Permanent Magnets
Poles, bar magnets, the Earth's field and the instruments that measure them.
Poles & Bar Magnets
Magnets pictured as pairs of magnetic charges.
A small current loop and an electric dipole obey equations of the same shape. That suggests a handy picture: magnetic “poles” that behave like magnetic charges. It explains bar magnets beautifully.
The magnetic pole model
A small current loop makes a field (μ₀/4π)(2μ⃗/r³) on its axis and feels a torque μ⃗ × B⃗; an electric dipole makes (1/4πε₀)(2p⃗/r³) and feels p⃗ × E⃗. Swap p⃗ for μ⃗ and 1/4πε₀ for μ₀/4π and they match. So we may imagine:
- Magnetic charges (poles) of two kinds. A pole of strength m in a field B⃗ feels F⃗ = mB⃗: along B⃗ for a positive pole, against it for a negative one.
- A pole m produces a radial field B = (μ₀/4π)(m/r²).
- A magnetic dipole is −m and +m a small distance d apart, with moment md pointing from −m to +m.
- A current loop of area A carrying i is equivalent to a dipole with md = iA, along the loop’s area vector.
A solenoid as two poles
Replace each turn of a solenoid by a tiny dipole, choosing the separation so the N of one touches the S of the next. Inside they cancel in pairs, leaving one S pole and one N pole of strength m at the two ends, a solenoid’s length apart. This gives the field outside the solenoid correctly. (Inside, the real field runs from S to N, opposite to the pole picture: magnetic field lines are closed loops, never starting or ending like electric ones.)
A solenoid 12 cm long and 1.5 cm in radius has 300 turns carrying 5.0 A. Find the field on its axis 12 cm from its centre.
Why a bar magnet is a magnet
Atoms contain circulating electrons: tiny current loops. In a magnet these loops are lined up, all circulating the same way. Inside the bar, neighbouring loops cancel; at the surface nothing cancels them, leaving a net surface current. So a magnet is equivalent to a tightly wound solenoid. If the surface current is I per unit length and the cross-section is A, the pole strength is m = IA.
End effects put the poles slightly inside the bar. The distance between them is the magnetic length; magnetic length ÷ geometric length ≈ 0.84. Writing the magnetic length as 2l:
Torque and potential energy
In a uniform field B at angle θ, the forces ±mB on the poles form a couple of moment 2ml B sin θ:
Turning it slowly from θ₀ to θ takes work ∫MB sin θ dθ = MB(cos θ₀ − cos θ), stored as potential energy. Taking U = 0 at θ = 90°:
A magnet of moment 0.80 A m² can turn in a horizontal field of 5.0 × 10⁻⁵ T. How much work turns it slowly from along the field to 120° from it?
- 1Pole model: F⃗ = mB⃗ (36.1), B = (μ₀/4π)m/r² (36.2); N = positive pole.
- 2A current loop ≡ dipole with md = iA; a solenoid ≡ two poles at its ends.
- 3Bar magnets: aligned atomic current loops; m = IA (36.3); M = 2ml (36.4); magnetic length ≈ 0.84 × geometric length.
- 4Γ⃗ = M⃗ × B⃗; U = −M⃗·B⃗ (36.5).
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The Field of a Bar Magnet
Two poles, two simple positions, one general formula.
With the pole picture, the field of a bar magnet is just the field of two point poles. Two positions have neat formulas, and far away every magnet becomes a dipole.
End-on position
A point on the magnet’s axis, at distance d from its centre (magnetic length 2l, pole strength m). The north pole gives μ₀m/4π(d − l)² away from the magnet; the south pole gives μ₀m/4π(d + l)² toward it:
A magnet has magnetic length 8.0 cm and pole strength 15 A m (150 CGS units, where 1 CGS unit = 0.1 A m). Find B on its axis 15 cm from its centre.
