The Workbook

\(M = m \times 2l = 10 \times 0.1 = 1\) A m\(^2\). Axial: \(B = 10^{-7} \times \dfrac{2 \times 1}{0.125} = 1.6 \times 10^{-6}\) T. Equatorial: \(B = 10^{-7} \times \dfrac{1}{0.125} = 8 \times 10^{-7}\) T.

\(\tau = MB\sin\theta = 1 \times 0.2 \times 0.5 = 0.1\) N m. \(U = -MB\cos\theta = -0.2 \times 0.866 \approx -0.17\) J.

1. At \(A\), on the axis beyond the N pole, the field points away from the magnet, along \(\vec M\). At \(E\), on the equatorial line, the field runs from the N pole round to the S pole, opposite to \(\vec M\).

2. \(\dfrac{B_A}{B_E} = \dfrac{2M/r^3}{M/r^3} = 2\).

\(T = 2\pi\sqrt{\dfrac{I}{MB_H}} = 2\pi\sqrt{\dfrac{4 \times 10^{-6}}{1 \times 4 \times 10^{-5}}} = 2\pi\sqrt{0.1} \approx 2\) s.

(i) Cut across: each piece has the same pole strength \(m\) but half the length, so \(M' = m \times l = \dfrac{M}{2}\). (ii) Cut along: each piece has half the cross-section, so half the pole strength \(\dfrac{m}{2}\), and the full length: \(M' = \dfrac{m}{2} \times 2l = \dfrac{M}{2}\).

Declination: the angle between the magnetic meridian and the geographic meridian. Dip (inclination): the angle the total field makes with the horizontal. Horizontal component \(B_H\): the horizontal part of the field, \(B\cos\delta\). The three fix the field completely.

1. \(B_V = B_H\tan\delta = 4 \times 10^{-5} \times 0.75 = 3 \times 10^{-5}\) T.

2. \(B = \sqrt{B_H^2 + B_V^2} = \sqrt{16 + 9} \times 10^{-5} = 5 \times 10^{-5}\) T. Check: \(B_H = B\cos 37^\circ = 5 \times 0.8 = 4\).

At the magnetic equator the field is horizontal, so the dip is \(0^\circ\) and a freely pivoted needle stays horizontal. At the magnetic poles the field is vertical, the dip is \(90^\circ\) and the needle stands vertical.

In a plane at angle \(\theta\) to the meridian only \(B_H\cos\theta\) lies in the plane, while \(B_V\) is unchanged: \(\tan\delta' = \dfrac{B_V}{B_H\cos\theta} = \dfrac{\tan\delta}{\cos\theta} = \dfrac{0.75}{0.5} = 1.5\). The apparent dip is larger than the true dip.

The compass needle settles \(10^\circ\) east of geographic north, because it aligns with the magnetic meridian. A compass responds only to the horizontal component \(B_H\); the vertical component only tries to tip it, which the pivot prevents.

1. With the N pole pointing north, the magnet's field on its equatorial line points south, opposite to \(B_H\). Where the two are equal in size the net field is zero: a neutral point.

2. \(\dfrac{\mu_0}{4\pi}\dfrac{M}{d^3} = B_H\): \(d^3 = \dfrac{10^{-7} \times 3.2}{4 \times 10^{-5}} = 8 \times 10^{-3}\), so \(d = 0.2\) m.

Now the axial field beyond each pole points south, opposite to \(B_H\), so the neutral points lie on the axis. \(\dfrac{\mu_0}{4\pi}\dfrac{2M}{d^3} = B_H\): \(d^3 = \dfrac{10^{-7} \times 3.2}{4 \times 10^{-5}} = 8 \times 10^{-3}\), so \(d = 0.2\) m.

Magnetisation \(\vec M\) is the magnetic moment per unit volume. \(\vec H = \dfrac{\vec B}{\mu_0} - \vec M\) is the part of the field due to free currents alone. \(\chi = \dfrac{M}{H}\). Since \(B = \mu_0(H + M) = \mu_0(1 + \chi)H\), \(\mu_r = 1 + \chi\).

