Magnetic Field of a Solenoid

What a Solenoid Is

A solenoid is a long wire wound in the form of a helix in which the neighbouring turns are closely spaced. Because each turn is so nearly circular, NCERT treats each one as a circular current loop, and the net field is the vector sum of the fields due to all the turns. Enamelled wire is used so that adjacent turns stay electrically insulated from one another.

The phrase that does all the work here is long solenoid: the length is large compared with the radius. Only then is the interior field genuinely uniform and the exterior field negligible. The clean formula below is an idealisation valid for such a long, tightly wound coil, built from many fields on the axis of a circular loop.

The Field Pattern

For a single section of the winding, the loops of neighbouring turns produce fields that cancel in the gaps and reinforce along the axis. So at the interior mid-point the field is uniform, strong and axial. At an exterior mid-point it is weak and, for a long solenoid, also parallel to the axis with no perpendicular component. As the solenoid is made longer it behaves like a long cylindrical current sheet, and the field outside approaches zero.

The external field lines emerge from one end, loop around the outside, and re-enter at the other end — exactly the pattern of a bar magnet. One end of the solenoid behaves as a north pole and the other as a south pole, which is why a current-carrying solenoid is the closest electrical analogue of a permanent magnet.

Deriving B = μ₀nI from Ampere's Law

The derivation rests on Ampere's circuital law, $\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I_e$, applied to the idealised long solenoid. As the solenoid is made longer it appears like a long cylindrical metal sheet; the field outside approaches zero and is taken as zero, while the field inside becomes everywhere parallel to the axis.

Following NCERT, consider a rectangular Amperian loop $abcd$ with one long side $ab$ deep inside the solenoid (parallel to the axis) and the opposite side $cd$ outside it.

Along $cd$ the field is zero, as argued above. Along the two transverse sections $bc$ and $ad$ the field component along the path is zero (the field there is either zero outside or perpendicular to the path). These three sides therefore make no contribution. Only the side $ab$, of length $L = h$ deep inside the solenoid, contributes. Let the field along $ab$ be $B$.

Let $n$ be the number of turns per unit length, so the total number of turns threaded by the loop is $nh$. The enclosed current is $I_e = I(nh)$, where $I$ is the current in the solenoid. Ampere's circuital law then gives:

$$BL = \mu_0 I_e \qquad\Rightarrow\qquad Bh = \mu_0 I(nh)$$ $$\boxed{\,B = \mu_0 n I\,}$$

The direction of the field is given by the right-hand rule. Notice what the result does not contain: there is no radius and no distance from the axis, which is precisely why the interior field of a long solenoid is uniform across its whole cross-section. The solenoid is, for this reason, the standard device for obtaining a uniform magnetic field in the laboratory.

Field at the End and Outside

The result $B = \mu_0 n I$ describes the deep interior. At the end (mouth) of the solenoid, a point on the axis sees turns extending in only one direction instead of both, so it receives half the contribution of the deep interior. The axial field at the end of a long solenoid is therefore exactly half the interior value:

$$B_{\text{end}} = \frac{\mu_0 n I}{2}$$

Outside the solenoid, the field is taken as zero for an ideal long solenoid. The closely wound turns behave like a current sheet whose internal field is intense and whose external field cancels; the longer and tighter the winding, the better this approximation holds.

Contrast: The Toroid

A toroid is an endless solenoid — a straight solenoid bent into a closed ring with no ends at all. For a point $P$ at distance $r$ from the centre, draw a circular Amperian path of radius $r$ concentric with the ring. By symmetry the field is everywhere tangential with constant magnitude, so $\oint \mathbf{B}\cdot d\mathbf{l} = B(2\pi r)$. With $N$ total turns each carrying current $i$, the threaded current is $Ni$, and Ampere's law gives:

$$B(2\pi r) = \mu_0 N i \qquad\Rightarrow\qquad B = \frac{\mu_0 N i}{2\pi r}$$

Two contrasts with the straight solenoid are worth fixing. First, the toroidal field uses the total number of turns $N$, whereas the solenoid formula uses turns per unit length $n$. Second, the toroidal field falls off as $1/r$ across the core — it is not uniform — and it is confined entirely within the core, with essentially no field outside the windings.

The Solenoid as an Electromagnet

A solenoid with an air core gives the modest field $B = \mu_0 n I$. Inserting a core of soft iron — a ferromagnetic material with a large relative permeability $\mu_r$ — multiplies this field dramatically. The iron becomes strongly magnetised by the solenoid's field and adds its own much larger contribution, so the field becomes:

$$B = \mu_0 \mu_r n I$$

The factor $\mu_r$ can run into the hundreds or thousands, which makes the iron-cored solenoid a practical electromagnet capable of lifting heavy loads. Soft iron is chosen deliberately: it magnetises strongly when current flows but loses almost all of that magnetism the instant the current is switched off, so the electromagnet can be turned on and off at will. Its strength depends on the current, the turns per unit length, and the core's permeability.

Field Summary Table