4.3 Ampère's Circuital Law, the Solenoid and Forces between Currents

In an electrical substation, thick copper bars carry the current of a whole town. Engineers bolt them down firmly — not because of their weight, but because if a fault shorts the circuit, tens of thousands of amperes surge through them for a fraction of a second, and neighbouring bars carrying opposite currents fling themselves apart with forces of hundreds of newtons per metre. Every parallel pair of current-carrying wires attracts or repels in this way; until 2019 that force was how the ampere itself was defined.

To find such forces we need the field of a long straight wire. The Biot–Savart law of Section 4.2 can give it, but the integration is not easy. For highly symmetric currents there is a much quicker route, Ampère's circuital law, which plays the same role for magnetic fields that Gauss's law (Section 1.3) plays for electric fields: it turns an integral over a whole circuit into one line of algebra. In this section we state Ampère's law, use it for a long straight wire (outside and inside) and for a long solenoid, and then find the force between two parallel currents and the definition of the ampere.

Ampère's Circuital Law

Instead of adding up the field produced by each piece of the current, Ampère's law looks at the field along a closed path. Pick any closed curve and walk round it. At each short step $d\mathbf l$ take the component of $\mathbf B$ along your step and multiply by the step length; add up all the steps.

The law is always true for steady currents, but it is only useful for finding $\mathbf B$ when symmetry lets us choose a loop on which $B$ is constant and either parallel or perpendicular to $d\mathbf l$; then $\oint\mathbf B\cdot d\mathbf l$ becomes simply $B$ times a length. (For currents that change in time Maxwell added a second term — the displacement current — in Chapter 8.) The comparison with Gauss's law is close: there, a closed surface, the flux of $\mathbf E$ and the enclosed charge; here, a closed curve, the circulation of $\mathbf B$ and the enclosed current.

The Field of a Long Straight Wire

Take a very long straight wire carrying a steady current $I$. By symmetry the field cannot depend on position along the wire or on the angle round it, and the Biot–Savart law shows that every element's $d\mathbf B$ is perpendicular to the plane containing the wire and the point, so the field has no component towards or away from the wire, nor along it. The field lines are therefore circles centred on the wire, and $B$ depends only on the distance $r$. Choose as Amperian loop the field-line circle of radius $r$. On it $\mathbf B\parallel d\mathbf l$ and $B$ is constant: $$\oint\mathbf B\cdot d\mathbf l=B\,(2\pi r)=\mu_0I\quad\Longrightarrow\quad B=\frac{\mu_0I}{2\pi r}.$$ Inside a thick wire of radius $R$ with the current spread uniformly over its cross-section, a circle of radius $r\lt R$ encloses only the fraction $\pi r^2/\pi R^2$ of the current, so $B(2\pi r)=\mu_0I\,r^2/R^2$.

The Solenoid

A solenoid is a long wire wound as a tight helix of many turns, usually on a cylindrical former. Each turn is nearly a circular loop, and the fields of neighbouring turns add. Between neighbouring wires the fields of the two wires nearly cancel, while inside the coil the fields of all the turns point the same way along the axis. For a long, tightly wound solenoid the result is a strong, nearly uniform field inside, parallel to the axis, and a very weak field outside (the same field lines that are crowded together inside spread out over the whole of the space outside). In the idealised limit of an infinitely long solenoid, the field outside is zero.

Now apply Ampère's law to a rectangular loop $abcd$: side $ab$ of length $h$ runs inside, parallel to the axis; side $cd$ runs outside; sides $bc$ and $da$ cross the winding at right angles to the axis. Along $ab$, $\mathbf B\parallel d\mathbf l$, contributing $Bh$. Along $bc$ and $da$, $\mathbf B$ is perpendicular to $d\mathbf l$ inside and zero outside, contributing nothing; along $cd$ the field is zero. So $\oint\mathbf B\cdot d\mathbf l=Bh$. If the solenoid has $n$ turns per unit length, the loop is threaded by $nh$ turns, each carrying $I$: $$Bh=\mu_0\,(nh)\,I\quad\Longrightarrow\quad B=\mu_0nI.$$ The answer does not depend on where inside the loop's side $ab$ lies — the field is the same everywhere inside, not just on the axis.

The field does not depend on the radius of the solenoid, only on the number of turns per metre and the current. That is why solenoids are the standard way to produce a known, uniform magnetic field in the laboratory, and — wound with superconducting wire — the main magnet of an MRI scanner.

Force between Two Parallel Currents and the Ampere

Two long parallel wires $a$ and $b$, a distance $d$ apart, carry currents $I_a$ and $I_b$. Wire $a$ produces at wire $b$ the field $B_a=\dfrac{\mu_0I_a}{2\pi d}$, perpendicular to wire $b$. By Result 4.2, a length $L$ of wire $b$ therefore feels the force $$F_{ba}=I_bLB_a=\frac{\mu_0I_aI_b}{2\pi d}\,L.$$ For the direction, take both currents along $+\hat{\mathbf k}$, with $b$ at distance $d$ along $+\hat{\mathbf i}$ from $a$. The field of $a$ at $b$ is $\hat{\mathbf k}\times\hat{\mathbf i}=+\hat{\mathbf j}$, and the force on $b$ is along $\hat{\mathbf k}\times\hat{\mathbf j}=-\hat{\mathbf i}$ — towards $a$. The same argument shows that $b$ pulls $a$ with an equal and opposite force, as Newton's third law requires. Reversing one current reverses both forces.

Exercises