Physics · Magnetism And Matter · NEET
Both formulas share the same base (μ₀/4π)·(m/r³); axial just carries an extra factor of 2. On the axis, the fields of the two poles point the same way and partly add, giving 2m/r³. On the equatorial line, the two pole fields point in opposite slanted directions, so only their components along the magnet survive and they partially cancel, giving m/r³. Same r, so B_axial / B_eq = 2. This is only exact for a SHORT magnet (r much larger than magnet length).
Axial field points in the SAME direction as the magnetic moment m (from S to N inside the magnet, i.e. along m). Equatorial field points OPPOSITE to m (anti-parallel to m). This is the mirror of the electric dipole, where the axial E is along p and equatorial E is opposite to p — the analogy is exact, just swap p→m and 1/4πε₀ → μ₀/4π.
They are the SHORT-magnet (point-dipole) approximation, valid when the distance r is much larger than the magnet length 2l, so r³ dominates. For a magnet of length 2l the exact axial field is (μ₀/4π)·2mr/(r²−l²)² and equatorial is (μ₀/4π)·m/(r²+l²)^{3/2}. For NEET, if the question says 'short bar magnet' or gives only r, use the simple 1/r³ forms.
Yes. Both a magnetic dipole's axial and equatorial fields decrease as 1/r³ (unlike a monopole's 1/r²). So doubling the distance makes each field 1/8 of its value. Only the constant differs by the factor 2, not the distance dependence — a common trap is thinking axial falls slower.
In the standard short-magnet formulas r is measured from the CENTRE of the magnet to the observation point. For a finite magnet the exact formulas also use the centre distance r, with the half-length l appearing separately. Do not measure from a pole unless the problem is a single-pole (pole-strength) question.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For a short bar magnet at the same distance r, B_axial : B_eq = 2 : 1. Both fall as 1/r³; only the numerical factor differs.
Yes. Replace p with m and 1/4πε₀ with μ₀/4π: E_axial = (1/4πε₀)(2p/r³) becomes B_axial = (μ₀/4π)(2m/r³), and the equatorial forms match too. This is the 'electrostatic analog'.
Opposite to the magnetic moment m (anti-parallel), because on the perpendicular bisector the surviving components of the two pole fields point back along −m.
1/r³ for a dipole (short magnet). A 1/r² law would apply only to a single isolated pole, which does not exist — magnets are dipoles.
It means the length 2l is negligible compared with the distance r (r ≫ 2l), so the magnet behaves as a point dipole and the simple (μ₀/4π)(2m/r³) and (μ₀/4π)(m/r³) forms are exact.
