Magnetic Fields due to Electric Current

What the chapter is about

• Oersted first showed that a current produces a magnetic field; Gauss, Henry, Faraday and Maxwell established that electric and magnetic fields are partners. Motors, generators, communication systems and computers all follow from it.
• Two parallel hanging wires attract when the currents are in the same direction and repel when they are opposite. The forces are not electrostatic — the wires stay electrically neutral throughout, so the origin must be magnetic.
• Right hand thumb rule: grab the wire with the right hand, thumb along the current; the curled fingers give the direction of the circular magnetic field lines.
• Notation: $\odot$ means out of the plane of the paper, $\otimes$ means into it.

10.2 Magnetic force and the Lorentz force

• Magnetic force on a charge $q$ moving with velocity $\vec{v}$ in a field $\vec{B}$: $\vec{F}_m = q(\vec{v}\times\vec{B})$. For an electron $\vec{F}_m = -e(\vec{v}\times\vec{B})$.
• Magnitude $F_m = qvB\sin\theta$, direction perpendicular to the plane containing $\vec{v}$ and $\vec{B}$.
• Lorentz force (both fields present): $\vec{F} = q[\vec{E} + (\vec{v}\times\vec{B})] = \vec{F}_e + \vec{F}_m$.
• Consequences: (i) if $\vec{v}$ is parallel (or antiparallel) to $\vec{B}$, $F_m = 0$; (ii) if the charge is stationary, $F_m = 0$ even when $\vec{B} \neq 0$.
• Because $\vec{F}_m \perp \vec{v}$, $\vec{F}_m \cdot \vec{v} = 0$: the magnetic force never does work. It can change the direction of motion but never the speed or the kinetic energy.
• The electric force depends only on $\vec{E}$ and $q$; the magnetic force depends on $q$, on the speed, and on the angle between $\vec{v}$ and $\vec{B}$.
• Unit of $B$: from $B = F/qv$, 1 tesla = 1 N s / (C m). $1\ \text{T} = 10^4$ gauss.

10.3 Cyclotron motion

• In a uniform field perpendicular to its velocity a charged particle moves in a circle, the magnetic force supplying the centripetal force: $qvB = mv^2/R$, hence $mv = p = qBR$ (the cyclotron formula).
• Radius $R = mv/qB$; period $T = 2\pi m/qB$; cyclotron frequency $f_c = qB/2\pi m$.
• The key fact: $T$ and $f_c$ do not depend on the speed or on the radius. The particle takes the same time for every semicircle however fast it goes.

10.3.1 The cyclotron accelerator

• Invented by Lawrence and Livingston in 1934 to study nuclear structure; accelerates charged particles to MeV-GeV energies for particle-matter interaction studies and for treating certain tumours.
• Construction: two hollow semicircular metal chambers, the dees $D_1$ and $D_2$, placed in a vacuum chamber, with a uniform magnetic field (up to about 1.5 T) perpendicular to their plane, and a high-frequency alternating voltage (up to 10000 V, about 10 MHz) applied across the gap between them. An ion source sits at O in the gap. The electric and magnetic fields are perpendicular to each other — crossed fields.
• Working: the magnetic field bends the ion into a semicircle inside a dee (there is no electric field inside a dee); each time it crosses the gap the alternating field accelerates it, adding $qV$ to its kinetic energy; the radius grows each time, so the path is a spiral. At the largest radius it is deflected out through an exit slit.
• Resonance condition: $f_a = f_c$, so that the polarity of the dees reverses exactly as the ion reaches the gap. The interval after which the field must be reversed is $t = T/2 = \pi m/qB$.
• Final energy: $\text{K.E.} = \dfrac{q^2B^2R_{\text{exit}}^2}{2m} = 2\pi^2 m f^2 R^2$, proportional to the square of the outermost radius.
• Accelerators in India: the 2 GeV electron Synchrotron at RRCAT, Indore, and the 8-10 MeV Microtron at Savitribai Phule Pune University.

10.4 Helical motion

• If the velocity has a component $\vec{v}_\parallel$ along $\vec{B}$, then for that component $F_m = qv_\parallel B \sin 0^\circ = 0$, so it is unaffected and the particle drifts steadily along the field.
• The perpendicular component $\vec{v}_\perp$ produces the circular motion as before.
• The two together give a helix: circular motion in the plane perpendicular to $\vec{B}$ superposed on uniform translation along $\vec{B}$.
• Radius $r = mv_\perp / qB$; period $T = 2\pi m/qB$ (unchanged); pitch $= v_\parallel T$.

