Chapter 5: Magnetism and Matter

Cutting a bar magnet is fundamentally different from cutting an electric dipole, because the magnetism of a magnet arises from the alignment of countless tiny atomic current loops (atomic dipoles) spread throughout its volume — not from separated 'magnetic charges'.

(i) Cut transverse to its length: Each of the two shorter pieces is itself a complete magnet. On the piece that contains the original north end, the freshly exposed cut face behaves as a new south pole; on the other piece, the exposed cut face behaves as a new north pole. So each piece has both an N pole and an S pole.

(ii) Cut along its length: Again, each thinner piece is a complete magnet, with both a north pole and a south pole.

In either case one obtains two magnets, each having a north pole and a south pole. An isolated single pole (a magnetic monopole) can never be produced this way.

In a uniform magnetic field the forces on the north and south poles of a dipole are equal in magnitude and opposite in direction, so the net force is zero; only a torque $$\vec{\tau} = \vec{m}\times\vec{B}$$ acts. This is the situation for the magnetised needle placed in the uniform field.

An iron nail placed near a bar magnet sits in the magnet's own field, which is non-uniform (it grows stronger closer to the magnet). The nail is unmagnetised iron, but the magnet's field induces a magnetic moment in it: the nail end facing the magnet's north pole becomes an induced south pole, and the far end becomes an induced north pole.

Because the field is non-uniform, the induced south pole (which is nearer the magnet's north pole) lies in a stronger field than the induced north pole (which is farther away). The attractive force on the near (induced S) pole therefore exceeds the repulsive force on the far (induced N) pole, giving a net force of attraction. The two poles also form a couple, so the nail experiences a torque as well.

No — not every magnetic configuration has a distinct north and south pole. Recognisable N and S poles appear only when the source has a net non-zero magnetic (dipole) moment, as a bar magnet does.

Consider a toroid (a solenoid bent round into a closed ring). Its magnetic field is entirely confined within the body of the toroid, forming closed loops; there is no field outside and no point from which the field lines start or end. Hence a toroid has no north or south pole at all.

The same is true of the field of a long straight current-carrying conductor: the field lines are concentric circles around the wire, with no poles anywhere.

Step 1 — Are both bars magnetised? Bring an end of bar A close to an end of bar B and try the various pole-to-pole combinations. If in any orientation a repulsion is observed, then both bars are magnetised — because repulsion can occur only between two like poles, which requires both bars to be magnets. (An unmagnetised iron bar is only ever attracted to a magnet, never repelled.) If every relative orientation gives only attraction, then exactly one of the two bars is unmagnetised.

Step 2 — If only one is magnetised, which one? Use the fact that in a bar magnet the field is strongest at the two ends (poles) and weakest near the middle, whereas a magnet attracts a piece of plain iron equally well anywhere along its length.

Take one bar, say A, and bring one of its ends first near an end of B and then near the middle of B:

• If A is pulled strongly near B's end but feels no force near the middle of B, then B is the magnet (its field vanishes at the neutral central region).
• If A is attracted with no change from B's end to B's middle, then A is the magnet (A's pole attracts the plain iron bar B equally everywhere).

This locates the magnetised bar using nothing but the two bars themselves.

The needle P at O has its magnetic moment $$\vec{m}_P$$ pointing vertically upward. P sets up a magnetic field whose form depends on direction:

On the axis of P (vertically above or below O), the field is parallel to $$\vec{m}_P$$:

$$\vec{B}_P = \dfrac{\mu_0}{4\pi}\dfrac{2\vec{m}_P}{r^3}$$

On the normal bisector of P (horizontally to the left or right of O), the field is anti-parallel to $$\vec{m}_P$$:

$$\vec{B}_P = -\dfrac{\mu_0}{4\pi}\dfrac{\vec{m}_P}{r^3}$$

A needle Q is in equilibrium only when its moment $$\vec{m}_Q$$ is along $$\vec{B}_P$$ (parallel ⇒ stable) or opposite to it (anti-parallel ⇒ unstable). If $$\vec{m}_Q$$ is perpendicular to $$\vec{B}_P$$, a torque $$\tau = m_Q B_P\sin\theta$$ acts and there is no equilibrium.