Broadside-on position
A point on the perpendicular bisector, at distance d from the centre: each pole is √(d² + l²) away. The components along the bisector cancel and those parallel to the axis add, giving a field parallel to the magnet, pointing from N toward S:
For the same magnet, find B at 15 cm on its perpendicular bisector.
Magnetic scalar potential
As with gravity and electrostatics, define a potential whose fall gives the field:
For a dipole, a point at distance r ≫ l and angle θ from the axis is about r − l cos θ from N and r + l cos θ from S, so
The general dipole field
Differentiating V along r and along θ gives the two components:
Find the field of a dipole of moment 2.0 A m² at 50 cm from it, in a direction at 30° to its axis.
- 1End-on: B = (μ₀/4π)2Md/(d² − l²)² → (μ₀/4π)2M/d³.
- 2Broadside-on: B = (μ₀/4π)M/(d² + l²)^(3/2) → (μ₀/4π)M/d³, parallel to the axis.
- 3V = (μ₀/4π)m/r; dipole V = (μ₀/4π)M cos θ/r².
- 4Dipole: B = (μ₀/4π)(M/r³)√(1 + 3cos²θ), tan α = ½ tan θ.
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Earth's Magnetism
The planet as a tilted bar magnet.
The Earth is a magnet. Its field is roughly that of a giant bar magnet at the centre, tilted from the spin axis, and three numbers describe it at any place.
A dipole inside the Earth
Near the surface the field is roughly that of a dipole of moment about 8.0 × 10²² J/T at the centre, its axis about 11.5° from the rotation axis. Where that axis meets the surface are the geomagnetic north pole(near the geographic north pole) and the geomagnetic south pole. There the field is vertical: a freely hung magnet stands straight up and down (north end down at the geomagnetic north pole). That is how the geomagnetic poles are defined.
A hung magnet’s north end is attracted to the dipole’s south pole, so the imaginary bar magnet inside the Earth has its south pole toward the geomagnetic north.
The elements of the Earth’s field
Declination
The geographic meridian at a point P is the vertical plane through P and the geographic poles; the magnetic meridian is the vertical plane through P and the geomagnetic poles. The Earth’s field at P lies in the magnetic meridian. The angle between the two meridians is the declination. A compass needle settles in the magnetic meridian, so navigators must correct its “north” by the declination.
Inclination or dip
The angle the field makes with the horizontal, in the magnetic meridian, is the dip δ. In the magnetic northern hemisphere the field slopes downward, so a freely hung needle’s north end dips.
Horizontal component
BH is the field’s horizontal component, pointing to magnetic north. Declination and dip fix the field’s direction; with BH they fix its size too:
At a place, BH = 3.0 × 10⁻⁵ T and the dip is 50°. Find the total field.
Measuring dip
A dip circle is a magnetic needle pivoted at the centre of a vertical circular scale (0°–0° horizontal, 90°–90° vertical), in a glass case that turns about a vertical axis over a horizontal scale.
- Level the base. Turn the case until the needle stands vertical (90°–90°): the scale’s plane is now perpendicular to the magnetic meridian, so BH can’t turn the needle and only BV acts.
- Turn the case through exactly 90°: the scale now lies in the magnetic meridian, the needle lies along B⃗, and its ends read the dip.
Errors and their cures: read both ends of the needle (off-centre pivot); rotate the circle by 180° (0–0 line not horizontal); turn the needle over on its bearing (magnetic axis ≠ geometric axis); remagnetize it the other way (centre of mass off the pivot). Averaging all 16 readings gives the true dip.
Apparent dip
If the scale’s plane is at angle α to the meridian, only BH cos α lies in it, while BV is unchanged, so the needle shows an apparent dip δ′:
With the dip circle at 60° to the magnetic meridian, the needle shows 45°. Find the true dip.
Apparent dips in two perpendicular vertical planes are 40° and 55°. Find the true dip.
Neutral points
Near a magnet there may be points where the magnet’s horizontal field exactly cancels BH. There a compass needle feels no horizontal field and stays in any position: a neutral point.