Diamagnetic: \(\chi\) small and negative, about \(-10^{-5}\); bismuth, copper, water. Paramagnetic: small and positive, about \(10^{-4}\) to \(10^{-3}\); aluminium, oxygen. Ferromagnetic: large and positive, \(10^{2}\) to \(10^{3}\) or more; iron, nickel, cobalt.

\(M = \chi H = 3 \times 10^{-4} \times 10^{4} = 3\) A/m. \(B = \mu_0(H + M) = 4\pi \times 10^{-7} \times 10003 \approx 1.26 \times 10^{-2}\) T, barely more than in vacuum.

Curie's law: for a paramagnet \(\chi = \dfrac{C}{T}\). Halving \(T\) doubles \(\chi\): \(4 \times 10^{-4}\).

1. \(R\) is the retentivity: the field \(B\) left in the sample when \(H\) is reduced to zero. \(C\) is the coercivity: the reverse \(H\) needed to bring \(B\) to zero.

2. The area of the loop is the energy dissipated as heat in unit volume of the material in one complete cycle of magnetisation.

3. A fat loop: high retentivity keeps the magnet strong, and high coercivity makes it hard to demagnetise.

An electromagnet must magnetise strongly and switch off cleanly, and it cycles many times a second. Soft iron has high permeability, low retentivity and low coercivity, so it loses little energy per cycle and keeps no magnetism when the current stops. A permanent magnet must keep its magnetism against stray fields, which needs the high retentivity and coercivity of steel.

A diamagnetic rod is weakly repelled: it moves towards the region where the field is weaker. Set along the field, it turns perpendicular to it. A superconductor expels the field completely, with \(\chi = -1\): it is a perfect diamagnet and can float above a magnet.

1. The magnet's axial field at the needle is perpendicular to \(B_H\), and the needle sets along the resultant: \(\dfrac{\mu_0}{4\pi}\dfrac{2M}{d^3} = B_H\tan\theta\).

2. \(M = \dfrac{B_H\tan\theta\,d^3}{2 \times 10^{-7}} = \dfrac{4 \times 10^{-5} \times 1 \times 8 \times 10^{-3}}{2 \times 10^{-7}} = 1.6\) A m\(^2\).

In the tan B position the needle sits on the magnet's equatorial line, where the field is \(\dfrac{\mu_0}{4\pi}\dfrac{M}{d^3}\). The same reading then implies \(M = \dfrac{4 \times 10^{-5} \times 8 \times 10^{-3}}{10^{-7}} = 3.2\) A m\(^2\), twice the Q20 value, because the equatorial field is only half the axial field. A given magnet therefore deflects less in tan B.

\(T \propto \dfrac{1}{\sqrt M}\) for the same \(I\) and \(B_H\): \(\dfrac{M_1}{M_2} = \dfrac{T_2^2}{T_1^2} = \dfrac{1}{4}\).

Sum position: \(T_s^2 \propto \dfrac{1}{M_1 + M_2}\); difference position: \(T_d^2 \propto \dfrac{1}{M_1 - M_2}\). So \(\dfrac{M_1 + M_2}{M_1 - M_2} = \dfrac{T_d^2}{T_s^2} = \dfrac{16}{4} = 4\), giving \(3M_1 = 5M_2\) and \(\dfrac{M_1}{M_2} = \dfrac{5}{3}\).

Treat the magnet as two poles \(\pm m\) a distance \(2l\) apart with \(r \gg l\). On the axis the poles give \(\dfrac{\mu_0m}{4\pi}\left[\dfrac{1}{(r - l)^2} - \dfrac{1}{(r + l)^2}\right] \approx \dfrac{\mu_0}{4\pi}\dfrac{2m \times 2l}{r^3} = \dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}\). On the equatorial line the two pole fields, each \(\dfrac{\mu_0m}{4\pi r^2}\), add only their components along the axis, giving \(\dfrac{\mu_0}{4\pi}\dfrac{m \times 2l}{r^3} = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}\). The axial field is twice the equatorial one.