Physics · Magnetism And Matter · NEET
Yes, for the same short magnet at the same distance r. Axial field is B_axial = (μ₀/4π)(2m/r³) and equatorial field is B_eq = (μ₀/4π)(m/r³). Divide them: B_eq / B_axial = 1/2. So the equatorial (side) point always feels half the field of the axial (end) point at equal distance. This '2:1 ratio' is a favourite NEET one-liner.
It points anti-parallel to the magnetic dipole moment m (opposite to the direction from S to N inside the magnet). At an equatorial point, the fields from the N-pole and S-pole add up so that the horizontal component survives and points from the N-side back toward the S-side. So on the equatorial line the field arrow is opposite to the magnet's own moment vector. On the axial line it is parallel to m — that sign flip is the key difference.
It is r cubed (r³) for a magnetic dipole. B_eq = (μ₀/4π)(m/r³). A single monopole would give 1/r², but a bar magnet is a dipole (two poles), so the far field falls off as 1/r³. If you wrote 1/r² you treated the magnet as one pole — that is wrong. This 1/r³ dependence is the same idea as the electric dipole field.
No. 'Equatorial line' of a bar magnet just means the perpendicular-bisector line — the line through the centre, perpendicular to the magnet's axis. It is also called the 'broadside-on' position. It has nothing to do with the geographic or magnetic equator of the Earth. Don't mix the two — NEET sometimes uses this wording to confuse you.
When r is much larger than the magnet's length (r >> l), i.e. a 'short' magnet or a point far away. Then B_eq = (μ₀/4π)(m/r³). For a magnet of half-length l the exact equatorial field is B = (μ₀/4π) · m/(r² + l²)^(3/2); setting l → 0 gives the simple 1/r³ form. NEET numericals almost always assume the short-magnet (dipole) case unless the length is comparable to r.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the magnetic field at a point on the perpendicular bisector (broadside-on line) of a short bar magnet. For distance r from the centre, B = (μ₀/4π)(m/r³), directed anti-parallel to the magnetic moment m.
For the same short magnet at the same distance, B_axial : B_eq = 2 : 1. The axial field is twice the equatorial field, and they point in opposite senses relative to m.
Because a bar magnet acts as a magnetic dipole (two equal and opposite poles), not a single pole. A dipole's field falls off as 1/r³ at large distances, just like an electric dipole.
Tesla (T). With μ₀/4π = 10⁻⁷ T·m/A, m in A·m², and r in metres, B = (10⁻⁷ × m / r³) tesla.
Not necessarily. The equatorial line is just the perpendicular bisector. A neutral point is where the magnet's field cancels an external field (like Earth's); for a bar magnet with N pointing north, neutral points appear on the equatorial line, but the two ideas are defined differently.