Q1 and Q2 lie on the normal bisector, where $$\vec{B}_P$$ points vertically (downward). Their moments point horizontally — i.e. perpendicular to $$\vec{B}_P$$ — so a non-zero torque acts on them.

The system is not in equilibrium in configurations PQ1 and PQ2.

Equilibrium occurs when $$\vec{m}_Q$$ lies along $$\vec{B}_P$$ (parallel ⇒ stable, lowest energy) or opposite to $$\vec{B}_P$$ (anti-parallel ⇒ unstable, highest energy). Recall that $$\vec{m}_P$$ points up; on P's axis $$\vec{B}_P$$ is parallel to $$\vec{m}_P$$ (points up), while on P's normal bisector $$\vec{B}_P$$ is anti-parallel to $$\vec{m}_P$$ (points down).

Q3: on the normal bisector ⇒ $$\vec{B}_P$$ points down; $$\vec{m}_{Q3}$$ points down ⇒ parallel ⇒ stable.

Q6: on the axis (below P) ⇒ $$\vec{B}_P$$ points up; $$\vec{m}_{Q6}$$ points up ⇒ parallel ⇒ stable.

Q4: on the axis (above P) ⇒ $$\vec{B}_P$$ points up; $$\vec{m}_{Q4}$$ points down ⇒ anti-parallel ⇒ unstable.

Q5: on the normal bisector ⇒ $$\vec{B}_P$$ points down; $$\vec{m}_{Q5}$$ points up ⇒ anti-parallel ⇒ unstable.

Therefore: (i) stable equilibrium in PQ3 and PQ6; (ii) unstable equilibrium in PQ5 and PQ4.

The potential energy of needle Q in P's field is $$U = -\vec{m}_Q\cdot\vec{B}_P$$. It is most negative (lowest) when $$\vec{m}_Q$$ is parallel to $$\vec{B}_P$$ — that is, in a stable configuration. The two stable configurations are PQ3 and PQ6, so the lowest energy is one of these.

To choose between them, compare the strength of $$\vec{B}_P$$ at the two locations, both at the same distance $$r$$ from O:

• Q3 is on the normal bisector: $$B_P = \dfrac{\mu_0}{4\pi}\dfrac{m_P}{r^3}$$
• Q6 is on the axis: $$B_P = \dfrac{\mu_0}{4\pi}\dfrac{2m_P}{r^3}$$

The axial field is twice the equatorial field at the same distance. For a stable configuration $$U = -m_Q B_P$$, so the larger value of $$B_P$$ gives the more negative (lower) energy.

Hence PQ6 has the lowest potential energy among all the configurations shown.

No. Magnetic field lines give the direction along which a small compass needle aligns, but they do not represent the lines of force on a moving charged particle.

The force on a charge $$q$$ moving with velocity $$\vec{v}$$ in a field $$\vec{B}$$ is

$$\vec{F} = q\,\vec{v}\times\vec{B}$$

Because of the cross product, $$\vec{F}$$ is always perpendicular to $$\vec{B}$$ (and to $$\vec{v}$$). The force is therefore never directed along the field line. For this reason it is misleading to call magnetic field lines 'lines of force'.

Gauss's law of magnetism states that the net magnetic flux through any closed surface is zero:

$$\oint_S \vec{B}\cdot d\vec{S} = 0$$

This holds precisely because isolated magnetic poles (monopoles) do not exist — there is no magnetic 'charge' to act as a source or sink of $$\vec{B}$$.

If magnetic monopoles did exist, the law would be modified — in analogy with Gauss's law of electrostatics — so that the right-hand side equals the enclosed magnetic charge:

$$\oint_S \vec{B}\cdot d\vec{S} = \mu_0\, q_m$$

where $$q_m$$ is the total magnetic (monopole) charge enclosed by the surface $$S$$.

A bar magnet exerts no torque on itself. A system cannot exert a net force or torque on itself through the field it produces — the internal forces between its parts cancel in pairs (Newton's third law). So a bar magnet feels no torque due to its own field.