10.5 Magnetic force on a wire carrying a current

• Straight wire, field perpendicular: the drifting electrons each feel $q(\vec{v}_d\times\vec{B})$; with $q = IL/v_d$ this gives $F_m = ILB$.
• If $\vec{B}$ is not perpendicular to the wire, $\vec{F}_m = I\vec{L}\times\vec{B}$, magnitude $ILB\sin\theta$, where $\vec{L}$ points along the current.
• Arbitrarily shaped wire: $d\vec{F}_m = I\,d\vec{l}\times\vec{B}$ and $\vec{F}_m = \displaystyle\int I\,d\vec{l}\times\vec{B}$.
• For uniform $\vec{B}$: $\vec{F}_m = I\left(\displaystyle\int d\vec{l}\right)\times\vec{B}$ — only the straight line joining the two ends matters.
• Diagram (Fig. 10.8): a vertical wire in the plane of the page carrying current $I$ upward, field $\vec{B}$ shown as dots coming out of the page, electrons drifting downward with velocity $\vec{v}_d$, and the resulting force $\vec{F}_m$ marked horizontally, perpendicular to both.

10.6 Force on a closed circuit in a magnetic field

• Extending the wire result to a closed circuit C: $\vec{F}_m = I\left(\displaystyle\oint_C d\vec{l}\right)\times\vec{B}$ for uniform $\vec{B}$.
• The sum of the length vectors right round a closed circuit is zero.
• Therefore the net force on a closed circuit placed in a uniform magnetic field is zero. This is a standard one-mark question — the answer is simply "zero".
• Careful: zero net force does not mean zero net torque. A current loop in a uniform field usually experiences a couple.

10.7 Torque on a current loop

• Take a rectangular loop abcd, sides $l_1$ (ab, cd) and $l_2$ (bc, da), carrying current $I$ in a uniform field $\vec{B}$, with ab and cd perpendicular to $\vec{B}$. Let $\theta$ be the angle between the normal $\hat{n}$ to the plane of the loop and $\vec{B}$.
• Forces on bc and da: $F = Il_2 B\sin(90^\circ - \theta)$ each. They are equal, opposite and act along the same line, so they cancel completely and produce no torque.
• Forces on ab and cd: $F = Il_1 B$ each. They are equal and opposite but do not act along the same line, so they form a couple. Moment arm of each about the central axis is $\dfrac{1}{2}l_2\sin\theta$.
• Torque $\tau = 2 \times \left(Il_1 B \times \dfrac{1}{2}l_2\sin\theta\right) = Il_1 l_2 B\sin\theta = IAB\sin\theta$.
• For a flat coil of $N$ turns: $\tau = NIAB\sin\theta$. This holds for a planar coil of any shape in a uniform field.
• This is the principle of an electric motor. To keep the coil rotating past the position where its plane is parallel to $\vec{B}$, a commutator with carbon brushes reverses the current every half rotation, so the torque keeps turning the coil the same way.

10.7.1 Moving coil galvanometer

• Principle: when a coil carrying a current is suspended in a uniform magnetic field, a torque acts on it, tending to rotate it about the axis of suspension so that the magnetic flux through the coil becomes maximum.
• Construction and working, as the board wants it: (i) a coil of several turns, suspended or pivoted so that it can rotate freely about a fixed axis in a radial uniform magnetic field; (ii) a soft iron cylindrical core makes the field radial and strong; (iii) current through the coil produces $\tau = NIAB\sin\theta$, and since the field is radial $\theta = 90^\circ$, so $\tau = NIAB$; (iv) a spring provides a counter torque, so a steady current gives a steady deflection $\phi$; (v) the restoring torque of the spring is $K\phi$, with $K$ the torsional constant.
• At equilibrium $K\phi = NIAB$, so $\phi = \left(\dfrac{NAB}{K}\right)I$, i.e. $\phi \propto I$. A galvanometer therefore has a linear (uniform) scale.
• Diagram: a rectangular coil of $N$ turns wound on a frame, suspended by a phosphor bronze strip between the concave pole pieces of a permanent magnet, with a soft iron cylindrical core between the poles; a spring at the bottom, a pointer or mirror to read the deflection, and terminals for the current.

