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Physics · Ch 12 — Magnetism

Magnetic Field due to a Bar Magnet at a Point along its Axis and at a Point along its Equator

12.3.1

Magnetic Field due to a Bar Magnet at a Point along its Axis and at a Point along its Equator

Consider a bar magnet of magnetic dipole length 2l2l and magnetic dipole moment m⃗\vec m, and let PP be a point along its axis at distance rr from the centre OO of the dipole, with OS=ON=lOS=ON=l, so that NP=r−lNP=r-l and SP=r+lSP=r+l. Rather than computing the exact field by combining the individual pole contributions, this section uses the electrostatic analogy (valid for r≫lr\gg l, i.e. points genuinely far from the magnet compared with its own size) to write down the magnetic result directly from the already-known electric-dipole result.

The electric field of an electric dipole of moment p=2qlp=2ql, at a distance rr along its axis (for r≫lr\gg l), is Ea=14πε02pr3E_a=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}, and the electric field on its equatorial line is Eeq=−14πε0pr3E_{eq}=-\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3} (the minus sign showing it is anti-parallel to p⃗\vec p). Using the analogy of Table 12.1 (replace electric charge qq by magnetic pole strength qmq_m, and the electrostatic constant 14πε0\tfrac{1}{4\pi\varepsilon_0} by the magnetic constant μ04π\tfrac{\mu_0}{4\pi}), the axial magnetic field of a bar magnet at distance rr (r≫lr\gg l) becomes Ba=μ04π2mr3B_a=\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3} (Eq. 12.3), directed along m⃗\vec m, and the equatorial magnetic field is Beq=−μ04πmr3B_{eq}=-\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3} (Eq. 12.4), the minus sign showing B⃗eq\vec B_{eq} is directed opposite to m⃗\vec m. A direct and useful consequence, for the same distance rr from the magnet's centre, is that the axial field is always exactly twice the equatorial field: Baxis=2BeqB_{axis}=2B_{eq} (Eq. 12.5). …

Figure 12.2aFig. 12.2(a): Magnetic field at a point along the axis of the magnet

What this figure shows. A bar magnet of dipole length 2l2l (poles marked S and N, centre O) lies along a horizontal line. A point P is marked further along this same line, outside the magnet, at distance rr from the centre O, so that NP=r−lNP=r-l and SP=r+lSP=r+l. The magnetic field B⃗a\vec B_a at P is drawn as an arrow along the same axis line, pointing in the same direction as the magnet's own moment m⃗\vec m (away from the magnet, along the extension of the N-pole end), illustrating that the axial field of a bar magnet is always parallel to i …

Figure 12.2bFig. 12.2(b): Magnetic field along the equatorial point

What this figure shows. The same bar magnet (centre O, poles S and N) is shown with a point P now on its perpendicular bisector (the equatorial line), at distance rr from the centre O, so that P is equidistant from both poles. The magnetic field B⃗eq\vec B_{eq} at P is drawn as an arrow parallel to the bar but pointing in the OPPOSITE sense to the magnet's own moment m⃗\vec m (i.e. from the N-pole side towards the S-pole side), illustrating that the equatorial field of a bar magnet is always anti-parallel to its dipole mome …

Table T12.1Table 12.1: The Electrostatic Analogue

Quantity | Electrostatics | Magnetism

Basic physical quantity | Electrostatic charge | Magnetic pole

Field | Electric Field E⃗\vec E | Magnetic Field B⃗\vec B

Constant | 14πε0\dfrac{1}{4\pi\varepsilon_0} | μ04π\dfrac{\mu_0}{4\pi}

Dipole moment | p⃗=q(2l⃗)\vec p=q(2\vec l), along (-ve) to (+ve) charge | m⃗=qm(2l⃗)\vec m=q_m(2\vec l), along S to N pole

Force | F⃗=qE⃗\vec F=q\vec E | F⃗=qmB⃗\vec F=q_m\vec B

Energy (in external field) of a dipole | U=−p⃗⋅E⃗U=-\vec p\cdot\vec E | U=−m⃗⋅B⃗U=-\vec m\cdot\vec B

Coulomb's law | F=14πε0q1q2r2F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2} | No analogous law as magnetic monopoles do not exist

Axial field for a short dipole | 14πε02pr3\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3} along p⃗\vec p | μ04π2mr3\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3} along m⃗\vec m …

Misc Ex.1Example 12.1: Magnetic field of a short dipole on its axis and equator

Worked out. A short magnetic dipole has magnetic moment m=0.5m=0.5 A m2^2. At a distance r=20r=20 cm =0.2=0.2 m from its centre, the axial field is Ba=μ04π2mr3=10−7×2×0.5(0.2)3=1.25×10−5B_a=\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}=10^{-7}\times\dfrac{2\times0.5}{(0.2)^3}=1.25\times10^{-5} T, and the equatorial field is Beq=μ04πmr3=10−7×0.5(0.2)3=6.25×10−6B_{eq}=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}=10^{-7}\times\dfrac{0.5}{(0.2)^3}=6.25\times10^{-6} T -- confirming Ba=2BeqB_a=2B_{eq} at the same distance, exactly as Eq. 12.5 predicts. …

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