Field of a bar magnet on its axis and on its equator (qualitative)
At far points, B(axial) = (μ₀/4π)(2m/r³) along m, and B(equatorial) = (μ₀/4π)(m/r³) opposite to m. Both fall as 1/r³.
The two special lines and the results
The axial line of a bar magnet is the line through its two poles. The equatorial line is the line through its centre, perpendicular to the axis. For a short magnet, or at a far point where r is much larger than the magnet's length, the results are as follows. On the axis, B = (μ₀/4π)(2m/r³), pointing along m. On the equator, B = (μ₀/4π)(m/r³), pointing opposite to m. At equal distances, the axial field is twice the equatorial field. Both fall as 1/r³, much faster than the 1/r² field of a single charge. The syllabus asks for these results qualitatively: know the formulas, the directions and the ratio, but you need not derive them.
The electrostatic analogue
| Quantity | Electric dipole | Magnetic dipole |
|---|---|---|
| Constant | 1/ε₀ | μ₀ |
| Dipole moment | p | m |
| Far field on the axis | (1/4πε₀)(2p/r³), along p | (μ₀/4π)(2m/r³), along m |
| Far field on the equator | (1/4πε₀)(p/r³), opposite to p | (μ₀/4π)(m/r³), opposite to m |
| Torque in a uniform field | p × E | m × B |
| Potential energy in a uniform field | −p·E | −m·B |
How to show the directions on a diagram
- Draw the magnet with S on the left and N on the right. Draw m as an arrow from S to N.
- Mark a point P on the axis beyond N. Draw B at P pointing away from the magnet, along m.
- Mark a point Q on the axis beyond S. Draw B at Q pointing towards the magnet, again along m.
- Mark a point R on the equatorial line. Draw B at R parallel to the magnet, pointing from the N side towards the S side, opposite to m.
- Label r for each point, measured from the centre of the magnet.
Worked example: axial and equatorial fields
A short bar magnet has magnetic moment 0.50 A m². Find the magnetic field (a) on its axis and (b) on its equatorial line, both 10 cm from its centre. (μ₀/4π = 10⁻⁷ T m A⁻¹)
- (a) Formula: B(axial) = (μ₀/4π)(2m/r³).
- B = 10⁻⁷ × 2 × 0.50/(0.10)³ T = 10⁻⁷ × 1.0/10⁻³ T = 1.0 × 10⁻⁴ T, along m.
- (b) Formula: B(equatorial) = (μ₀/4π)(m/r³).
- B = 10⁻⁷ × 0.50/10⁻³ T = 5.0 × 10⁻⁵ T, opposite to m.
- Check: the axial value is twice the equatorial value.
(a) 1.0 × 10⁻⁴ T along m; (b) 5.0 × 10⁻⁵ T opposite to m
Worked example: changing the distance
On the axis of a short magnet, the field 10 cm from its centre is 4.0 × 10⁻⁵ T. Find the field (a) on the axis at 20 cm and (b) on the equatorial line at 20 cm.
- B ∝ 1/r³, so doubling r divides B by 2³ = 8.
- (a) B(axial, 20 cm) = 4.0 × 10⁻⁵/8 T = 5.0 × 10⁻⁶ T.
- (b) At the same distance, the equatorial field is half the axial field.
- B(equatorial, 20 cm) = 5.0 × 10⁻⁶/2 T = 2.5 × 10⁻⁶ T.
(a) 5.0 × 10⁻⁶ T; (b) 2.5 × 10⁻⁶ T
A smartphone compass app reads tiny magnetic fields with a sensor. Hold a magnet near the phone and the reading jumps, but move it a little away and the effect drops fast, because a dipole's field falls as 1/r³.
1 marks: “What is the ratio of the magnetic field on the axis to that on the equatorial line of a short bar magnet, at the same distance?” 2 : 1.
2 marks: “Write the expressions for the magnetic field of a short bar magnet at a far point (a) on its axis and (b) on its equatorial line. State their directions.” (a) (μ₀/4π)(2m/r³), along m. (b) (μ₀/4π)(m/r³), opposite to m.
2 marks: “How does the field on the axis of a short bar magnet change if the distance from its centre is doubled? Explain.” B ∝ 1/r³, so doubling r makes the field 1/8 of its earlier value.
Axis: 2, along m. Equator: 1, against m. Both fall as 1/r³.