A ferry crosses the Hooghly from Howrah to Babughat. The boat turns, rocks and swings round in the current. A passenger holds a small compass on her palm. Whatever the boat does, the red end of the needle swings back to the same direction, towards the north. Nothing touches the needle. What turns it?
Poles
Sprinkle iron filings on a sheet of paper laid over a bar magnet. The filings crowd near the two ends and hardly gather in the middle. The two ends are the poles of the magnet. Hang the magnet from a thread and it settles along a north-south line. The end that points north is the north pole, N; the other end is the south pole, S. Like poles repel each other and unlike poles attract.
Each pole has a pole strength \(m\), measured in A m. The N pole carries \(+m\) and the S pole \(-m\), and the two are always equal in size. The poles do not sit exactly at the ends; they lie a little inside. The distance between them, \(2l\), is the magnetic length. It is about five-sixths of the length a ruler measures.
The magnetic moment of a bar magnet is the vector \(\vec M = m \times 2\vec l\). Here \(m\) is the pole strength and \(2\vec l\) is the vector from the S pole to the N pole. So \(\vec M\) points from S to N inside the magnet, and its size is \(M = 2ml\).
The unit A m² is the unit of the moment \(M = NIA\) of a current loop. The chapter Magnetic Force on Currents and the Magnetic Dipole met it there. This is no accident; the end of this lesson shows that a magnet is a set of current loops. A magnet of pole strength \(10\,\text{A m}\) and magnetic length \(0.1\,\text{m}\) has \(M = 10 \times 0.1 = 1\,\text{A m}^2\).
Cutting a magnet
Break a bar magnet in two, and each piece is a complete magnet with its own N and S. New poles appear at the break. A north pole alone, a magnetic monopole, has never been found, however small the piece. Each pole has the same strength over the same cross-section. So the pole strength goes with the area of the cross-section, and the moment goes with the volume.
Fig 1.1 shows the two cuts. Cut across, each half has the same cross-section, so the same pole strength \(m\), but half the magnetic length: its moment is \(m \times l = M/2\). Cut along, each half has half the cross-section, so pole strength \(m/2\), and the full length: its moment is \((m/2) \times 2l = M/2\). Both halves have half the volume, so both have half the moment.
A bar magnet has a pole strength of \(5\,\text{A m}\) and a magnetic length of \(8\,\text{cm}\). Find its magnetic moment, in A m².
A bar magnet of moment \(M\) and pole strength \(m\) is cut across its length into three equal pieces. What does each piece have?
The field of one pole
A useful model treats each pole as a point source of field, like a point charge. A pole of strength \(m\) makes a field \(B = \dfrac{\mu_0}{4\pi}\dfrac{m}{r^2}\) at a distance \(r\). The field points away from an N pole and towards an S pole. Here \(\mu_0/4\pi = 10^{-7}\,\text{T m/A}\). The model is not the true physics, since single poles do not exist. It still gives the right field outside a magnet, and it is quick.
The line through the two poles is the axial line of the magnet. The line through its centre, perpendicular to the axis, is the equatorial line. The field on each is the sum of the fields of the two poles.
\[ B_N = \dfrac{\mu_0}{4\pi}\dfrac{m}{(r-l)^2} \]
Why. a point on the axis at a distance \(r\) from the centre, beyond the N pole, is \(r - l\) from N. The N field there points away from N.
\[ B_S = \dfrac{\mu_0}{4\pi}\dfrac{m}{(r+l)^2} \]
Why. the same point is \(r + l\) from S, and the S field points towards S, the opposite way along the axis.
\[ B = \dfrac{\mu_0 m}{4\pi}\left[\dfrac{1}{(r-l)^2} - \dfrac{1}{(r+l)^2}\right] \]
Why. the nearer N pole wins, so subtract.
\[ B = \dfrac{\mu_0}{4\pi}\dfrac{4mlr}{(r^2-l^2)^2} \]
Why. put the two terms over one denominator, \((r - l)^2(r + l)^2 = (r^2 - l^2)^2\).
\[ B = \dfrac{\mu_0}{4\pi}\dfrac{2Mr}{(r^2-l^2)^2} \]
Why. \(4ml = 2M\). This is exact for the pole model.
\[ B_{axial} \approx \dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3} \]
Read it. for a short magnet \(l \ll r\), so \((r^2 - l^2)^2 \approx r^4\). The field points from S to N, along \(\vec M\).
