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⚛️ NEET Physics❓ Question

Why is a bar magnet's axial field exactly double its equatorial field? And do field lines start at the N pole?

Farhan ShaikhDropper Sample

15 Sept 2026 · 1 min read

Revising Magnetism and Matter without paying for any extra test series, just NCERT + PYQs, so I want the basics to stick.

For a short bar magnet with magnetic moment mm, NCERT gives at a distance rr (much larger than the length):

Axial: B=μ04π2mr3B = \dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}

Equatorial: B=μ04πmr3B = \dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}

I just mugged these and the ratio 2:1. I'd like to know where the 2 comes from, and the directions. Which one is along the dipole moment and which one is opposite? I once wrote them the other way in a test and lost a full question.

Second doubt: field lines. NCERT figure shows lines emerging from the north pole and entering the south pole, but I read somewhere that inside the magnet they go from S to N and form closed loops. So they don't "start" at the N pole? Also if I cut a bar magnet in half do I get a separate N and S pole? A friend said yes and I said that is wrong. Who's right?

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7 answers & replies

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  • [+]Neel Chatterjee12thSample· 1mo ago

    Think of poles of strength ±qm\pm q_m separated by 2l2l, so m=qm⋅2lm = q_m\cdot2l.

    On the axis at distance rr from the centre: N-pole's field and S-pole's field point opposite ways, and the nearer pole wins: B=μ0qm4π[1(r−l)2−1(r+l)2]≈μ04π4qmlr3=μ04π2mr3B = \frac{\mu_0 q_m}{4\pi}\left[\frac{1}{(r-l)^2}-\frac{1}{(r+l)^2}\right]\approx\frac{\mu_0}{4\pi}\frac{4q_ml}{r^3}=\frac{\mu_0}{4\pi}\frac{2m}{r^3} On the equator both fields have the same magnitude and their components perpendicular to the axis cancel, while the axis-parallel components add. For r≫lr\gg l that gives μ04πmr3\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}. Same math as the electric dipole where Eaxial=2EequatorialE_{axial}=2E_{equatorial}.

  • [+]Tanvi Deshmukh12thSample· 1mo ago

    Directions (I have this in a table!!):

    • Axial: field is along the dipole moment (the direction S→N through the magnet), same sense as m⃗\vec m.
    • Equatorial: field is opposite to m⃗\vec m.

    Picture a compass placed on the equatorial line: its needle ends up pointing the way opposite to the magnet's own N-S arrow. Axial: the needle lines up with the arrow.

  • [+]Harsh Gupta12thSample· 1mo ago

    Axial field is μ04πmr3\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3} and equatorial is double that, because on the equator two poles contribute so you get twice. That's the logic I use.

    • [+]Aman YadavDropperSample· 1mo ago

      Opposite. Axial is 2m/r32m/r^3 and equatorial is m/r3m/r^3. Counting poles doesn't work. On the axis one pole is near and one is far, so the two fields oppose but the nearer one wins. On the equator both poles are equally far, but their sideways parts cancel and only the axis-parallel parts add. Do the algebra like Neel instead of inventing a rule.

      • [+]Harsh Gupta12thSample· 1mo ago

        Fair, I just tried the numbers and it's axial 2m, equatorial m. My "two poles = double" argument was nonsense. Noted. Still funny though.

  • [+]Nisha Verma12thSample· 1mo ago

    Field lines: they are closed loops. Outside the magnet they go N→S, inside S→N. So they don't start or end anywhere (consistent with Gauss's law for magnetism: net flux through any closed surface is zero, there are no isolated monopoles).

    Cutting a bar magnet: both halves become full magnets, each with its own N and S pole. You get two smaller magnets, you can't isolate a single pole. Your friend is right and that is exactly the point of "monopoles don't exist".

    • [+]Farhan ShaikhDropperSample· 1mo agoAuthor

      Ok that settles the argument with my friend 😄 (I owe him chai). Neel's derivation also finally showed me where the 2 comes from. Thanks all.

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