Magnetic Field due to a Current Element, Biot-Savart Law

Statement of the law

• Take a conductor $XY$ carrying a steady current $I$, and an infinitesimal element $d\vec{l}$ of it. Let $P$ be a point whose position vector from the element is $\vec{r}$, and $\theta$ the angle between $d\vec{l}$ and $\vec{r}$.
• Biot-Savart's law: the magnitude of the field $d\vec{B}$ at $P$ is proportional to the current $I$, to the element length $dl$, to $\sin\theta$, and inversely proportional to $r^2$; its direction is perpendicular to the plane containing $d\vec{l}$ and $\vec{r}$.
• Vector form — the form the board asks for: $$d\vec{B} = \dfrac{\mu_0}{4\pi}\,\dfrac{I\,d\vec{l}\times\vec{r}}{r^{3}}$$
• Magnitude form: $$\lvert d\vec{B}\rvert = \dfrac{\mu_0}{4\pi}\,\dfrac{I\,dl\,\sin\theta}{r^{2}}$$
• $\mu_0 = 4\pi\times10^{-7}\ \text{T m A}^{-1}$ is the permeability of free space; the expressions hold when the medium is vacuum. SI unit of $B$ is the tesla; the cgs unit is the gauss, $1\ \text{T} = 10^{4}\ \text{G}$.
• Direction by the right-hand screw rule: look at the plane containing $d\vec{l}$ and $\vec{r}$ and turn from the first vector to the second — anticlockwise means the result points towards you, clockwise means away.
• The total field is obtained by integrating this vector expression over the whole conductor.

Diagram in words

• Draw a curved wire $XY$ with an arrow for the current $I$. Mark a short segment $d\vec{l}$ on it, and from that segment draw $\vec{r}$ to the field point $P$, with the angle $\theta$ between them.
• At $P$ draw the crossed circle symbol for $d\vec{B}$ going into the plane of the page, since $d\vec{l}\times\vec{r}$ is perpendicular to the plane of the diagram.

Comparison with Coulomb's law — the two-mark and five-mark staple

Similarities

• Both are long-range laws, falling off as the inverse square of the distance from source to field point.
• The principle of superposition applies to both fields.
• Both are linear in their source: $\vec{E}$ in the charge $q$, $\vec{B}$ in the element $I\,d\vec{l}$.

Differences

• The last point is worth stating explicitly: along the line of $d\vec{l}$, $\theta = 0$ and $\lvert d\vec{B}\rvert = 0$. That is why straight portions of wire in line with the point of observation contribute nothing.

A connection to remember

$$\mu_0\varepsilon_0 = \dfrac{1}{c^{2}}$$

• Dividing the two constants of proportionality, $\dfrac{\mu_0/4\pi}{1/4\pi\varepsilon_0} = \mu_0\varepsilon_0 = \dfrac{10^{-7}}{9\times10^{9}} = \dfrac{1}{\left(3\times10^{8}\right)^{2}} = \dfrac{1}{c^{2}}$. Since the speed of light is fixed, choosing $\mu_0 = 4\pi\times10^{-7}$ fixes $\varepsilon_0$.

Field of a long straight wire

• Applying the law (or, more quickly, Ampere's circuital law) to an infinitely long straight wire at perpendicular distance $r$: $$B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2I}{r}$$
• The lines are concentric circles centred on the wire, in planes perpendicular to it — cylindrical symmetry, so $B$ depends on $r$ alone. Direction by the right-hand thumb rule: grasp the wire with the right hand, thumb along the current, and the curled fingers give the sense of $\vec{B}$.
• Inside a thick wire of radius $a$ carrying current spread uniformly over the cross-section, $B = \dfrac{\mu_0 I r}{2\pi a^2}$ for $r<a$, so $B\propto r$ inside and $B\propto 1/r$ outside, peaking at the surface.

Field at the centre of a circular loop, from the law

• For a loop of radius $r$, every element $d\vec{l}$ is tangential and $\vec{r}$ is radial, so $d\vec{l}\perp\vec{r}$ and $\sin\theta = 1$ everywhere: $$dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{r^{2}}$$
• All contributions point the same way (all into the page, or all out of it), so they add arithmetically: $$B = \displaystyle\int dB = \dfrac{\mu_0 I}{4\pi r^{2}}\int dl = \dfrac{\mu_0 I}{4\pi r^{2}}\times 2\pi r = \dfrac{\mu_0 I}{2r}$$
• For an arc subtending $\phi$ radians at the centre, only that fraction of the circumference contributes: $B = \dfrac{\mu_0 I\phi}{4\pi r}$. A semicircle therefore gives $\dfrac{\mu_0 I}{4r}$ — half the full-loop value, with the direction still set by the right-hand rule.

Two parallel wires: field at a point between them

• Wires $A$ and $B$, distance $d$ apart, carrying equal currents $I$ in the same direction; $P$ lies on the line joining them, at distance $x$ from $A$.
• The two contributions are oppositely directed at $P$ (one out of the page, one into it), so $$B_P = \dfrac{\mu_0 I}{2\pi}\left[\dfrac{1}{x} - \dfrac{1}{d-x}\right] = \dfrac{\mu_0 I}{2\pi x}\left[\dfrac{d-2x}{d-x}\right]$$
• Graph: $B_P\to+\infty$ as $x\to 0$, falls through zero at $x = d/2$ (exact cancellation by symmetry), and tends to $-\infty$ as $x\to d$ — a hyperbola-like branch on each side of the midpoint.
• If the currents are antiparallel instead, the two fields at a midway point are in the *same* sense and add: each is $\dfrac{\mu_0}{2\pi}\cdot\dfrac{2I}{d}$, so $B_{\text{net}} = \dfrac{2\mu_0 I}{\pi d}$.
• For unequal currents in opposite senses, the null point lies between the wires, nearer the smaller current; for currents in the same sense the null point also lies between them. Outside the pair the two fields reinforce and no null point exists.

Worked numericals from the board papers

• Current $10\ \text{A}$ east to west in a long wire on a table; field at $10\ \text{cm}$ due north: $B = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2I}{r} = \dfrac{10^{-7}\times 2\times 10}{10^{-1}} = 2\times10^{-5}\ \text{T}$, directed downwards (into the table) by the right-hand thumb rule.
• Proton at $4\times10^{6}\ \text{m s}^{-1}$ parallel to a wire carrying $4\ \text{A}$, at $0.2\ \text{m}$ from it, moving opposite to the current: $B = \dfrac{\mu_0 I}{2\pi x} = 2\times10^{-7}\times\dfrac{4}{0.2} = 4\times10^{-6}\ \text{T}$; then $F = qvB = 1.6\times10^{-19}\times4\times10^{6}\times4\times10^{-6} = 25.6\times10^{-19}\ \text{N}$, directed away from the wire.

Results to carry away

• $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\vec{r}}{r^{3}}$, magnitude $\dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^{2}}$.
• Long straight wire: $\dfrac{\mu_0 I}{2\pi r}$; thick wire inside: $\dfrac{\mu_0 I r}{2\pi a^{2}}$.
• Loop centre $\dfrac{\mu_0 I}{2R}$; semicircle $\dfrac{\mu_0 I}{4R}$; arc of angle $\phi$: $\dfrac{\mu_0 I\phi}{4\pi R}$.
• Two similarities and two differences with Coulomb's law, as tabled above.