10.8 Magnetic dipole moment

• A current carrying coil behaves as a magnetic dipole with moment $\vec{m}$ (also written $\vec{\mu}$), directed along the normal $\hat{n}$ to the plane of the coil, sense given by the right-hand thumb rule.
• Magnitude $m = NIA$; SI unit A m$^2$, equivalently J/T.
• Torque on it in a uniform field: $\tau = mB\sin\theta$, or in vector form $\vec{\tau} = \vec{m}\times\vec{B}$ — the exact analogue of $\vec{\tau} = \vec{p}\times\vec{E}$ for an electric dipole.
• For a single loop, $\vec{m} = I\vec{A}$.

10.9 Magnetic potential energy of a dipole

• A dipole freely suspended in a field has potential energy on account of its orientation: $U = -\vec{m}\cdot\vec{B} = -mB\cos\theta$, analogous to $U = -\vec{p}\cdot\vec{E}$.
• Case (i): $\theta = 0^\circ$ gives $U = -mB$, the minimum — $\vec{m}$ parallel to $\vec{B}$, the stable orientation.
• Case (ii): $\theta = 180^\circ$ gives $U = +mB$, the maximum — $\vec{m}$ antiparallel to $\vec{B}$.

10.10 Biot-Savart law

• Statement: the magnitude of the field $dB$ at a point P due to a current element $I\,d\vec{l}$ is $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$, where $r$ is the distance of P from the element and $\theta$ the angle between $d\vec{l}$ and $\vec{r}$.
• Vector form: $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\vec{r}}{r^3}$. The direction of $d\vec{B}$ is that of $d\vec{l}\times\vec{r}$.
• $\mu_0 = 4\pi\times 10^{-7}$ T m/A $\approx 1.26\times 10^{-6}$ T m/A.
• It is still an inverse square law: $\vec{r}$ in the numerator cancels one power of the $r^3$ below.
• Biot-Savart law plays the role in magnetostatics that Coulomb's law plays in electrostatics.
• Long straight wire, perpendicular distance $R$: integrating over the whole wire (substitute $l = R\tan\theta$) gives $B = \dfrac{\mu_0 I}{2\pi R}$. Half of the wire alone (semi-infinite) gives $B = \dfrac{\mu_0 I}{4\pi R}$.

10.11 Force between two long parallel wires

• Wire 1 produces $B_1 = \dfrac{\mu_0 I_1}{2\pi d}$ at wire 2. The Lorentz force law then gives, on a length $L$ of wire 2, $F_{21} = I_2 B_1 L = \dfrac{\mu_0 I_1 I_2 L}{2\pi d}$.
• Force per unit length: $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$.
• Parallel currents attract; antiparallel currents repel. $\vec{F}_{21} = -\vec{F}_{12}$, consistent with Newton's third law.
• Definition of the ampere: two infinitely long parallel wires 1 m apart in vacuum, each carrying the same constant current, experience a force of $2\times10^{-7}$ N per metre of length; that current is 1 ampere.
• Note that for infinitely long wires the total force would be infinite, which is why the result is always quoted per unit length.

10.12 Field due to a current in a circular arc

• On an arc, $d\vec{l}$ is always perpendicular to $\vec{r}$, so $\sin\theta = 1$ and every element contributes in the same sense at the centre.
• $B = \dfrac{\mu_0 I \theta}{4\pi r}$ at the centre, with the subtended angle $\theta$ in radians.
• Semicircle: $B = \dfrac{\mu_0 I}{4r}$. Full circle: $B = \dfrac{\mu_0 I}{2r}$.
• Straight sections in line with the centre contribute nothing, because $\sin 0^\circ = \sin 180^\circ = 0$. That is the whole trick in the "two straight sections and one arc" problems.

10.13 Axial field of a circular current loop

• Loop of radius $R$ in the x-y plane, centre at the origin, point P on the z axis at distance $z$; $r^2 = R^2 + z^2$ and every $d\vec{l}$ is perpendicular to $\vec{r}$.
• Components of $d\vec{B}$ perpendicular to the axis cancel in pairs from diametrically opposite elements; only the z components survive, with $dB_z = dB\cos\theta$ and $\cos\theta = R/\sqrt{R^2+z^2}$.
• Result: $B_z = \dfrac{\mu_0 I R^2}{2(R^2+z^2)^{3/2}}$.
• Setting $z = 0$: $B_0 = \dfrac{\mu_0 I}{2R}$, and for $N$ turns $B_0 = \dfrac{\mu_0 N I}{2R}$.

