The bar magnet
Poles and field lines
Place a bar magnet under a sheet of paper and sprinkle iron filings on top. Tap the paper gently. The filings arrange themselves into curves that run from one end of the magnet to the other.
The filings crowd most thickly near the two ends. These are the poles. Hang the magnet from a thread and it turns until one end points roughly north. That end is the north pole, N. The other is the south pole, S.
Bring two magnets together. Like poles repel each other, and unlike poles attract.
Now cut the magnet in half. Do you get a lone N and a lone S? No. Each piece is a complete magnet, with its own N and S pole. Cut again, and again: the same thing happens. An isolated magnetic pole, a monopole, has never been found.
Now put a small compass near the magnet. The needle settles along a definite direction. Move the compass a little in that direction, again and again, and join the points. You have drawn a magnetic field line.
The tangent to a field line at any point gives the direction of the magnetic field B at that point. That is the direction a compass needle points.
The closer together the lines, the stronger the field. Near the poles they are packed tightly. Far away they spread out, and the field is weak.
Two field lines never cross. If they did, a compass placed at the crossing would have to point in two directions at once.
Outside the magnet, the lines leave the north pole and enter the south pole. But they do not stop there. Inside the magnet they carry on, from S back to N. Every magnetic field line is a closed loop.
Compare the electric dipole: its lines begin on the positive charge and end on the negative charge. Magnetic field lines have no beginning and no end.
A bar magnet is like a solenoid
Here are the iron filings around a bar magnet, and here are the filings around a solenoid carrying a current. The two patterns are almost the same. Ampère suggested why: a bar magnet behaves like a solenoid.
Cut the solenoid in half, and each half is a weaker solenoid with two ends of its own, just like a cut magnet. Hang it freely, and it settles along north and south, just like a magnet.
Each turn of the solenoid is a current loop. A loop of radius a carrying current I has magnetic moment I times πa². A solenoid of length 2l with n turns per unit length has n × 2l turns, so its total magnetic moment is m = n(2l)I(πa²).
The magnetic moment m points along the axis, from the S end to the N end. Its unit is ampere square metre, A m², which is the same as joule per tesla, J T⁻¹.
Far from the solenoid, at a distance r along its axis, the field works out to B = (μ₀/4π) × 2m/r³. Measure the field of a bar magnet on its axis, and you get exactly the same formula.
On the equator, the line through the middle at right angles to the axis, the field is half as large and points opposite to m: B = −(μ₀/4π) × m/r³.
These are the electric dipole formulas with p replaced by m, E by B, and 1/ε₀ by μ₀. That is why a small bar magnet is called a magnetic dipole.
Worked example. A bar magnet 5 cm long has magnetic moment 0.40 A m². Find the field 50 cm from its centre. 50 cm is ten times the length of the magnet, so the dipole formulas can be used.
On the equator: B = 10⁻⁷ × 0.40 ÷ (0.5)³. Now 0.5³ is 0.125, so B = 3.2 × 10⁻⁷ tesla.
On the axis, the field is twice as large: 6.4 × 10⁻⁷ tesla. The dipole field falls off fast, as 1/r³. Go twice as far, and it drops to one-eighth.
Try it: compass and field lines
- Drag the compasses around. Each needle lines up along the field line through its centre, its red end pointing along B.
- Tick “Iron filings”: the filings show the same pattern as the field lines.
- Switch to “As a solenoid”. The field outside is the same; inside, B runs from S to N in both.
- Move a compass far out along the axis, then onto the equator at the same distance. The field on the axis is about twice as strong.
- Cut the magnet in two. Each half has its own N and S.
A magnetic dipole in a uniform field
Place a compass needle in a uniform magnetic field B, with its moment m at an angle θ to the field. The N end is pushed along B, and the S end is pushed the opposite way.
The two forces are equal and opposite, so the net force is zero. But they do not act along the same line. They form a couple, and the couple turns the needle.
The torque is τ = m × B. Its magnitude is mB sin θ. It is largest when the needle is at right angles to the field, and zero when the needle is along the field.
Let the needle go. It swings past the field direction, comes back, and swings again. For small angles, sin θ is almost equal to θ, so I d²θ/dt² = −mBθ. That is simple harmonic motion.
Compare with the SHM equation: ω² = mB/I. So the period is T = 2π√(I/mB). A stronger field, or a stronger magnet, makes the needle swing faster.
Worked example. A needle has m = 6.7 × 10⁻² A m² and moment of inertia I = 7.5 × 10⁻⁶ kg m². It makes 10 complete oscillations in 6.70 s, so its period is 0.670 s.
Rearrange the formula: B = 4π²I/(mT²). That is 4π² × 7.5 × 10⁻⁶, divided by 6.7 × 10⁻² × 0.670². The answer is about 0.01 tesla.
Turning the needle against the torque takes work, so it stores potential energy: U = −m·B = −mB cos θ. We take U = 0 when the needle is at 90° to the field.
At θ = 0°, U = −mB, its lowest value. This is stable equilibrium: nudge the needle and it comes back. At θ = 180°, U = +mB, the highest value. This is unstable: the slightest nudge, and the needle flips round.
So turning a magnet from along the field to opposite the field takes work 2mB. For m = 1.5 J T⁻¹ in a field of 0.22 T, that is 2 × 1.5 × 0.22 = 0.66 joule.
Try it: a needle in a uniform field
- Drag the needle to any angle and let go. It oscillates about the field direction.
- Watch the torque arrow: biggest at 90°, zero along the field.
- Compare the measured period with T = 2π√(I/mB). With the starting values it is about 0.67 s, as in the worked example.
- Make B four times larger. The period halves.
- Set the needle at 180° and release it. It is in equilibrium, but at the top of the energy curve: unstable.
Find the moment
A short bar magnet placed with its axis at 30° to a uniform magnetic field of 0.25 T experiences a torque of 4.5 × 10⁻² N m. What is its magnetic moment?
- 0.18 J T⁻¹
- 0.21 J T⁻¹
- 0.36 J T⁻¹
- 0.72 J T⁻¹
Answer
C. τ = mB sin θ, so m = τ/(B sin θ) = 4.5 × 10⁻² / (0.25 × sin 30°) = 4.5 × 10⁻² / 0.125 = 0.36 J T⁻¹. (0.18 forgets sin θ; 0.21 uses cos 30° instead of sin 30°.)
A stronger field
A compass needle oscillates in a uniform magnetic field with period T. The field is made 4 times stronger. What is the new period?
- 4T
- 2T
- T/2
- T/4
Answer
C. T = 2π√(I/mB), so T ∝ 1/√B. Making B four times larger divides the period by √4 = 2: the new period is T/2. A stronger field gives a stronger restoring torque, so the needle swings faster.
Summary
- Field of a bar magnet, far away:
axis: B = (μ₀/4π)(2m/r³); equator: B = −(μ₀/4π)(m/r³) - Torque in a uniform field:
τ = m × B, |τ| = mB sin θ - Small oscillations:
T = 2π√(I/mB) - Potential energy:
U = −m·B = −mB cos θ
- Every magnet has a north and a south pole; cutting it gives two magnets. Isolated poles (monopoles) do not exist.
- Magnetic field lines are closed loops: out of N, into S outside the magnet, and S to N inside. They never cross; their tangent gives B and their crowding gives its strength.
- A bar magnet behaves like a solenoid of moment m = n(2l)I(πa²). Far away, its field is the electric dipole field with p → m, E → B, 1/ε₀ → μ₀.
- In a uniform field a dipole feels a torque but no net force. θ = 0° is stable (U = −mB), θ = 180° is unstable (U = +mB).