One element of a current-carrying wire does, in general, exert a force on another element of the same wire. The field produced by one current element acts on a different element, so there is an internal force (and torque) between elements of the same wire. No element, however, exerts a force on itself. For the special case of a straight wire, the net force on any element due to the rest of the wire works out to be zero.

Yes. A magnetic moment does not require a net charge — it requires charges in motion (current loops).

In a system the total (net) electric charge can be zero, yet the individual charges may still circulate and constitute current loops, each with its own magnetic moment. If these elementary moments do not all cancel, the system possesses a net magnetic moment despite its zero net charge.

A real example is a paramagnetic material: its atoms are electrically neutral (net charge zero), but the orbital and spin motions of their electrons give each atom a net magnetic dipole moment.

The work done by an external torque in turning the magnet from an initial orientation $$\theta_1$$ to a final orientation $$\theta_2$$ equals the change in its potential energy ($$U = -mB\cos\theta$$):

$$W = U(\theta_2) - U(\theta_1) = -mB\cos\theta_2 - (-mB\cos\theta_1) = mB(\cos\theta_1 - \cos\theta_2)$$

Initially the magnet is aligned with the field, so $$\theta_1 = 0^\circ$$ and $$\cos\theta_1 = 1$$:

$$W = mB(1 - \cos\theta_2)$$

Given: $$m = 1.5\,\mathrm{J\,T^{-1}}$$, $$B = 0.22\,\mathrm{T}$$, so $$mB = 1.5\times 0.22 = 0.33\,\mathrm{J}$$.

(i) Moment normal to the field ($$\theta_2 = 90^\circ$$, $$\cos 90^\circ = 0$$):

$$W = mB(1 - 0) = mB = 0.33\,\mathrm{J}$$

(ii) Moment opposite to the field ($$\theta_2 = 180^\circ$$, $$\cos 180^\circ = -1$$):

$$W = mB\big(1 - (-1)\big) = 2mB = 2\times 0.33 = 0.66\,\mathrm{J}$$

The torque on the magnet in the field is

$$\tau = mB\sin\theta$$

with $$mB = 1.5\times 0.22 = 0.33\,\mathrm{J}$$.

(i) Moment normal to the field ($$\theta = 90^\circ$$):

$$\tau = mB\sin 90^\circ = 0.33\times 1 = 0.33\,\mathrm{N\,m}$$

This torque is a maximum, and it tends to rotate the magnet back into alignment with the field.

(ii) Moment opposite to the field ($$\theta = 180^\circ$$):

$$\tau = mB\sin 180^\circ = 0.33\times 0 = 0$$

The torque is zero — this anti-parallel orientation is an equilibrium position (an unstable one).

The magnetic moment of a closely wound solenoid (or coil) of $$N$$ turns, each of cross-sectional area $$A$$, carrying current $$I$$ is

$$m = N I A$$

Given: $$N = 2000$$, $$A = 1.6\times 10^{-4}\,\mathrm{m^2}$$, $$I = 4.0\,\mathrm{A}$$.

$$m = 2000\times 4.0\times (1.6\times 10^{-4})$$

$$m = 8000\times 1.6\times 10^{-4}$$

$$m = 1.28\,\mathrm{J\,T^{-1}}$$

The moment is directed along the axis of the solenoid.

Force. The applied field is uniform. In a uniform field the forces on the two ends (poles) of the dipole are equal in magnitude and opposite in direction, so the net force is zero.

$$F = 0$$

Torque. The torque on the dipole is

$$\tau = mB\sin\theta$$

Given: $$m = 1.28\,\mathrm{J\,T^{-1}}$$ (from part a), $$B = 7.5\times 10^{-2}\,\mathrm{T}$$, $$\theta = 30^\circ$$.

$$\tau = (1.28)(7.5\times 10^{-2})\sin 30^\circ$$

$$\tau = (1.28)(7.5\times 10^{-2})(0.5)$$

$$\tau = 4.8\times 10^{-2}\,\mathrm{N\,m}$$