The Field of a Bar Magnet

Two poles, two simple positions, one general formula.

11 minUpdated 22 October 2026end-on positionbroadside-on positionmagnetic scalar potentialdipole field

Lesson at a glanceShow the key points

End-on: (μ₀/4π)2Md/(d² − l²)² (36.6) → (μ₀/4π)2M/d³ (36.7).
Broadside-on: (μ₀/4π)M/(d² + l²)^(3/2) (36.8) → (μ₀/4π)M/d³ (36.9).
V = (μ₀/4π)m/r (36.12); dipole V = (μ₀/4π)M cos θ/r² (36.13).
Dipole: B = (μ₀/4π)(M/r³)√(1 + 3cos²θ) (36.14), tan α = ½ tan θ (36.15).

With the pole picture, the field of a bar magnet is just the field of two point poles. Two positions have neat formulas, and far away every magnet becomes a dipole.

End-on position

A point on the magnet’s axis, at distance d from its centre (magnetic length 2l, pole strength m). The north pole gives μ₀m/4π(d − l)² away from the magnet; the south pole gives μ₀m/4π(d + l)² toward it:

B = \frac{\mu_0 m}{4\pi}\left[\frac{1}{(d-l)^2} - \frac{1}{(d+l)^2}\right]

End-on(36.6–36.7)

B = \frac{\mu_0}{4\pi}\,\frac{2Md}{(d^2 - l^2)^2} ;\xrightarrow{d \gg l}; \frac{\mu_0}{4\pi}\,\frac{2M}{d^3}

Worked example

On the axis

A magnet has magnetic length 8.0 cm and pole strength 15 A m (150 CGS units, where 1 CGS unit = 0.1 A m). Find B on its axis 15 cm from its centre.

Broadside-on position

A point on the perpendicular bisector, at distance d from the centre: each pole is √(d² + l²) away. The components along the bisector cancel and those parallel to the axis add, giving a field parallel to the magnet, pointing from N toward S:

Broadside-on(36.8–36.9)

B = \frac{\mu_0}{4\pi}\,\frac{M}{(d^2 + l^2)^{3/2}} ;\xrightarrow{d \gg l}; \frac{\mu_0}{4\pi}\,\frac{M}{d^3}

Worked example

On the bisector

For the same magnet, find B at 15 cm on its perpendicular bisector.

Interactive

The field of a bar magnet

Model the magnet as a north pole (+15 A m) and a south pole 8 cm apart. Drag the compass: it lines up with the field. On the axis (end-on) and on the perpendicular bisector (broadside-on) the field has simple formulas; far away every magnet looks like a dipole, B = (μ₀/4π)(M/r³)√(1 + 3cos²θ).

B from the two poles82.4µT

dipole formula71.1µT

positionend-on

distance from centre15.0cm

Magnetic scalar potential

As with gravity and electrostatics, define a potential whose fall gives the field:

Magnetic scalar potential(36.10–36.12)

V(\vec r_2) - V(\vec r_1) = -\int_{\vec r_1}^{\vec r_2}\vec B\cdot d\vec r, \qquad B_l = -\frac{dV}{dl}, \qquad V = \frac{\mu_0}{4\pi}\,\frac{m}{r}

For a dipole, a point at distance r ≫ l and angle θ from the axis is about r − l cos θ from N and r + l cos θ from S, so

Dipole potential(36.13)

V = \frac{\mu_0}{4\pi}\,\frac{M\cos\theta}{r^2}

The general dipole field

Differentiating V along r and along θ gives the two components:

B_r = -\frac{\partial V}{\partial r} = \frac{\mu_0}{4\pi}\,\frac{2M\cos\theta}{r^3}, \qquad B_\theta = -\frac1r\frac{\partial V}{\partial\theta} = \frac{\mu_0}{4\pi}\,\frac{M\sin\theta}{r^3}

Dipole field(36.14–36.15)

B = \frac{\mu_0}{4\pi}\,\frac{M}{r^3}\sqrt{1 + 3\cos^2\theta}, \qquad \tan\alpha = \tfrac12\tan\theta

α is the angle between B⃗ and the radial line

Worked example

Off the axis

Find the field of a dipole of moment 2.0 A m² at 50 cm from it, in a direction at 30° to its axis.

What to remember

1End-on: B = (μ₀/4π)2Md/(d² − l²)² → (μ₀/4π)2M/d³.
2Broadside-on: B = (μ₀/4π)M/(d² + l²)^(3/2) → (μ₀/4π)M/d³, parallel to the axis.
3V = (μ₀/4π)m/r; dipole V = (μ₀/4π)M cos θ/r².
4Dipole: B = (μ₀/4π)(M/r³)√(1 + 3cos²θ), tan α = ½ tan θ.

