would be no gravitational force inside.

### Example: using plane symmetry

Express the electric field anywhere in space for a very large plane with a uniformly distributed charge $$Q$$, with charge density $$\sigma$$ on area $$A$$ $$\sigma = \frac{Q}{A} \quad \left[\frac{C}{m^2}\right] \nonumber$$

If the very large plane is $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \parallelsum$$ $$xy$$-plane, there is symmetry along the $$x,y$$-axis $$\Longrightarrow$$ physical characteristics will not depend on $$x$$ or $$y$$ $$\Longrightarrow \vec E$$ can only have a $$z$$-component $$\Longrightarrow$$ $$\vec E \perp$$ the plane.

As the Gaussian surface, we choose a closed cylinder through the plane

The Gaussian surface takes advantage of symmetry

1. The rounded wall is $$\perp$$ to $$\vec E \Longrightarrow \phi_\text{wall}=0$$.
2. The flat end plates are $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \parallelsum$$ to $$\vec E$$.
3. Both end plates are distance $$d$$ from the plane.

For both end plates $$\vec E$$ is pointed away from the plate. Let the surface of each end plate be $$A$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \phi &= \phi_\text{wall} + \phi_\text{top} + \phi_\text{bottom} \\ &= 0 + 2 \oiint_S \vec E \cdot d\vec A = 2 \oiint_S E\,dA \\ &= 2EA \end{align*}

Apply $$\eqref{eq:gauss}$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \phi &= \frac{\sum_i Q_i}{\varepsilon_0} \quad \Rightarrow \quad 2E\bcancel{A} = \frac{\sigma \bcancel{A}}{\varepsilon_0} \end{align*}

So, the electric field is independent of distance $$d$$ $$\shaded{ E = \frac{\sigma}{2\varepsilon_0} } \tag{E-field plate}$$

While this is for an infinitely large plane, but also holds if the distance $$d$$ is small compared to the linear size of the plane.

### Example: two charged planes

Two very large plates, with surface charge densities ($$Q/A$$), $$+\sigma$$ and $$-\sigma$$, a distance $$d$$ apart

Sum the electric field vectors using the superposition principle $$\shaded{ E = \frac{\sigma}{\varepsilon_0} } \tag{E-field plates}$$

Picture

Note that, when you get to the edge of the plates, the symmetry argument is no longer true.

### Example: using cylindrical symmetry (lec.12)

A very long cylinder with radius $$R$$ with uniform charge distribution throughout the whole cylinder with positive density $$\rho$$. What is the electric field inside and outside the cylinder?

Let’s start with outside the cylinder, $$r\leq R$$. The gaussian surface is a cylinder with radius $$r$$ and length $$l$$. The ends of the gaussian surface are flat perpendicular to the axis of symmetry

Symmetry arguments:

1. On the curved gaussian surface, the distance $$r$$ is the same everywhere, so the electric field is the same.
2. Given that the curved gaussian surface is a cylinder, the electric field must everywhere be radial, perpendicular to the axis of symmetry.
3. The electric flux through the flat end surfaces must be zero, because $$\vec E\perp d\vec A$$
4. On the curved gaussian surface, $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec E\parallelsum d\vec A$$

To apply Gauss’ law, we only have to take the curved surface into account. The curved area is of the gaussian surface is $$l\,2\pi\,r$$ and the volume of the enclosed charge is $$l\,\pi R^2$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} & \left(\vec E \parallelsum d\vec A\right) \\ \Rightarrow E\,A_\rm{curved} &= \frac{V_\rm{encl.}\,\rho}{\varepsilon_0} \\ \Rightarrow E\,\bcancel{l}\,2\cancel{\pi}\,r &= \frac{\bcancel{l}\,\cancel{\pi} R^2\,\rho}{\varepsilon_0} \\ \Rightarrow \vec E &= \frac{R^2\,\rho}{2\,\varepsilon_0\,r} \hat r \end{align*}

Inside the cylinder, $$r\leq R$$, a similar gaussian surface, but inside the cylinder

In applying Gauss’ law, the first term is the same, but and the volume of the enclosed charge is $$l\,\pi r^2$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} & \left(\vec E \parallelsum d\vec A\right) \\ \Rightarrow E\,A_\rm{curved} &= \frac{V_\rm{encl.}\,\rho}{\varepsilon_0} \\ \Rightarrow E\,\bcancel{l}\,2\cancel{\pi}\,\bcancel{r} &= \frac{\bcancel{l}\,\cancel{\pi} r^\bcancel{2}\,\rho}{\varepsilon_0} \\ \Rightarrow \vec E &= \frac{r\,\rho}{2\,\varepsilon_0} \hat r \end{align*}

Plotting $$E(r)$$

If it were a solid conductor, the charge inside would be zero