Chapter 5: Magnetism and Matter (Magnetism)

$\tau =mB\sin \theta$, so

$$m=\frac{\tau }{B\sin \theta }=\frac{4.5\times 10^{-2}}{0.25\times \sin 30^{\circ }}=\frac{4.5\times 10^{-2}}{0.125}=0.36{\text{J T}}^{-1}$$

$m=0.36$ J T⁻¹.

$U=-mB\cos \theta$, with $mB=0.32\times 0.15=0.048$ J.

(a) Stable equilibrium: $\vec{m}$ along $\vec{B}$ ($\theta =0^{\circ }$). The energy is the lowest possible: $U=-mB=-0.048$ J. A small twist produces a torque that brings it back.

(b) Unstable equilibrium: $\vec{m}$ opposite to $\vec{B}$ ($\theta =180^{\circ }$). The energy is the highest: $U=+mB=+0.048$ J. A small twist makes it turn right round.

(a) m parallel to B, U = −0.048 J. (b) m antiparallel to B, U = +0.048 J.

A current-carrying solenoid produces a magnetic field just like a bar magnet's: one end behaves as a north pole and the other as a south pole; it points north-south when freely suspended; it attracts and repels other magnets; and its field lines outside have the same pattern. Its magnetic moment lies along its axis, in the direction given by the right-hand rule.

$$m=NIA=800\times 3.0\times 2.5\times 10^{-4}=0.60{\text{J T}}^{-1}$$

It has the same field pattern and poles as a bar magnet; $m=0.60$ J T⁻¹ along its axis.

(The book prints "Exercise 5.5"; it means the solenoid of Exercise 5.3, with $m=0.60$ J T⁻¹.)

$$\tau =mB\sin \theta =0.60\times 0.25\times \sin 30^{\circ }=7.5\times 10^{-2}\text{N m}$$

$\tau =7.5\times 10^{-2}$ N m.

$mB=1.5\times 0.22=0.33$ J. The magnet starts at ${\theta }_1=0^{\circ }$.

(a) $W=mB(\cos {\theta }_1-\cos {\theta }_2)$:

(i) ${\theta }_2=90^{\circ }$: $W=0.33(1-0)=0.33$ J.

(ii) ${\theta }_2=180^{\circ }$: $W=0.33(1-(-1))=0.66$ J.

(b) $\tau =mB\sin \theta$:

(i) $\tau =0.33\times \sin 90^{\circ }=0.33$ N m (it tends to turn the magnet back into line with the field).

(ii) $\tau =0.33\times \sin 180^{\circ }=0$.

(a) (i) 0.33 J, (ii) 0.66 J. (b) (i) 0.33 N m, (ii) zero.

(a) $m=NIA=2000\times 4.0\times 1.6\times 10^{-4}=1.28$ J T⁻¹, along the axis.

(b) In a uniform field the forces on the two "poles" are equal and opposite, so the net force is zero. The torque is

$$\tau =mB\sin \theta =1.28\times 7.5\times 10^{-2}\times \sin 30^{\circ }=4.8\times 10^{-2}\text{N m}$$

It tends to turn the axis of the solenoid into line with the field.

(a) 1.28 J T⁻¹. (b) Force zero; torque $4.8\times 10^{-2}$ N m.

$r=0.10$ m, $r^3=10^{-3}$ m³.

(a) Axial point:

$$B=\frac{{\mu }_0}{4\pi }\frac{2m}{r^3}=10^{-7}\times \frac{2\times 0.48}{10^{-3}}=9.6\times 10^{-5}\text{T}$$

along the magnetic moment (from S to N of the magnet).

(b) Equatorial point:

$$B=\frac{{\mu }_0}{4\pi }\frac{m}{r^3}=10^{-7}\times \frac{0.48}{10^{-3}}=4.8\times 10^{-5}\text{T}$$

opposite to the magnetic moment (from N to S, parallel to the magnet's axis).

(a) $9.6\times 10^{-5}$ T along the moment (S→N). (b) $4.8\times 10^{-5}$ T opposite to the moment (N→S).