10.14 Magnetic lines for a current loop

• Diagram (Fig. 10.20): closed loops that pass through the ring, crowded and nearly straight along the axis inside, spreading out and returning round the outside — exactly the pattern of a short bar magnet.
• Right hand thumb rule for a loop: curl the fingers of the right hand along the current in the wire; the stretched thumb points along the field on the axis, i.e. towards the north face.
• The face from which the field lines emerge behaves as the north pole, the other face as the south pole.
• For $z \gg R$: $B_z = \dfrac{\mu_0 I R^2}{2z^3} = \dfrac{\mu_0}{4\pi}\dfrac{2IA}{z^3} = \dfrac{\mu_0}{4\pi}\dfrac{2m}{z^3}$, with $m = IA$. So at large distances a current loop is a magnetic dipole. $\vec{B}_z$ and $\vec{m}$ point the same way.
• By the electrostatic analogue, on the perpendicular bisector $\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{\vec{m}}{x^3}$ for $x \gg R$.
• Fundamental difference between an electric and a magnetic dipole: an electric dipole can be split into two isolated charges, but a magnetic dipole cannot be split into isolated poles — no magnetic monopole has ever been found.

10.15 Ampere's circuital law

• Statement: the line integral of the magnetic induction $\vec{B}$ around any closed path in free space equals $\mu_0$ times the total current flowing through the area bounded by the path: $\displaystyle\oint \vec{B}\cdot d\vec{s} = \mu_0 I$.
• The closed path is called the Amperian loop; $I$ is the net current it encircles.
• Since $\vec{B}\cdot d\vec{l} = B\,dl\cos\theta$, the law is an integral of the tangential component of $\vec{B}$ round the loop.
• Sign convention: curl the right palm along the loop with the fingers in the direction of integration; a current parallel to the stretched thumb is positive, one opposite to it negative. For the standard figure with $I_1, I_2$ out of the page, $I_3$ into the page and $I_4$ outside the loop: $\displaystyle\oint B\cos\theta\,dl = \mu_0(I_1 + I_2 - I_3)$. $I_4$ contributes nothing, as its contributions cancel over the full loop.
• Ampere's law plays the role in magnetostatics that Gauss' law plays in electrostatics; it can be derived from the Biot-Savart law and is at its most useful when the current distribution is symmetric.
• Long straight wire: take a circle of radius $r$ round the wire, on which $B$ is constant and tangential, so $\vec{B}\cdot d\vec{l} = B r\,d\theta$ and $\displaystyle\oint\vec{B}\cdot d\vec{l} = B(2\pi r) = \mu_0 I$, giving $B = \dfrac{\mu_0 I}{2\pi r}$.

10.16 Solenoid

• A solenoid is a long, closely wound helical coil whose diameter is much smaller than its length.
• Inside, the field lines are dense, parallel and uniform along the axis; outside they are widely spaced, so the field there is weak. In an ideal solenoid (infinite length, square cross-section wire, very closely wound) the field inside is uniform and along the axis, and outside it is exactly zero.
• Derivation: take a rectangular Amperian loop abcd with side ab of length $L$ inside and parallel to the axis, cd outside, bc and da perpendicular to the axis. Then $\displaystyle\int_{ab} + \int_{bc} + \int_{cd} + \int_{da} = \mu_0 I$. The bc and da integrals vanish because $\vec{B} \perp d\vec{l}$ there; the cd integral vanishes because $B = 0$ outside. So $BL = \mu_0 I$.
• With $n$ turns per unit length and current $i$, the current enclosed is $I = nLi$, so $BL = \mu_0 nLi$ and $B = \mu_0 n i$.
• Valid for a real solenoid too at points near the middle, but not near its ends.

10.16 Toroid

• A toroid is a solenoid of finite length bent into a hollow circular tube, like the inner tube of a tyre.
• The field lines inside are concentric circles; the direction of $\vec{B}$ at a point is along the tangent to the circle through that point, fixed by the direction of the current.
• Derivation: take a circular Amperian loop of radius $R$ along the central axis of the tube. Here $\vec{B}$ and $d\vec{l}$ are in the same direction, so $\theta = 0^\circ$, $\cos\theta = 1$, and $\displaystyle\oint\vec{B}\cdot d\vec{l} = B\oint dl = B(2\pi R)$. The net enclosed current is $Ni$, so $B(2\pi R) = \mu_0 Ni$ and $B = \dfrac{\mu_0 N i}{2\pi R}$.
• Since $B \propto 1/R$, the field is not constant over the cross-section of a toroid — unlike a solenoid. For an ideal toroid, $B = 0$ everywhere outside.

Electrostatic - magnetostatic analogy