On the equatorial line, a point at a distance \(r\) from the centre is \(\sqrt{r^2 + l^2}\) from each pole. Each pole gives a field \(\dfrac{\mu_0}{4\pi}\dfrac{m}{r^2 + l^2}\). The parts of the two fields across the axis cancel. The parts parallel to the axis add, and both point from the N pole towards the S pole. Each part is the whole field times \(l/\sqrt{r^2 + l^2}\).
\[ B_{eq} = \dfrac{\mu_0}{4\pi}\dfrac{2ml}{(r^2 + l^2)^{3/2}} = \dfrac{\mu_0}{4\pi}\dfrac{M}{(r^2 + l^2)^{3/2}} \]
For a short magnet \(l \ll r\), and this becomes
\[ B_{eq} \approx \dfrac{\mu_0}{4\pi}\dfrac{M}{r^3} \]
So the equatorial field is half the axial field at the same distance, and it points against \(\vec M\).
Put the compass on the axis, at \(0°\) or \(180°\), \(14\,\text{cm}\) from the centre. The needle points along \(\vec M\) on both sides, and the field is \(74.8\,\mu\text{T}\). Now move it to \(90°\) or \(270°\). The needle turns round to point against \(\vec M\), and the field is \(35.7\,\mu\text{T}\). The ratio is \(2.1\), not quite 2, because the poles are \(3.2\,\text{cm}\) apart. The short-magnet formulas give \(72.9\) and \(36.4\,\mu\text{T}\), exactly in the ratio 2. Farther out, the two answers meet.
A bar magnet has a pole strength of \(10\,\text{A m}\) and a magnetic length of \(0.1\,\text{m}\). Treat it as a short magnet. Find (a) its moment and (b) the field at \(0.5\,\text{m}\) on its axis. Find (c) the field at \(0.5\,\text{m}\) on its equatorial line, and (d) the ratio of the two fields. Give the directions.
(a) \(M = m \times 2l = 10 \times 0.1 = 1\,\text{A m}^2\).
(b) \(B_{axial}\) \(= 10^{-7} \times 2 \times 1/0.5^3\) \(= 2 \times 10^{-7}/0.125\) \(= 1.6 \times 10^{-6}\,\text{T}\), along \(\vec M\), away from the magnet beyond N.
(c) \(B_{eq}\) \(= 10^{-7} \times 1/0.125\) \(= 8 \times 10^{-7}\,\text{T}\), against \(\vec M\), parallel to the magnet from its N end towards its S end.
(d) \(B_{axial}/B_{eq}\) \(= (2M/r^3)/(M/r^3)\) \(= 2\) at equal distances.
Values used: \(\mu_0/4\pi = 10^{-7}\,\text{T m/A}\).
At a point at a distance \(r\) whose line from the centre makes an angle \(\theta\) with \(\vec M\), a short magnet gives \(B = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}\sqrt{1 + 3\cos^2\theta}\). The field makes an angle \(\alpha\) with that line, where \(\tan\alpha = \tfrac12\tan\theta\). At \(\theta = 0\) this is the axial field, and at \(90°\) the equatorial one. A current loop, far away, has exactly this field, as the chapter Sources of the Magnetic Field shows for its axis.
A short magnet has a moment of \(0.5\,\text{A m}^2\). Find the field it makes at \(20\,\text{cm}\) from its centre on its axis, in microtesla. Take \(\mu_0/4\pi = 10^{-7}\,\text{T m/A}\).
A bar magnet lies along the east-west line with its N pole to the east. Which way does its field point at a place due north of its centre?
A magnet in a uniform field
Put a bar magnet at an angle \(\theta\) to a uniform field \(\vec B\). The N pole feels a force \(mB\) along \(\vec B\), and the S pole an equal force against \(\vec B\). The net force is zero. The two forces act along different lines, so they form a couple. The distance between their lines is \(2l\sin\theta\).
The torque of the couple is \(mB \times 2l\sin\theta = MB\sin\theta\). It turns \(\vec M\) towards \(\vec B\), so in vector form \(\vec\tau = \vec M \times \vec B\). Turning the magnet against this torque stores energy \(U = -\vec M\cdot\vec B = -MB\cos\theta\). These are the torque and energy of a current loop in the chapter Magnetic Force on Currents and the Magnetic Dipole. A magnet is a magnetic dipole of the same kind.
