Field on the axis and on the equator.

A compass probe round a bar magnet reading B on the axis and on the equator to scale, falling as 1/r³.

Find the equatorial distance where B equals the axial field at 20 cm (≈15.9 cm) and show the equatorial field points opposite to m.

Thinking the equatorial field points along m.

• The far field of a bar magnet of moment m follows from the electric dipole's field by swapping E → B, p → m and 1/(4πε₀) → μ₀/4π.
• On the axis, at distance r much larger than the magnet's size l: B_A = (μ₀/4π)(2m/r³), along m.
• On the equatorial line (the normal bisector), for r ≫ l: B_E = −(μ₀/4π)(m/r³), half the axial size at the same distance and pointing opposite to m.
• Both fall off as 1/r³: doubling the distance cuts the field to one-eighth.
• The dipole analogy runs through the table: 1/ε₀ ↔ μ₀, p ↔ m, torque p × E ↔ m × B, energy −p·E ↔ −m·B.
• Example 5.2: a needle Q near a needle P is in equilibrium only when m_Q is parallel or antiparallel to P's field at Q. Parallel is stable, antiparallel unstable, and the lowest energy comes with Q on P's axis, pointing along P's field there, where that field is twice its equatorial size.

Teaching this idea? Put the simulation on your own page or LMS with this code.