Physics · Magnetism And Matter · NEET
On the AXIS it is 2m/r³ (with the μ0/4π factor): B_axial = (μ0/4π)(2m/r³). The 'm/r³' form (with a factor of 1) is the EQUATORIAL field, on the perpendicular bisector. So B_axial = 2 × B_equatorial at the same distance. This exactly mirrors the electric dipole: E_axial = 2p/r³, E_equatorial = p/r³ (times 1/4πε0). Remember: the point on the LINE of the dipole feels the stronger, doubled field.
Along the axis both poles push the field the SAME way (their contributions add up along the axis), and being on the line means the 1/r² fall-off works fully in that direction. On the equator the two pole fields point in opposite axial senses; only their components along the magnet survive and they are weaker by a geometry factor. Working the vectors out gives axial = 2m/r³ and equatorial = m/r³, so the ratio is exactly 2:1 for a short magnet.
Use B = (μ0/4π)(2m/r³) only when r ≫ l (the observation point is far compared to the magnet's half-length). If r is comparable to l, use the exact result B = (μ0/4π) · 2mr/(r² − l²)². In NEET numericals the phrase 'short bar magnet' or 'far point' is your signal to use the 2m/r³ form.
Outside the magnet on the axis, B points along m, i.e. from the S pole to the N pole and continues outward beyond the N pole. So it is parallel to the magnetic moment vector. On the equatorial line the field is ANTI-parallel to m. Getting the sign wrong is a common trap in superposition problems where two magnets' axial and equatorial fields must be added.
They have the same far-field shape because a magnet is equivalent to a dipole. A loop's axial field is B = μ0 I R² / [2(x² + R²)^{3/2}], and for x ≫ R this becomes (μ0/4π)(2m/x³) with m = I·πR². So far away, a current loop and a short bar magnet give the identical 2m/r³ axial field. Close up (x comparable to R) use the loop formula, not the dipole one.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Exact: B = (μ0/4π) · 2m·r / (r² − l²)², where m is magnetic moment, 2l is the magnet length, r is distance from the centre. For a short magnet (r ≫ l): B = (μ0/4π) · (2m/r³).
Along the axis both poles' contributions add in the same direction, while on the equator only weaker perpendicular components survive. The vector algebra gives a 2:1 ratio, exactly like an electric dipole (2p/r³ versus p/r³).
Outside the magnet on its axis, B is parallel to the magnetic moment m — pointing from S to N and outward beyond the N pole. This is opposite to the equatorial field, which is anti-parallel to m.
Treat the N and S poles as +qm and −qm separated by 2l. Add their fields at an axial point at distance r: the near pole gives (μ0/4π)qm/(r−l)² and the far pole (μ0/4π)qm/(r+l)² in the opposite sense. Subtracting and using m = qm·2l gives B = (μ0/4π)·2mr/(r²−l²)², which reduces to 2m/r³ for r ≫ l.
Yes, far away. A loop of moment m = IπR² gives axial field (μ0/4π)(2m/x³) for x ≫ R, identical to a short bar magnet. Close to the loop, use B = μ0IR²/[2(x²+R²)^{3/2}].