The magnet of the last example, \(M = 1\,\text{A m}^2\), is held at \(30°\) to a uniform field of \(0.2\,\text{T}\). Find (a) the torque on it, (b) its potential energy and (c) the work needed to turn it there from the stable position.
(a) \(\tau = MB\sin\theta = 1 \times 0.2 \times 0.5 = 0.1\,\text{N m}\), turning \(\vec M\) towards \(\vec B\).
(b) \(U\) \(= -MB\cos\theta\) \(= -0.2 \times 0.866\) \(= -0.17\,\text{J}\).
(c) At \(\theta = 0\), \(U = -0.2\,\text{J}\). So \(W\) \(= U(30°) - U(0°)\) \(= -0.173 + 0.2\) \(= 0.027\,\text{J}\).
Now the compass on the ferry. Its needle is a small magnet on a vertical pivot. The Earth's field makes a torque \(MB\sin\theta\) on it, where \(\theta\) is the angle between the needle and the field. The torque turns the needle until \(\theta = 0\), the stable position of least energy. The boat's turning does not enter, because the torque depends only on the needle's angle to the field.
A magnet of moment \(2\,\text{A m}^2\) rests along a uniform field of \(0.05\,\text{T}\). How much work is needed to turn it slowly until its moment points against the field, in joules?
Inside a magnet
Compare the field of a bar magnet with the field of a solenoid, a long coil carrying a current. Outside, the two are the same shape. Field lines leave one end, spread round and return to the other end. The end where the lines leave behaves as an N pole.
Fig 1.2 suggests the answer to what a magnet is. In the 1820s Ampère proposed that every magnet is full of tiny circulating currents. We now know what they are. Each electron in an atom circles its nucleus, a current loop with a moment of about \(10^{-23}\,\text{A m}^2\). The chapter Magnetic Force on Currents and the Magnetic Dipole worked out that moment. Each electron also spins, which gives it a moment of the same size.
In most materials these tiny moments point every way and cancel. In a magnet a large share of them point the same way. Their currents cancel inside, where neighbouring loops run opposite ways. They add round the surface, like the current in the turns of a solenoid. So a magnet is a solenoid whose current is made by its atoms.
A bar magnet has a moment of \(1\,\text{A m}^2\). A solenoid of the same size has \(500\) turns, each of area \(1\,\text{cm}^2\). What current must it carry to have the same moment?
A solenoid's moment is \(M = NIA\), so \(I\) \(= M/NA\) \(= 1/(500 \times 10^{-4})\) \(= 20\,\text{A}\).
That is a large current for a thin coil, and it would heat the wire. A steel magnet keeps the same moment with no battery at all.
A loop of current has no end. So the field lines of a magnet, like those of a loop, never start or stop anywhere. Inside the magnet they run from S to N and close up. For the same reason a magnet has no single pole: cut it anywhere and each piece is still a set of loops. This is Gauss's law for magnetism. The chapter Electric Flux and Gauss's Law gives the idea of flux that it uses.
The net magnetic flux through any closed surface is zero: \(\oint \vec B \cdot d\vec A = 0\). Every field line that enters a closed surface also leaves it, so there is no isolated magnetic pole.
Which statements about a bar magnet are true? Choose all that apply.
In 1269 a French scholar, Petrus Peregrinus, was with an army besieging the town of Lucera in southern Italy. He wrote a letter to a friend about the lodestone, a natural magnet. He shaped one into a sphere and traced the lines along which a needle pointed. They met at two points, which he named the poles. He noted that like poles repel. He also found that a broken lodestone gives two complete magnets, each with two poles. His letter is the oldest known account of these facts.
A flat fridge magnet from a Mumbai souvenir shop is a sheet of rubber filled with a magnetic powder. It is not one magnet but many narrow strips, magnetised N, S, N, S side by side, a few millimetres apart. On the back face the fields of the strips add; on the front face they largely cancel. So the magnet grips the steel door strongly and hardly pulls a pin on its printed side.
- Writing the magnetic moment as pole strength times half the length.\(M = m \times 2l\), with \(2l\) the whole distance between the poles.
- Giving the equatorial field the direction of \(\vec M\).On the axis the field is along \(\vec M\); on the equatorial line it points against \(\vec M\).
- Thinking that cutting a magnet across halves its pole strength.Pole strength goes with the cross-section. A cut across keeps \(m\) and halves the length; a cut along halves \(m\).
- Saying that field lines start on the N pole and end on the S pole.Magnetic field lines are closed. Inside the magnet they run from S back to N.