Physics · Ch 4 — Moving Charges and Magnetism

The Solenoid

4.7

The Solenoid

The Solenoid: A Source of Uniform Magnetic Field

A solenoid is a long wire wound into a tight helix. For it to be considered "long," its length must be much greater than its radius. This geometry is crucial because it produces a highly uniform magnetic field inside, similar to how a parallel-plate capacitor produces a uniform electric field.

Why the Field Inside is Uniform and Outside is Negligible

Structure: Each turn of the solenoid acts like a circular current loop. The net magnetic field is the vector sum of the fields from all these loops.
Cancellation between turns: Between two neighbouring turns, the magnetic field lines from adjacent loops point in opposite directions and cancel out (see Fig. 4.15a).
Inside the solenoid: At interior points (like point P in Fig. 4.15b), the contributions from all turns add up, giving a strong, uniform field directed along the axis.
Outside the solenoid: At exterior points (like point Q), the field is very weak. For an ideal, infinitely long solenoid, the field outside is exactly zero. For a real long solenoid, it is negligible.

Deriving the Magnetic Field Inside Using Ampere's Circuital Law

We use Ampere's law to find the magnetic field $B$ inside a long solenoid.

Choose an Amperian loop: Consider a rectangular loop abcd (Fig. 4.16). Side ab of length $h$ lies inside the solenoid, parallel to the axis. Side cd lies outside, where the field $B = 0$ . Sides bc and ad are perpendicular to the axis.
Evaluate the line integral $\oint \mathbf{B} \cdot d\mathbf{l}$ :
- Along ab: $\mathbf{B}$ is parallel to $d\mathbf{l}$ , so the contribution is $B h$ .
- Along cd: $B = 0$ , so contribution is $0$ .
- Along bc and ad: $\mathbf{B}$ is perpendicular to $d\mathbf{l}$ (field is axial, path is radial), so the dot product is $0$ .
- Therefore, $\oint \mathbf{B} \cdot d\mathbf{l} = B h$ .
Calculate the enclosed current $I_e$ :
- Let $n$ be the number of turns per unit length.
- The number of turns passing through the loop is $n h$ .
- If the current in the solenoid is $I$ , the total current enclosed is $I_e = I (n h)$ .
Apply Ampere's law: $\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_e$

B h = \mu_0 I (n h)

Cancelling $h$ gives the final result:

B = \mu_0 n I

- **$B$** is the magnitude of the magnetic field inside the solenoid.
- **$\mu_0$** is the permeability of free space ($4\pi \times 10^{-7} \, \text{T m A}^{-1}$).
- **$n$** is the number of turns per unit length (turns/m).
- **$I$** is the current in the solenoid (A). …

Figure 4.15(a) The magnetic field due to a section of the solenoid which has been stretched out for clarity. Only the exterior semi-circular part is shown. Notice how the circular loops between neighbouring turns tend to cancel. (b) The magnetic field of a finite solenoid.

Fig. 4.15 — (a) The magnetic field due to a section of the solenoid which has been stretched out for clarity. Only the exterior semi-circular part is shown. Notice how the circular loops between neighbouring turns tend to cancel. (b) The magnetic field of a finite solenoid.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

What Figure 4.15 Shows

The figure has two panels, (a) and (b), that together explain how the magnetic field of a solenoid arises from its individual turns and how it behaves inside and outside the coil.

Panel (a) shows a short, stretched-out section of the solenoid. Five turns are represented as small circles: each circle has a ⊙ on its top half (current coming out of the page) and a ⊗ on its bottom half (current going into the page). Around each turn, magnetic field lines form closed loops. Between neighbouring turns, the field lines from adjacent loops point in opposite directions and cancel — this is why the field in the gaps is nearly zero. Two points are marked: P inside the solenoid (where the field is strong and uniform) and Q outside (where the field is weak). The field lines exit the ends of the section, showing that the net field inside points along the axis.

Panel (b) shows the entire finite solenoid as an oval shape. The top edge is a row of ⊙ symbols, the bottom edge a row of ⊗ symbols — this represents the current flowing around the solenoid. Inside the solenoid, the field lines are straight, parallel, and equally spaced along the axis through point P, indicating a uniform field. Outside, the field lines bow outward and loop back around the ends through point Q, forming a pattern identical to that of a bar magnet (a dipole). The field outside is much weaker than inside.

Physical Idea Taught

The figure teaches two key ideas:

Cancellation between turns: In a tightly wound solenoid, the magnetic fields of adjacent turns cancel in the region between them, so the net field there is negligible. This is why the interior field is uniform and parallel to the axis.
Solenoid as an electromagnet: The overall field pattern is that of a bar magnet — one end acts as a north pole, the other as a south pole. Inside, the field is strong and uniform; outside, it is weak and resembles a dipole field.

Key Formula Developed from This Figure

Using Ampere's circuital law with a rectangular loop (as shown in Fig. 4.16), the textbook derives the magnetic field inside a long solenoid (length >> radius):

B = \mu_0 n I

where: …

Figure 4.16The magnetic field of a very long solenoid. We consider a rectangular Amperian loop abcd to determine the field.

Fig. 4.16 — The magnetic field of a very long solenoid. We consider a rectangular Amperian loop abcd to determine the field.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

The figure shows a very long solenoid drawn as a horizontal capsule. The top wall of the solenoid has a row of ⊙ symbols (current coming out of the page), and the bottom wall has a row of × symbols (current going into the page). Inside the solenoid, several horizontal blue arrows point to the right, representing the uniform axial magnetic field $\mathbf{B}$ . The field is labelled 'B' at the left side, and a point P is marked inside. Outside the solenoid, above it, the field is essentially zero.

A dashed rectangular Amperian loop $abcd$ is overlaid:

Side $ab$ lies inside the solenoid, along the field direction (bottom edge, left to right).
Side $dc$ lies outside above the solenoid.
The vertical sides $bc$ and $ad$ are transverse (perpendicular to the axis).
The interior length between the top corners is labelled $h$ .

At the right side, a stylised open right hand shows the curl direction (the right-hand rule for the field direction).

Physical idea taught by the figure

The figure is used to apply Ampere’s circuital law to an idealised long solenoid. The key assumptions are:

The field inside is uniform and parallel to the axis.
The field outside is zero (for a very long solenoid).
Along the transverse sides $bc$ and $ad$ , the magnetic field component is zero (because $\mathbf{B}$ is axial and these sides are perpendicular to it).

Thus, only side $ab$ (length $h$ ) contributes to the line integral $\oint \mathbf{B} \cdot d\mathbf{l}$ .

Key formula derived from this figure

From Ampere’s law:

\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enclosed}}

The only non-zero contribution is along $ab$ :

B h = \mu_0 I_{\text{enclosed}}

The enclosed current is the current through the solenoid times the number of turns inside the loop. If $n$ is the number of turns per unit length and $I$ is the current in each turn, then the number of turns inside the loop is $n h$ , so:

I_{\text{enclosed}} = I (n h)

Substituting:

B h = \mu_0 I (n h)

Cancelling $h$ (since $h \neq 0$ ):

\boxed{B = \mu_0 n I}

Symbol meanings: …

Start your 14-day free trial to unlock the full solution →

← Previous · 4.6Example 4.7 — Figure 4.13 shows a long straight wire of a circular cross-section (ra…Next · 4.7 →Example 4.8 — A solenoid of length 0.5m has a radius of 1cm and is made up of 500 tu…