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Sources of the Magnetic Field · Lesson 5 of 6

The Solenoid

14 min read3 ideas7 checks2 figures to playBy Asheshanand Acharya
What you will learn 0 / 3
Field inside a long solenoid

Many loops in a row

Wind an insulated wire in a tight helix round a long tube, turn beside turn. The coil is a solenoid. Each turn is a loop, and on its axis each loop gives a field along the axis. Close to one loop its field is strong; further along it fades as \(1/x^3\). Inside a long solenoid, the fields of the many turns add up to a strong field.

Between two neighbouring turns, their fields run opposite ways and nearly cancel. Inside, they run the same way and add. So inside a long, closely wound solenoid the field lines are straight, parallel and evenly spaced. Outside, the lines spread out from one end and return to the other, very thinly.

A solenoid filling with fieldA solenoid 40 centimetres long seen from the side, carrying current so that the field inside points to the right. Sliders set the number of turns, the current and the radius. As turns or current grow, the straight lines inside crowd together and B equals mu nought n I, read live. Changing the radius changes nothing inside. A chip puts in an iron core of relative permeability 500, which multiplies the field 500 times. n = 100 turns/m, B = 0.25 mT
SimA solenoid \(40\,\text{cm}\) long. Add turns, raise the current and change the radius. The inside field lines crowd as turns or current grow, and \(B = \mu_0 nI\) is read live. The radius changes nothing inside. Choose the iron core to multiply the field by \(500\).

Play with the figure above. Doubling the turns on the same \(40\,\text{cm}\) doubles the field, and so does doubling the current. Change the radius from \(2\) to \(6\,\text{cm}\) and the reading does not move. Only the number of turns per metre and the current set the field inside.

Proof with a rectangle

A long, closely wound solenoid has the symmetry Ampere's law needs. Take the solenoid as endless, with \(n\) turns per metre, the turns per unit length, each carrying \(I\). Cut it along its axis, as in Fig 5.1. The turns come out of the page along the top and go back in along the bottom. The field inside then points along the axis, to the right.

The Amperian rectangle of a solenoidA long solenoid cut along its axis: the turns come out of the page along the top and go back in along the bottom, so the field inside points to the right. A rectangle abcd has its side ab inside, parallel to the field, and its side cd outside, where the field is zero. The sides bc and da cross the field at right angles. The turns threading the rectangle along its top edge, n L of them, each carry the current I. a b c d B B = 0 outside; n L turns pass through abcd
Fig 5.1A long solenoid cut along its axis. The rectangle abcd has side ab, of length \(L\), inside and along the field, and side cd outside, where the field is zero. The sides bc and da cross the field at right angles. The \(nL\) turns along the top of the rectangle each thread it once.
Deriving \(B = \mu_0 n I\) step 1 of 5
1

\[ \oint\vec B\cdot d\vec l = \int_a^b + \int_b^c + \int_c^d + \int_d^a \]

Why. split the rectangle abcd of Fig 5.1 into its four sides.

RememberField inside a long solenoid
\[ \begin{gathered} B = \mu_0 n I \\ n = \dfrac{N}{\ell} \end{gathered} \]
Holds whenwell inside a long, closely wound solenoid, far from its ends; the field is uniform and along the axis, its sense given by the right-hand grip on the turns.
Symbols\(B\) field inside, \(n\) turns per unit length, \(N\) total turns, \(\ell\) length of the solenoid, \(I\) current
Units\(B\) in T when \(n\) is in turns per metre and \(I\) in A

The radius is nowhere in the result. A fat solenoid and a thin one with the same turns per metre and the same current hold the same field. The fat one holds more flux, because its area is larger, but not a stronger field. For the sense, curl the fingers of the right hand along the current in the turns. The thumb points along the field inside, towards the end that acts as a north pole.

ExampleThree solenoids

(a) A long solenoid has \(1000\) turns per metre and carries \(2\,\text{A}\). Find the field inside. (b) A solenoid \(0.5\,\text{m}\) long has \(2000\) turns and carries \(5\,\text{A}\). Find the field inside. (c) A solenoid \(0.2\,\text{m}\) long has \(400\) turns. What current gives \(5\times10^{-3}\,\text{T}\) inside?

Try it on paper first. The solution is one tap away when you are ready.
Checkpoint · numericUnanswered

A solenoid \(25\,\text{cm}\) long has \(500\) turns and carries \(3\,\text{A}\). What is the field well inside it, in millitesla?

mT± 2 %
CheckpointUnanswered

Two long solenoids have the same turns per metre and the same current. One has twice the radius of the other. How do the fields inside compare?

Checkpoint · numericUnanswered

A long solenoid must give \(0.01\,\text{T}\) inside with a current of \(4\,\text{A}\). How many turns per metre does it need?

turns/m± 2 %
Checkpoint · select all that applyUnanswered

A long solenoid has \(N\) turns on a length \(\ell\) and carries \(I\). Which changes double the field inside?