- 1Earth ≈ dipole of 8 × 10²² J/T, tilted ≈ 11.5°; its south pole points to geomagnetic north.
- 2Elements: declination, dip δ, horizontal component B_H = B cos δ.
- 3Dip circle in the meridian reads δ; apparent dip cot δ′ = cot δ cos α; cot²δ′ + cot²δ″ = cot²δ (36.16).
- 4Neutral point: the magnet's horizontal field cancels B_H.
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Galvanometers
Measuring current with a compass or a twisting coil.
Two ways to turn a magnetic effect into a current reading: let a compass needle balance a coil’s field against the Earth’s, or let a coil twist against a spring.
The tangent law
A compass needle normally lies along BH. Add a horizontal field B perpendicular to BH: the needle turns to lie along the resultant √(B² + BH²), at angle θ from BH where tan θ = B/BH.
The tangent galvanometer
A vertical circular coil (n turns, radius r) on a levelled base, with a small compass needle at its centre carrying a light aluminium pointer at right angles to it, over a scale marked 0°–90° in each quadrant. A mirror under the pointer removes parallax.
Turn the base until the coil’s plane contains the needle (so it lies in the magnetic meridian) and the pointer reads 0°–0°. Pass the current: the coil’s field μ₀ni/2r at the centre is horizontal and perpendicular to the meridian, so by the tangent law the needle deflects by θ with
The needle should be short compared with the coil, since μ₀ni/2r holds only at the centre.
Sensitivity. Differentiating i = K tan θ: di/i = 2dθ/sin 2θ, so dθ = ½ sin 2θ (di/i). A given fractional change in current moves the needle most when sin 2θ = 1, i.e. near 45°.
A tangent galvanometer has 50 turns of radius 10 cm, where BH = 3.5 × 10⁻⁵ T. What current gives a deflection of 45°?
The moving-coil galvanometer
A rectangular coil of many turns, wound on a soft-iron core, hangs between the poles of a strong permanent magnet by a fine phosphor-bronze strip (from a torsion head), with a phosphor-bronze spring below. A small mirror on the strip reflects a light beam onto a scale. Current enters and leaves through the strip and spring.
The pole pieces are curved (cylindrical) so the field at the coil’s sides always lies in the coil’s plane: the deflecting torque is niAB whatever the angle. The twisted strip pushes back with kθ. At equilibrium:
The soft-iron core strengthens B and so raises the sensitivity (why soft iron does this comes in the next chapter).
A galvanometer coil has 50 turns of area 1.0 cm × 2.0 cm in a field of 0.25 T, on a strip with k = 5.0 × 10⁻⁷ N m/rad. What deflection does 10 µA give?
Protecting it with a shunt
A galvanometer is delicate; too much current damages it. A small shunt resistance Rs in parallel diverts most of the current, leaving
A galvanometer coil of 50 Ω gives full-scale deflection at 5.0 mA. What shunt lets a total of 1.0 A pass at full scale?
- 1Tangent law: B = B_H tan θ (36.20).
- 2Tangent galvanometer: i = K tan θ, K = 2rB_H/μ₀n (36.17); most sensitive near 45°.
- 3Moving-coil: niAB = kθ, i ∝ θ (36.18); sensitivity nAB/k.
- 4Shunt: i_g = R_s i/(R_s + R_g) (36.19).
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Magnetometers
M/B_H from deflection, MB_H from oscillation.
Two simple instruments, one measuring M/BH and one measuring MBH, together pin down both the strength of a magnet and the Earth’s horizontal field.
The deflection magnetometer
A small compass with a pointer over a 0°–90° scale (with a mirror against parallax), in a box mounted on a wooden frame with two long arms carrying metre scales that read distance from the needle’s centre. Its job is to measure M/BH for a bar magnet, in one of two standard positions.
