Magnetic Field of a Straight Solenoid

The solenoid. A solenoid is a long, cylindrical coil formed by winding an insulated wire

into many closely spaced, essentially circular turns, one after another along the length of a

cylinder. When a current $I$ flows through the wire, each individual turn produces its own

circular-loop-like field (Section 4.5), and, provided the turns are closely and uniformly spaced,

these individual fields overlap and add up to a single, much stronger, and remarkably UNIFORM field

running along the solenoid's own axis -- the practical reason a solenoid, rather than a single loop,

is the standard device used whenever a strong, uniform magnetic field is needed over an extended

region (an electromagnet, an MRI machine, a relay, and, as a further extension, this is essentially

how a straight solenoid becomes a toroid when its two ends are bent around to meet).

Idealisation and assumptions. An exact Biot-Savart calculation for a real, finite solenoid is

considerably more involved than the derivation below suggests. The standard, and very good,

approximation treats the solenoid as effectively INFINITELY long (or, equivalently, considers a

field point deep inside the solenoid, far from either end) and its turns as densely and evenly

wound, so that the winding can be idealised as a smooth surface current rather than a set of

discrete loops. Under this idealisation, two results, both confirmed experimentally to good

accuracy for a real, sufficiently long solenoid, are simply asserted (and then exploited via

Ampere's law): the field INSIDE such a solenoid, away from its ends, is essentially UNIFORM,

directed along the axis; the field OUTSIDE the solenoid, away from its ends, is essentially ZERO.

Applying Ampere's circuital law. Let $n$ be the number of turns per unit length of the solenoid,

carrying current $I$. Choose a rectangular Amperian loop $PQRS$: one side, $PQ$ of length $L$, lying

along the solenoid's axis INSIDE the coil (where the field is the uniform $B$ to be found); the

opposite side, $RS$, lying OUTSIDE the solenoid (where $B=0$ by the assumption above); and the two

remaining sides, $QR$ and $SP$, drawn perpendicular to the axis, one inside and one outside. Along

$PQ$ (inside, parallel to $\vec{B}$), the contribution to $\oint\vec{B}\cdot d\vec{l}$ is simply

$B L$. Along $RS$ (outside, where $B=0$), the contribution is zero. Along the two perpendicular

sides $QR$ and $SP$, the field (wherever it is nonzero, i.e. only over the small portion actually

inside the solenoid) is PERPENDICULAR to $d\vec{l}$ there, so $\vec{B}\cdot d\vec{l}=0$ along both of

these sides too. The full loop integral therefore reduces to just the one nonzero side:

The current enclosed by this rectangular loop is the current threading through the $N=nL$ turns that

lie within the length $L$ of the loop's side $PQ$, i.e. $I_{\text{enc}} = (nL)I$. Ampere's law then

gives directly

Reading the result. The field inside a long solenoid depends only on the number of turns PER

UNIT LENGTH, $n$, and the current $I$ -- notably, it does NOT depend on the solenoid's own radius, …