The Bar Magnet

What the board asks here

• One mark: properties of field lines, torque and force on a dipole aligned with the field, why poles cannot be separated.
• Three marks: the four properties of magnetic field lines; work done and torque numericals.
• Five marks: show that a current-carrying solenoid behaves as a small bar magnet and derive its far axial field; magnetic moment of a coil; equilibrium of a pivoted magnet.

The bar magnet as a dipole

• Iron filings sprinkled on a glass sheet over a short bar magnet settle into a pattern with two centres of concentration. The pattern mimics the magnetic field lines and shows the magnet behaves as a magnetic dipole with a north pole and a south pole, exactly as an electric dipole has $+q$ and $-q$.
• Freely suspended, the pole pointing towards geographic north is the north pole, the other the south pole.
• Like poles repel, unlike poles attract.
• The poles cannot be separated. Break the magnet and each piece is a complete magnet with both poles, only weaker. Isolated magnetic poles — monopoles — do not exist. This is the content of Gauss's law for magnetism, which is why "monopoles do not exist" is the correct explanation of "poles cannot be separated" in the assertion-reason question.
• A current-carrying solenoid produces a filings pattern of the same shape.

Magnetic field lines — the four properties (three-mark answer)

• Magnetic field lines are continuous closed curves. Outside the magnet they run from the north pole to the south pole; inside the magnet they continue from the south pole to the north pole, closing each loop. This is unlike the electric dipole, whose lines begin on the positive charge and end on the negative charge or escape to infinity.
• The tangent drawn at any point on a field line gives the direction of the net magnetic field $\vec{B}$ at that point.
• The number of field lines crossing unit area measures the magnitude of $\vec{B}$: where the lines are crowded the field is strong, where they are widely separated the field is weak. In a bar magnet the field is strongest at the two ends and weakest at the middle.
• Two magnetic field lines can never intersect. If they did, two tangents could be drawn at the point of intersection and the direction of the field there would not be unique, which is impossible.

Field lines are deliberately not called "lines of force": the magnetic force on a moving charge is $q\vec{v}\times\vec{B}$, perpendicular to $\vec{B}$, so a field line does not point along the force. To map the field, move a small compass needle from point to point and record its orientation.

Bar magnet as an equivalent solenoid (the five-mark derivation)

Ampere's hypothesis is that all magnetic phenomena can be explained by circulating currents. A solenoid is a stack of identical circular current loops, each of which is a magnetic dipole, so a solenoid should behave as one magnet — and cutting a solenoid, like cutting a bar magnet, gives two smaller solenoids with the field lines still continuous.

Diagram to draw: a solenoid of length $2l$ and radius $a$ drawn horizontally, its centre marked $O$; an element of thickness $dx$ shaded at a distance $x$ to the right of $O$; the axis extended to an external point $P$ at a distance $r$ from $O$, so the element is $(r-x)$ from $P$; the current direction and the resulting axial $\vec{B}$ at $P$ marked with arrows.

Take $n$ turns per unit length and current $i$. The element of thickness $dx$ carries $n\,dx$ turns. Using the axial field of a circular loop,

$$dB = \dfrac{\mu_0\,n\,dx\,i\,a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{3/2}}$$

Integrating over the whole solenoid,

$$B = \dfrac{\mu_0 n i a^{2}}{2}\int_{-l}^{\,l}\dfrac{dx}{\left[(r-x)^{2}+a^{2}\right]^{3/2}}$$

For a far external point, $r \gg a$ and $r \gg l$, so $\left[(r-x)^{2}+a^{2}\right]^{3/2} \approx r^{3}$ and

$$B = \dfrac{\mu_0 n i a^{2}}{2r^{3}}\int_{-l}^{\,l} dx = \dfrac{\mu_0\, n\,i\,(2l)\,a^{2}}{2r^{3}}$$

$$B = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{r^{3}}, \qquad m = n\,(2l)\,i\,(\pi a^{2})$$

This is identical in form to the far axial field of a bar magnet, so a current-carrying solenoid behaves like a small bar magnet, and the magnetic moment of a bar magnet equals that of the equivalent solenoid producing the same field.

Magnetic moment of a coil

$$\vec{m} = N I \vec{A}, \qquad m = N I \pi a^{2}\ \text{for a circular coil of radius } a$$

• The direction is normal to the plane of the coil, given by the right-hand rule: curl the fingers along the current, and the thumb points along $\vec{m}$. For a coil lying in the $X$-$Y$ plane, $\vec{m}$ is along the $z$-axis.
• Worked value CBSE has set: $N = 5$, $i = 2\ \text{A}$, $a = 7\ \text{cm}$ gives $m = 5\times2\times\dfrac{22}{7}\times(7\times10^{-2})^{2} = 0.154\ \text{A m}^{2}$, directed perpendicular to the plane of the coil.