Ends and outside of a solenoid

The ends, and the outside

Stand at one end of a long solenoid, on its axis. Deep inside, the turns on both sides of a point contribute, half from each side. At the end, only the turns on one side exist. So the field at an end is half the field inside, \(\tfrac12\mu_0 nI\). This is the same halving as at the end of a long straight wire.

Outside an ideal, endless solenoid the field is zero. Take a rectangle lying wholly outside, with two long sides at different distances from the solenoid. It encloses no current, so the fields along its two long sides must be equal. The field outside is then the same at every distance. Far away it is certainly zero, so it is zero everywhere outside.

The field along the axis of a real solenoidA solenoid 20 centimetres long drawn above a graph of the field on its axis, as a fraction of mu nought n I. Chips make the solenoid as long as two radii, five radii or twenty radii. The long one is flat at nearly the full value inside, drops to one half at each end and falls to nearly zero a few radii outside. The short fat one never reaches the full value. 1020−10−200.51Ox (cm)B/μ0nI centre: 0.707 μ0nI, end: 0.447 μ0nI
Fig 5.2The field on the axis of a solenoid \(20\,\text{cm}\) long, as a fraction of \(\mu_0 nI\). Choose its length as \(2\), \(5\) or \(20\) radii. The long one is flat at nearly the full value, half at each end and nearly zero a little way outside. The short fat one reaches only \(0.71\) at its centre.

A real solenoid has ends, and Fig 5.2 shows its field along the axis. Far outside, the field is tiny. It grows as the end comes near and reaches about half the full value at the end face. Inside, it rises quickly and levels off at \(\mu_0 nI\) over the middle, where it is uniform. The pattern is the same about the centre. A solenoid as long as \(20\) radii has \(0.995\) of the ideal field at its centre and \(0.50\) at its ends.

Deeper

The whole profile has a formula. Take a point on the axis. Let \(\phi_1\) and \(\phi_2\) be the angles between the axis and the lines to the rims of the two end faces. Then \(B = \tfrac12\mu_0 nI(\cos\phi_1 + \cos\phi_2)\). Deep inside a long solenoid both angles are nearly zero and \(B = \mu_0 nI\). At an end one angle is \(90°\) and the other nearly zero, which gives the half.

Checkpoint · numericUnanswered

A long solenoid has \(1000\) turns per metre and carries \(2\,\text{A}\). What is the field on its axis at one end, in millitesla?

mT± 2 %
CheckpointUnanswered

Why is the field outside an ideal long solenoid zero?

The iron core and the mri magnet

An iron core

Slide a rod of soft iron into the solenoid. The field inside grows hundreds of times. The iron's own atoms line up with the field and add their fields to it; the chapter Magnetism and Matter explains how. For now one number describes the iron, its relative permeability \(\mu_r\). The field in a long solenoid filled with it is \(\mu_r\mu_0 nI\).

A solenoid of \(1000\) turns per metre with \(2\,\text{A}\) gives \(2.51\times10^{-3}\,\text{T}\) in air. With a core of \(\mu_r = 500\) it gives \(500\times 2.51\times10^{-3} = 1.26\,\text{T}\). That is the principle of every electromagnet, from the crane in a scrapyard to the relay in a car. Iron stops helping near \(2\,\text{T}\), when all its atoms are lined up.

Checkpoint · numericUnanswered

A solenoid gives \(2\,\text{mT}\) inside with an air core. A soft iron core of relative permeability \(400\) is slid in. What is the new field, in tesla?

T± 2 %
In the world
The magnet inside an MRI scanner

Hospital MRI scanners, including those at AIIMS in New Delhi, commonly work at \(1.5\,\text{T}\) or \(3\,\text{T}\). The patient lies inside a long solenoid wide enough for a person, so no iron core can fill it. For \(1.5\,\text{T}\) in air, \(nI = B/\mu_0 = 1.19\times10^6\) ampere-turns per metre. A copper coil of \(1000\) turns per metre would need about \(1200\,\text{A}\) and would melt. The coil is therefore made of superconducting wire, cooled by liquid helium, which carries such currents with no resistance and no heat.

Story · William Sturgeon, 1825
Iron inside a coil

William Sturgeon was an English shoemaker turned soldier and lecturer. In 1825 he varnished an iron bar bent into a horseshoe and wound bare copper wire round it. With the current on, the bar of about \(200\,\text{g}\) lifted about \(4\,\text{kg}\) of iron, many times its own weight. With the current off, it let go. It was the first electromagnet. A few years later Joseph Henry wound many turns of insulated wire on such cores and lifted hundreds of kilograms.

Exam traps
  1. Using the total number of turns \(N\) in place of \(n\).
    \(n = N/\ell\) is turns per metre; divide by the length first.
  2. Expecting a fatter solenoid to have a weaker field inside.
    \(B = \mu_0 nI\) has no radius; only turns per metre and current matter.
  3. Using the full \(\mu_0 nI\) at an end.
    On the axis at an end of a long solenoid the field is half the inside value.
  4. Multiplying by \(\mu_r\) without limit.
    Soft iron saturates near \(2\,\text{T}\); stronger fields need air cores and large currents.