Derivation: B-Field inside a Current-Carrying Coil

Current-carrying coil that generates a magnetic field.

Our goal is to determine the magnitude of the magnetic field $ \class{violet}{B} $ within a coil through which a current $ \class{red}{I} $ is flowing. The coil has length $l$ and $N$ turns.

To calculate the magnetic field inside, we use the fourth Maxwell equation of electrostatics in integral form:

Ampere's law of electrostatics

$$ \begin{align} \mu_0 \, \class{red}{I} ~=~ \oint_{\mathcal S} \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{s} \end{align} $$

Here $ \class{red}{I} $ is the current enclosed by any closed oriented path $ \mathcal{S} $. We choose $ \mathcal{S} $ so that the integral 1 is as easy as possible to compute. To do this, we form a rectangular loop around the coil as shown in Illustration 2.

The path is oriented counterclockwise and is composed of four straight paths $ \class{gray}{\mathcal{S}_{1}} $, $ \class{brown}{\mathcal{S}_{2}} $, $ \class{gray}{\mathcal{S}_{3}} $ and $ \class{brown}{\mathcal{S}_{4}} $. You put the path $ \class{brown}{\mathcal{S}_{4}} $ so that it is inside the coil and the path $ \class{brown}{\mathcal{S}_{2}} $ outside the coil and thus parallel to $ \class{brown}{\mathcal{S}_{4}} $. Thus the integral to be solved becomes:

$$ \begin{align} \mu_0 \, \class{red}{I} &~=~ \int_{\class{gray}{\mathcal{S}_{1}}} \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{s}_{1} ~+~ \int_{\class{brown}{\mathcal{S}_{2}}} \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{s}_{2} \\\\
&~+~ \int_{\class{gray}{\mathcal{S}_{3}}} \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{s}_{3} ~+~ \int_{\class{brown}{\mathcal{S}_{4}}} \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{s}_{4} \end{align} $$

The first and the third integral over $ \class{gray}{\mathcal{S}_{1}} $ and $ \class{gray}{\mathcal{S}_{3}} $ are zero, because the magnetic field is perpendicular to the respective path and the scalar product in the integral vanishes. The other two integrals remain:

$$ \begin{align} \mu_0 \, \class{red}{I} ~=~\int_{\class{brown}{\mathcal{S}_{2}}} \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{s}_{2} ~+~ \int_{\class{brown}{\mathcal{S}_{4}}} \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{s}_{4} \end{align} $$

We choose a rectangular loop such that the paths $ \class{gray}{\mathcal{S}_{1}} $ and $ \class{gray}{\mathcal{S}_{3}} $ are very long. Thus the integral over $ \class{brown}{\mathcal{S}_{2}} $ is omitted because it is so far away from the coil that the magnetic field is approximately zero at this distance. What remains is only the integral over the path $ \class{brown}{\mathcal{S}_{4}} $ inside the coil:

$$ \begin{align} \mu_0 \, \class{red}{I} ~=~ \int_{\class{brown}{\mathcal{S}_{4}}} \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{s}_{4} \end{align} $$

Since the coil usually has $ N $ turns, the current "passes through" the loop $ N $-times. The current contribution $ \class{red}{I} $ is $N$-times:

$$ \begin{align} \mu_0 \, N \, \class{red}{I} ~=~ \int_{\class{brown}{\mathcal{S}_{4}}} \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{s}_{4} \end{align} $$

We assume that the magnetic field $ \class{violet}{\boldsymbol{B}} ~=~ \class{violet}{B} \, \boldsymbol{\hat{e}}_{4} $ inside is constant. This allows us to pull it out of the integral:

$$ \begin{align} \mu_0 \, N \, \class{red}{I} ~=~ \class{violet}{B} \, \int_{\class{brown}{\mathcal{S}_{4}}} \boldsymbol{\hat{e}}_{4} \cdot \text{d}\boldsymbol{s}_{4} \end{align} $$

We oriented the coil so that the unit vector $ \boldsymbol{\hat{e}}_{4} $, which specifies the magnetic field direction, is parallel to the infinitesimal path $ \text{d}\boldsymbol{s}_{4} = \text{d}s_{4} \, \boldsymbol{\hat{e}}_{4} $. Thus the scalar product between the unit vectors is $ \boldsymbol{\hat{e}}_{4} \cdot \boldsymbol{\hat{e}}_{4} = 1 $:

$$ \begin{align} \mu_0 \, N \, \class{red}{I} &~=~\class{violet}{B} \, \int_{\class{brown}{\mathcal{S}_{4}}} \text{d}s_{4} \\\\
&~=~ \class{violet}{B} \, l \end{align} $$

Here, in the second step, we exploited the fact that the integration path $ \class{brown}{\mathcal{S}_{4}} $ runs along the coil and thus corresponds to the coil length $ l $. Rearrange 7 for the magnetic field magnitude $ \class{violet}{B} $:

Formula: Magnetic field inside a long coil

$$ \begin{align} \class{violet}{B} ~=~ \mu_0 \, \frac{N \, \class{red}{I}}{l} \end{align} $$

With this you can calculate the magnetic field of a long coil (long, so that boundary effects do not play a role) or rearrange the formula and calculate $ N $, $l$ or $\class{red}{I}$, depending on what you are looking for.