Far field of a bar magnet — the electrostatic analogue

Comparing the dipole equations with those of Chapter 1, the magnetic results follow from the electric ones by the replacements $\vec{E}\to\vec{B}$, $\vec{p}\to\vec{m}$, $\dfrac{1}{4\pi\varepsilon_0}\to\dfrac{\mu_0}{4\pi}$. For $r \gg l$:

$$\vec{B}_E = -\dfrac{\mu_0}{4\pi}\cdot\dfrac{\vec{m}}{r^{3}} \quad\text{(equatorial)}, \qquad \vec{B}_A = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2\vec{m}}{r^{3}} \quad\text{(axial)}$$

• At equal distances $B_A = 2B_E$, and the two are oppositely directed.

The dipole in a uniform magnetic field

Diagram to draw: a small magnetic needle of moment $\vec{m}$ inclined at $\theta$ to a uniform field $\vec{B}$ shown as parallel horizontal lines; equal and opposite forces on the two ends forming a couple; the angle $\theta$ marked between $\vec{m}$ and $\vec{B}$. The same arrangement, with the needle set oscillating, is used to measure either $B$ or $m$.

• Torque: $\vec{\tau} = \vec{m}\times\vec{B}$, magnitude $\tau = mB\sin\theta$. It is a restoring torque.
• Net force: zero. The two poles feel equal and opposite forces in a uniform field, so there is a couple but no translation.
• Potential energy, obtained by integrating the work done against the restoring torque:

$$U = \int \tau(\theta)\,d\theta = \int mB\sin\theta\,d\theta = -mB\cos\theta = -\vec{m}\cdot\vec{B}$$

Taking the constant of integration as zero fixes the zero of potential energy at $\theta = 90^\circ$, i.e. with the needle perpendicular to the field.

• Work done by an external agent turning the magnet from $\theta_1$ to $\theta_2$:

$$W = U_2 - U_1 = -mB(\cos\theta_2 - \cos\theta_1)$$

• Worked value CBSE has set: $m = 6\ \text{J T}^{-1}$, $B = 0.44\ \text{T}$, starting at $\theta_1 = 60^\circ$. Turning to $\theta_2 = 90^\circ$ gives $W = -6\times0.44\left(0-\dfrac{1}{2}\right) = 1.32\ \text{J}$; turning to $\theta_2 = 180^\circ$ gives $W = -6\times0.44\left(-1-\dfrac{1}{2}\right) = 3.96\ \text{J}$; and the torque in the final orientation is $\tau = mB\sin 180^\circ = 0$.

Pivoted magnet held at an angle

• A magnet on a frictionless pivot is held at $\theta$ to the field by a force $F$ applied perpendicular to its axis at distance $r$ from the pivot. Balancing the deflecting and restoring torques,

$$F\,r = mB\sin\theta$$

• Worked value CBSE has set: $m = 3\ \text{A m}^{2}$, $r = 0.10\ \text{m}$, $B = 0.25\ \text{T}$, $\theta = 30^\circ$ gives $F = \dfrac{3\times0.25\times0.5}{0.10} = 3.75\ \text{N}$.
• If $F$ is withdrawn the magnet is no longer in equilibrium: the restoring torque $mB\sin\theta$ makes it oscillate about the field direction and, because of damping, it finally comes to rest aligned along $\vec{B}$.

Standard results to carry into the exam

• $\vec{m} = NI\vec{A}$; for a solenoid $m = n(2l)i(\pi a^{2})$.
• $B_A = \dfrac{\mu_0}{4\pi}\dfrac{2m}{r^{3}}$, $B_E = \dfrac{\mu_0}{4\pi}\dfrac{m}{r^{3}}$ (opposite to $\vec{m}$), for $r \gg l$.
• $\tau = mB\sin\theta$, $F_{\text{net}} = 0$ in a uniform field.
• $U = -mB\cos\theta$; $W = -mB(\cos\theta_2 - \cos\theta_1)$; $W_{0^\circ\to 90^\circ} = mB$, $W_{0^\circ\to180^\circ} = 2mB$.
• Cutting a magnet transversely or lengthwise both give two magnets, each with an N and an S pole.
• Not every magnetic configuration has poles: a toroid, or a straight infinite conductor, has no net magnetic moment and no poles.
• To find which of two identical bars is the magnet using nothing else: bring the end of bar A to the middle of bar B. A magnet's field is strongest at its ends and weakest at the centre, so if there is no force at the middle of B, then B is the magnet. Repulsion in any orientation proves both are magnetised; attraction always means one is unmagnetised.