Tan-A position
Arms along magnetic east–west, pointer at 0°–0°. Lay the magnet along an arm, centre at distance d. The needle is end-on, so the magnet’s field (east or west) is perpendicular to BH; the tangent law gives
Tan-B position
Arms along magnetic north–south; the magnet lies across an arm, symmetric about it. Now the needle is broadside-on, the field again east–west:
Errors are removed by averaging 16 readings: both ends of the pointer (off-centre pivot), the magnet reversed end for end (magnetic centre ≠ geometric centre), turned upside down (magnetic axis ≠ geometric axis), and placed on the other arm (scale zero ≠ needle centre).
A magnet of magnetic length 8.0 cm, its centre 20 cm from the needle in Tan-A position, gives a deflection of 45°. Find M/BH.
Uses
- Comparing two magnets by the null method: put one magnet on each arm and adjust distances until the needle doesn’t deflect; their fields at the needle are then equal. In Tan-B with short magnets, M₁/M₂ = d₁³/d₂³. No angle is measured, so it is both easier and more sensitive.
- Testing the inverse-square law: in Tan-A, cot θ ∝ 2d/(d² − l²)², so plotting cot θ against (d² − l²)²/2d should give a straight line through the origin (in Tan-B, against (d² + l²)^(3/2)). Since these formulas come from B ∝ m/r² for a pole, the straight line verifies it.
- Comparing BH at two places: same magnet, same d: BH1/BH2 = tan θ₂/tan θ₁.
Two short magnets on opposite arms in Tan-B position give no deflection when they are 20 cm and 25 cm from the needle. Compare their moments.
The oscillation magnetometer
A glass-walled box with a mirror on its base and a reference line along the middle; a hanger on an unspun silk thread from a torsion head. Level it, turn the box so the reference line lies in the magnetic meridian (check with a compass), take the twist out of the thread with a brass bar, then hang the magnet, north end north, and set it swinging gently.
Displaced by θ, the magnet feels a restoring torque −MBH sin θ ≈ −MBHθ (ignoring the thread’s twist). With moment of inertia I, that is angular SHM:
A compass needle makes 20 oscillations a minute where the dip is 30°, and 24 a minute where the dip is 50°. Compare the total field at the two places.
Finding M and the horizontal field separately
With X = M/BH from a deflection magnetometer and Y = MBH from an oscillation magnetometer:
The magnet above (M/BH = 3.69 × 10⁴ A m²/T) has I = 2.9 × 10⁻⁵ kg m² and oscillates with T = 5.0 s. Find M and BH.
- 1Deflection magnetometer: M/B_H from Tan-A (36.21) or Tan-B (36.22) with the tangent law.
- 2Null method compares moments; cot θ plots verify the inverse-square law; tan θ ratios compare B_H.
- 3Oscillation magnetometer: T = 2π√(I/MB_H) (36.23).
- 4M = √(XY), B_H = √(Y/X).
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Gauss's Law for Magnetism
Magnetic field lines never start or end.
Electric field lines begin and end on charges. Magnetic field lines never begin or end at all, and that one fact is a law of nature.
From poles to a flux law
Coulomb’s law E = q/4πε₀r² leads to Gauss’s law ∮E⃗·dS⃗ = qinside/ε₀. In exactly the same way, the pole law B = (μ₀/4π)m/r² would give ∮B⃗·dS⃗ = μ₀minside, the net pole strength enclosed.
But no isolated magnetic pole has ever been found. Cut a magnet in half and you get two magnets, each with both poles. The smallest source of magnetic field is a dipole, whose net pole strength is zero. So the net pole strength inside any closed surface is zero, and:
Draw a closed surface round one charge of an electric dipole: field lines only leave it, so the flux is not zero. Draw one round one end of a bar magnet: lines leave through part of the surface and enter through another part, because inside the magnet they continue from S back to N. As many go in as come out.
- 1No isolated magnetic poles: the smallest magnetic source is a dipole.
- 2∮B⃗·dS⃗ = 0 for every closed surface (36.25).
- 3Magnetic field lines form closed loops; electric field lines start and end on charges.