Check yourself

Far from a short magnet, end-on and broadside-on fields at equal distances are in the ratio…

Doubling the distance from a short magnet changes its field by a factor of…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadMagnetic dipoleWikipedia

PreviousPoles & Bar Magnets Next Earth's Magnetism

The Field of a Bar Magnet

Two poles, two simple positions, one general formula.

11 minUpdated 22 October 2026end-on positionbroadside-on positionmagnetic scalar potentialdipole field

Lesson at a glanceShow the key points

End-on: (μ₀/4π)2Md/(d² − l²)² (36.6) → (μ₀/4π)2M/d³ (36.7).
Broadside-on: (μ₀/4π)M/(d² + l²)^(3/2) (36.8) → (μ₀/4π)M/d³ (36.9).
V = (μ₀/4π)m/r (36.12); dipole V = (μ₀/4π)M cos θ/r² (36.13).
Dipole: B = (μ₀/4π)(M/r³)√(1 + 3cos²θ) (36.14), tan α = ½ tan θ (36.15).

With the pole picture, the field of a bar magnet is just the field of two point poles. Two positions have neat formulas, and far away every magnet becomes a dipole.

End-on position

B = \frac{\mu_0 m}{4\pi}\left[\frac{1}{(d-l)^2} - \frac{1}{(d+l)^2}\right]

End-on(36.6–36.7)

B = \frac{\mu_0}{4\pi}\,\frac{2Md}{(d^2 - l^2)^2} ;\xrightarrow{d \gg l}; \frac{\mu_0}{4\pi}\,\frac{2M}{d^3}

Worked example

On the axis

A magnet has magnetic length 8.0 cm and pole strength 15 A m (150 CGS units, where 1 CGS unit = 0.1 A m). Find B on its axis 15 cm from its centre.

Broadside-on position

Broadside-on(36.8–36.9)

B = \frac{\mu_0}{4\pi}\,\frac{M}{(d^2 + l^2)^{3/2}} ;\xrightarrow{d \gg l}; \frac{\mu_0}{4\pi}\,\frac{M}{d^3}

Worked example

On the bisector

For the same magnet, find B at 15 cm on its perpendicular bisector.

Interactive

The field of a bar magnet

B from the two poles82.4µT

dipole formula71.1µT

positionend-on

distance from centre15.0cm

Magnetic scalar potential

As with gravity and electrostatics, define a potential whose fall gives the field:

Magnetic scalar potential(36.10–36.12)

V(\vec r_2) - V(\vec r_1) = -\int_{\vec r_1}^{\vec r_2}\vec B\cdot d\vec r, \qquad B_l = -\frac{dV}{dl}, \qquad V = \frac{\mu_0}{4\pi}\,\frac{m}{r}

For a dipole, a point at distance r ≫ l and angle θ from the axis is about r − l cos θ from N and r + l cos θ from S, so

Dipole potential(36.13)

V = \frac{\mu_0}{4\pi}\,\frac{M\cos\theta}{r^2}

The general dipole field

Differentiating V along r and along θ gives the two components:

B_r = -\frac{\partial V}{\partial r} = \frac{\mu_0}{4\pi}\,\frac{2M\cos\theta}{r^3}, \qquad B_\theta = -\frac1r\frac{\partial V}{\partial\theta} = \frac{\mu_0}{4\pi}\,\frac{M\sin\theta}{r^3}

Dipole field(36.14–36.15)

B = \frac{\mu_0}{4\pi}\,\frac{M}{r^3}\sqrt{1 + 3\cos^2\theta}, \qquad \tan\alpha = \tfrac12\tan\theta

α is the angle between B⃗ and the radial line

Worked example

Off the axis

Find the field of a dipole of moment 2.0 A m² at 50 cm from it, in a direction at 30° to its axis.

What to remember

1End-on: B = (μ₀/4π)2Md/(d² − l²)² → (μ₀/4π)2M/d³.
2Broadside-on: B = (μ₀/4π)M/(d² + l²)^(3/2) → (μ₀/4π)M/d³, parallel to the axis.
3V = (μ₀/4π)m/r; dipole V = (μ₀/4π)M cos θ/r².
4Dipole: B = (μ₀/4π)(M/r³)√(1 + 3cos²θ), tan α = ½ tan θ.

Check yourself

Far from a short magnet, end-on and broadside-on fields at equal distances are in the ratio…

Doubling the distance from a short magnet changes its field by a factor of…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadMagnetic dipoleWikipedia

PreviousPoles & Bar Magnets Next Earth's Magnetism