## Section40.5Magnetic Field of Magnets

Previously we have studied magnetic field by steady electric current as given by Biot-Savart law and Ampere's law. We study magnetic field of magnets mostly empirically by experiments. However, the current-loop equivalence of a magnetic dipole provides a way to study magnets theoretically as well.

For instance, to know the magnetic field of a tiny magnet at a point along the dipole, we just work out the magnetic field at the axis of a loop of current. Then, we substitute $\mu$ for $IA$ and take a limit where size of the loop becomes small. For a dipole $\mu$ pointed towards $z$ axis, we get magnetic field at a point on the $z$ axis to be

$$\vec B_P = \dfrac{\mu_0}{2\pi}\dfrac{\mu}{z^3}\, \hat u_z.\label{eq-dipole-field-at-axis}\tag{40.5.1}$$

Magnetic field of a magnetic dipole at a point that is not on the axis is more difficult to work out.

Here we will just cite the formula for magnetic field at a point P that is located at a spherical radial distance $r$ from the origin and line from origin to P makes an angle $\theta$ with respect to $z$ axis for a magnetic dipole placed at the origin in the $z$ axis direction.

$$\vec B_P = \dfrac{\mu_0\mu}{4\pi r^3}\left( 2\cos\theta\ \hat u_r + \sin\theta\ \hat u_{\theta}\right).\label{eq-dipole-field-at-arbitrary-point}\tag{40.5.2}$$

In this formula $\hat u_r$ and $\hat u_\theta$ are radial unit vector and a unit vector tangent to the great circle through P. This field is called dipole field due to its distance dependence of $1/r^3\text{.}$

The dipole field is symmetric about the direction of the dipole - you get the same picture when you rotate about the direction of the dipole. Therefore, we can just draw field lines in $xz$ plane as shown in Figure 40.5.3. You can rotate the field lines in your mind to figure out what it would look like in three dimensional space.

### Subsection40.5.1(Calculus) A Uniformly Magnetized Thin Bar Magnet

To figure out magnetic field around a thin bar magnet, we can superpose the magnetic field by each piece of the thin bar. To be specific, consider a bar magnet of length $L\text{,}$ which is unifomly magnetized in the direction of its length with magnetic dipole moment per unit length equal to $M\text{.}$

Let bar be placed along the $z$ axis with its magnetization pointed in the positive $z$ direction. We will find magnetic field from this magnet at a point on the $z$ axis that is far away from the magnet and on the positive $z$ axis.

Using Eq. (40.5.1) we can write the field by the piece of bar between $z'$ and $z'+dz'\text{.}$ Note that $z$ in the formula in Eq. (40.5.1) is the distance from the dipole at origin. Therefore, we will need to replace $z$ there by $z-z'\text{.}$ The dipole moment in $dz'$ is $d\mu = Mdz'\text{.}$ Therefore, $z$ component of the field at P is

\begin{align*} dB_z \amp = \dfrac{\mu_0}{2\pi}\dfrac{d\mu}{(z-z')^3}\\ \amp = \dfrac{\mu_0}{2\pi}\dfrac{ Mdz'}{(z-z')^3} \end{align*}

We will integrate this from $z'=-L/2$ to $z'=L/2$ to get the net field at P.

\begin{align*} B_z \amp = \dfrac{\mu_0 M }{2\pi} \int_{-L/2}^{L/2}\, \dfrac{ dz'}{(z-z')^3} \\ \amp = \dfrac{\mu_0 M }{2\pi} \left[ \dfrac{1}{2(z-z')^2} \right]_{-L/2}^{L/2} \\ \amp = \dfrac{\mu_0M}{\pi}\left[ \dfrac{1}{\left(2z-L \right)^2} - \dfrac{1}{\left(2z+L \right)^2} \right]. \end{align*}

Note that this formula is only good for point on the axis. The magnetic field at an arbitray point will be different from this formula.

### Subsection40.5.2(Calculus) A Uniformly Magnetized Sphere

Consider a sphere of radius R that is uniformly magnetized with magnetization $M\text{.}$ We choose $z$ axis to be the direction of the magnetization and use a spherical coordinate to write the answer here. Let $m$ denote the total dipole moment of the sphere, i.e.,

\begin{equation*} m = \dfrac{4}{3}\pi R^3 M. \end{equation*}

I will not give a derivation of the results. Instead, I will just quote the results and let it a challenge to a more ambitious student to prove them. The magnetic field at a distance $r$ from the center of the sphere and at an angle $\theta$ from the direction of the dipole is given by the following formula in spherical coordinates.

\begin{equation*} \vec B(r,\theta,\phi) = \left\{ \begin{array}{ll} \dfrac{2}{3}\mu_0 M\, \hat k\amp\ \ \ \ (r\lt R)\\ \dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}\left( 2\cos\theta\,\hat u_r + \sin\theta\,\hat u_{\theta}\right) \amp\ \ \ \ (r \gt R) \end{array} \right. \end{equation*}

The magnetic field outside is that of the total magnetic moment placed at the center of the sphere.