Physics · Ch 3 — Magnetism and Magnetic Effects of Electric Current

Magnetic Field at a Point Along the Equatorial Line Due to a Magnetic Dipole (Bar Magnet)

3.2.2

Magnetic Field at a Point Along the Equatorial Line Due to a Magnetic Dipole (Bar Magnet)

Now place the unit north pole at a point $C$ on the equatorial line (the perpendicular bisector of the magnet), at distance $r$ from the centre $O$ . Both poles are now equidistant from $C$ , at distance $r' = \sqrt{r^2+l^2}$ , so $B_N = B_S = \dfrac{\mu_0}{4\pi}\dfrac{q_m}{r^2+l^2}$ . Resolving each into components parallel and perpendicular to the magnet's axis, the two perpendicular components cancel by symmetry while the two parallel components -- both directed opposite to $\vec p_m$ (i.e. from N to S) -- add. Using $\cos\alpha = l/r'$ for the angle each vector makes with the axis, the net field works out to

B_{equatorial} = \frac{\mu_0}{4\pi}\,\frac{p_m}{(r^2+l^2)^{3/2}}

For a short magnet ( $r\gg l$ ), $(r^2+l^2)^{3/2}\approx r^3$ , giving the standard short-dipole equatorial-line result

$\boxed{B_{equatorial} = \frac{\mu_0}{4\pi}\,\frac{p_m}{r^3}}$ …

Figure 3.14Magnetic field at a point along the equatorial line due to a magnetic dipole

What this figure shows. A bar magnet NS of magnetic length 2l sits centred at O, and a test point C is marked on the perpendicular bisector of the magnet (the equatorial line) at distance r from O. Two equal-length line segments are drawn from N to C and from S to C, each of length r' = sqrt(r^2+l^2), with field vectors B_N and B_S drawn along each of those two lines; the diagram resolves both vectors into components parallel and perpendicular to the magnet's axis to show that the perpendicular components cancel while the components anti-parallel to t …

Misc Example 3.6Axial and equatorial field of a short bar magnet

Worked out. A short bar magnet of moment 0.5 J/T (i.e. 0.5 A m^2) is examined at r=0.1 m. Along the axial line, B_axial = (mu0/4pi)(2 p_m/r^3) = 10^-7 x 2 x 0.5 / (0.1)^3 = 10^-7 x 1 / 0.001 = 1x10^-4 T, directed from S to N (along the magnet's own moment). Along the equatorial line, B_eq = (mu0/4pi)(p_m/r^3) = 10^-7 x 0.5 / 0.001 = 0.5x10^-4 T, directed opposite to the magnet's moment (N to S). So at the same distance, the axial field is exactly twice the equatorial field, and the two point in opposite senses -- a result that holds for any shor …

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