Magnetic Dipole Field Calculator
Enter magnetic moment and distance to compute the on-axis and equatorial magnetic field of a dipole B = (μ₀/4π)·m/r³.
Input Data
Results
At a glance:A magnetic dipole (current loop, bar magnet) at distance r produces a field that falls as 1/r³. On the dipole axis the field is B_axis = (μ₀/4π)·(2m/r³) and in the equatorial plane B_eq = (μ₀/4π)·(m/r³), where m is the magnetic moment and μ₀ = 4π×10⁻⁷ T·m/A. Both point along (axis) or opposite (equator) the dipole moment. The general vector field is B = (μ₀/4π)·[3(m·r̂)r̂ − m]/r³. Far from any magnet its field looks dipolar; this tool computes the axial and equatorial magnitudes from m and r.
Formula
Axial: B_axis = (μ₀/4π)·(2m/r³)
Equatorial: B_eq = (μ₀/4π)·(m/r³)
$$B_{\text{axial}} = \frac{\mu_0 \mu}{2\pi r^3}, \quad B_{\text{eq}} = \frac{\mu_0 \mu}{4\pi r^3}, \quad \mu_0 = 4\pi \times 10^{-7}$$How to Use
- Enter the magnetic moment m (A·m²).
- Enter the distance r (m).
- The calculator returns the axial and equatorial fields (and μ₀).
Case Studies
Small bar magnet
m = 0.5 A·m², r = 0.1 m.
B_axis = 1e-7×2×0.5/0.001 = 1.0×10⁻⁴ T.
B_eq = 5.0×10⁻⁵ T (half).
FAQ
Why does the field fall as 1/r³?
A dipole is a pair of opposite sources; at large distance their fields nearly cancel, leaving the next-order (dipole) term ∝ 1/r³ (monopole would be 1/r², but magnetic monopoles don't exist).
Is the axis field twice the equatorial?
Yes, B_axis = 2·B_eq at the same r, because of the dipole geometry: the angular factor gives 2 on axis and 1 in magnitude at the equator.
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.