Magnetic Dipole Field

A bar magnet of moment $\vec m$ produces a field at far points whose form mirrors the electric dipole field, with the substitution $1/(4\pi\varepsilon_0) \to \mu_0/(4\pi)$ and $p \to m$. Two special directions — the axis and the equator — give the simplest closed forms.

Concept

For a dipole of moment $\vec m$, far from the magnet ($r \gg l$),

• Axial field (along $\vec m$): $\vec B_{\text{ax}} = \frac{\mu_0}{4\pi}\,\frac{2\vec m}{r^3}.$

• Equatorial field (perpendicular bisector, antiparallel to $\vec m$): $\vec B_{\text{eq}} = -\frac{\mu_0}{4\pi}\,\frac{\vec m}{r^3}.$

At equal distance the axial field is exactly twice the equatorial field in magnitude, and opposite in sign relative to $\vec m$.

Derivation

Place the magnet on the $x$-axis: $+q_m$ at $+l$, $-q_m$ at $-l$, moment $m = q_m(2l)$ along $+\hat x$.

Axial. Point $P$ at $x = r$ on the axis ($r > l$). The fields from the two poles (along $\hat x$) are

$B_N = \frac{\mu_0}{4\pi}\,\frac{q_m}{(r-l)^2},\qquad B_S = -\frac{\mu_0}{4\pi}\,\frac{q_m}{(r+l)^2}.$

Adding, $B_{\text{ax}} = \frac{\mu_0 q_m}{4\pi}\,\frac{(r+l)^2 - (r-l)^2}{(r^2-l^2)^2} = \frac{\mu_0}{4\pi}\,\frac{2m r}{(r^2-l^2)^2}.$

For $r \gg l$, $B_{\text{ax}} = \frac{\mu_0}{4\pi}\,\frac{2m}{r^3}.$

Equatorial. Point $P$ at distance $r$ on the perpendicular bisector. Each pole is at distance $\sqrt{r^2 + l^2}$, giving equal magnitudes

$B_N = B_S = \frac{\mu_0}{4\pi}\,\frac{q_m}{r^2 + l^2}.$

Perpendicular components cancel; axial components (opposite to $\vec m$) add. Using $\cos\theta = l/\sqrt{r^2+l^2}$,

$B_{\text{eq}} = \frac{\mu_0}{4\pi}\,\frac{2 q_m l}{(r^2+l^2)^{3/2}} = \frac{\mu_0}{4\pi}\,\frac{m}{(r^2+l^2)^{3/2}}.$

For $r \gg l$, $B_{\text{eq}} = \frac{\mu_0}{4\pi}\,\frac{m}{r^3}.$

General point. At polar angle $\theta$ from the axis, $B(r,\theta) = \frac{\mu_0}{4\pi}\,\frac{m\sqrt{1+3\cos^2\theta}}{r^3}.$

Worked Example

A short bar magnet has $m = 0.48\,A\,m^2$. At $r = 10$ cm on the axis,

$B_{\text{ax}} = 10^{-7}\cdot\frac{2(0.48)}{(0.10)^3} = 9.6\times 10^{-4}\,T \approx 9.6\,G.$

On the equator at the same distance,

$B_{\text{eq}} = 10^{-7}\cdot\frac{0.48}{(0.10)^3} = 4.8\times 10^{-4}\,T = 4.8\,G.$

The ratio $B_{\text{ax}}/B_{\text{eq}} = 2$ — independent of $m$ and $r$.

Common Confusions

• The equatorial field is opposite to $\vec m$, not zero.
• The dipole formulas are only valid for $r \gg l$. For short distances use the exact expressions with $(r^2 \pm l^2)$ terms.
• Don't drop the $4\pi$ in $\mu_0/4\pi = 10^{-7}\,T\,m/A$ — it makes calculations quick.

Key Takeaways

• $B_{\text{ax}} = (\mu_0/4\pi)\,2m/r^3$ along $\vec m$.
• $B_{\text{eq}} = (\mu_0/4\pi)\,m/r^3$ opposite to $\vec m$.
• Ratio is always 2:1 at the same distance.
• Same $1/r^3$ structure as the electric dipole, with $p\to m$ and $1/\varepsilon_0 \to \mu_0$.