The Solenoid

What a solenoid is

• A long wire wound in the form of a closely spaced helix, so that each turn can be treated as a circular loop; the net field is the vector sum of the fields of all the turns. Enamelled wire is used so that neighbouring turns are insulated from each other.
• "Long" means the length is large compared with the radius.
• Looking at a stretched-out section, the loop fields between two neighbouring turns point oppositely and cancel; along the inside they reinforce.
• At an interior mid-point the field is strong, uniform and along the axis. At an exterior mid-point it is weak, and still axial, with no perpendicular component. As the solenoid is made longer it approaches a cylindrical current sheet: the field outside tends to zero and the field inside becomes everywhere parallel to the axis. In the idealised long solenoid the outside field is taken as zero.

Deriving $B = \mu_0 n I$ by Ampere's circuital law

• Take a rectangular amperian loop $abcd$ near the middle of the solenoid, with the side $ab$ of length $h$ lying inside and parallel to the axis, and $cd$ well outside.
• Along $cd$ the field is zero, so that side contributes nothing.
• Along the transverse sides $bc$ and $da$ the field has no component along the path (the field is axial inside and negligible outside), so they contribute nothing either.
• Along $ab$ the field is $B$, parallel to the path, so that side contributes $Bh$. The relevant length of the loop is therefore $L = h$: $$\displaystyle\oint_{abcd}\vec{B}\cdot d\vec{l} = Bh + 0 + 0 + 0 = Bh$$
• With $n$ turns per unit length, the number of turns threading the rectangle is $nh$, so the enclosed current is $I_e = I(nh)$. Ampere's law $BL = \mu_0 I_e$ gives $$Bh = \mu_0 I\left(nh\right) \;\Rightarrow\; \boxed{\;B = \mu_0 n I\;}$$
• $n = N/L$ is in turns per metre. The field is independent of position inside — that is why a solenoid is the standard way to obtain a uniform magnetic field. Its direction is along the axis, given by the right-hand rule.
• A soft iron core inside the solenoid makes the field much larger.

Diagram in words

• Draw the solenoid as a row of circles (the turns) with dots on top and crosses on the bottom, or as a long coil in section. Inside, draw closely spaced straight parallel arrows along the axis; outside, draw the lines spreading out from one end, looping round and re-entering the other end — the bar-magnet pattern.
• Superimpose the rectangle $abcd$: $ab$ inside along the axis (length $h$), $cd$ outside and parallel to it, $bc$ and $da$ perpendicular. Label $B$ along $ab$ and mark $B=0$ along $cd$.

Polarity and the bar-magnet equivalence

• End rule: viewed from one face, if the current appears anticlockwise that face is a north pole; if clockwise, it is a south pole.
• So for an observer on the left of a solenoid who sees the current flowing clockwise, the left face is a south pole and the right face a north pole. Inside the solenoid the field lines then run from left to right (south to north inside, exactly as in a bar magnet), and outside they curve from the right face round to the left face.
• Each turn has magnetic moment $m' = IA$; the $N$ turns all carry the same current in the same sense, so their moments add: $$m = N m' = N I A$$
• Hence a current-carrying solenoid of $N$ turns and cross-section $A$ behaves as a bar magnet of magnetic moment $NIA$, with the polarity fixed as above. Unit $\text{A m}^2$.

Two coaxial solenoids

• Two long insulated solenoids $S_1$ (inner, $n_1$ turns per unit length) and $S_2$ (outer, $n_2$ turns per unit length) of equal length, wound one over the other, with the same current $I$ flowing through them in opposite senses.
• Inside, on the axis, the two fields are antiparallel, so they subtract: $$B_{\text{in}} = \mu_0 n_2 I - \mu_0 n_1 I = \mu_0 I\left(n_2 - n_1\right)$$ directed along the axis in the sense of whichever solenoid has the larger $nI$.
• Outside the combined system, $B = 0$, because the field of an ideal long solenoid vanishes outside, and that holds separately for each of the two.

Worked numerical

• Solenoid of length $0.5\ \text{m}$, radius $1\ \text{cm}$, $500$ turns, current $5\ \text{A}$: $n = 500/0.5 = 1000\ \text{turns m}^{-1}$; since length/radius $=50\gg1$ the long-solenoid formula applies, so $B = 4\pi\times10^{-7}\times10^{3}\times5 = 6.28\times10^{-3}\ \text{T}$.

Results to carry away

• $B = \mu_0 n I$ inside a long solenoid, uniform and axial; $B = 0$ outside.
• $n$ is turns per unit length, $n = N/L$.
• $m = NIA$; the solenoid is equivalent to a bar magnet.
• Coaxial pair with opposed currents: $B_{\text{in}} = \mu_0 I(n_2-n_1)$, $B_{\text{out}}=0$.