In the following, the magnetic field \(B(r)\) inside and outside a coaxial cable is derived as a function of the distance \(r\) from the symmetry axis.

A coaxial cable consists of an inner wire with radius \(R_1\), then the subsequent insulation layer ending at radius \(R_2\). Then follows the outer wire, which ends at radius \(R_3\). The outer wire is covered with a protective sheath. A current \(I_{\text i}\) flows in the inner wire and a current \(I_{\text e}\) in the outer wire.

To calculate the magnetic field, the 4th Maxwell equation of magnetostatics in integral form, Ampere's law, is used:

Formula anchor$$ \begin{align} \oint_{S} \boldsymbol{B} ~\cdot~ \text{d}\boldsymbol{s} ~=~ \mu_0 \, I \end{align} $$

Ampere's law relates the current \(I\) enclosed by any closed loop \(S\) to the magnetic field \(B\) along that loop. Here, of course, we choose an ampere loop that is adapted to the symmetry of the problem. So we take a circular ampere loop. We assume that the current in the inner and outer wire is constant. Then the magnetic field \(\text{d}\boldsymbol{s}\) always points parallel to the \(\text{d}\boldsymbol{s}\) element, therefore the scalar product in 1 becomes maximal. So we can replace the scalar product \(\boldsymbol{B}\cdot \text{d}\boldsymbol{s}\) of vectors with a product \(B \, \text{d}s\) of magnitudes:

Amperage law for magnitudes

Formula anchor$$ \begin{align} \oint_{S} B \, \text{d}s ~=~ \mu_0 \, I \end{align} $$

Since in the line integral in 2 at a fixed radius \(r\) of the ampere loop is integrated along the ampere loop and the current is homogeneous everywhere, the magnetic field must always be the same at the distance \(r\) around the wire, that is independent of the integration variable. Consequently, the magnitude \(B\) may be pulled out to the front of the integral:

Ampere's law with constant magnetic field

Formula anchor$$ \begin{align} B \oint_{S} \text{d}s ~=~ \mu_0 \, I \end{align} $$

The integral over the selected circular ampere loop corresponds to the circumference \(2\pi \, r\) of a circle of variable radius \(r\):

Ampere's law for a circular loop with constant magnetic field

Formula anchor$$ \begin{align} B \, 2\pi \, r ~=~ \mu_0 \, I \end{align} $$

The current \(I\) enclosed by the ampere loop depends, of course, on whether, for example, the ampere loop encloses only the inner wire or both the inner wire and the outer wire, or even only part of it. Therefore, different cases will be considered in the following.

Magnetic field in inner wire

To calculate the magnetic field in the inner wire, an ampere loop with radius \(r\) is chosen to be inside the inner wire (with radius \(R_1\)): \(r \leq R_1\). The enclosed current \(I\) can first be written as the product of the constant current density \(j_{\text i}\) in the inner wire and the area enclosed by the ampere loop \(A(r) = \pi \, r^2 \):

Enclosed current in the inner conductor

Formula anchor$$ \begin{align} I ~=~ j_{\text i} \, A(r) ~=~ j_{\text i} \, \pi \, r^2 \end{align} $$

However, the current density \(j_{\text i}\) is not known, so it is expressed by the current \(I_{\text i}\) of the inner wire. By definition, the current density is the current per cross-sectional area through which the current flows. \(I_{\text i}\) passes through the area \(\pi \, R_1^2\), thus:

Enclosed current via current in the inner conductor

Formula anchor$$ \begin{align} I ~=~ \frac{I_{\text i}}{\pi \, R_1^2} \, \pi \, r^2 ~=~ \frac{I_{\text i}}{R_1^2} \, r^2 \end{align} $$

Now the included current 6 has to be inserted into eq. 4 and rearranged for the magnetic field \(B\):

Formula anchor$$ \begin{align} B(r) ~=~ \frac{\mu_0 \, I_{\text i}}{2\pi \, R_1^2} \, r ~~~~ (r \leq R_1) \end{align} $$

We can therefore state: The magnetic field \(B(r)\) in the inner wire of the coaxial cable increases linearly with the radial distance \(r\).

Magnetic field in the insulation

To determine the magnetic field between the inner and outer wire, an ampere loop with radius \(r\) is now placed in the region between the inner and outer wire (in the insulation layer): \(R_1 \leq r \leq R_2\). The enclosed current \(I\) in this case is the current \(I_{\text i}\) of the inner wire. Thus Eq. 4 becomes:

Ampere law for the insulation layer

Formula anchor$$ \begin{align} B \, 2\pi \, r ~=~ \mu_0 \, I_{\text i} \end{align} $$

Let us rearrange Eq. 8 with respect to the magnetic field \(B\):

Formula anchor$$ \begin{align} B(r) ~=~ \frac{\mu_0 \, I_{\text i}}{2\pi} \, \frac{1}{r} ~~~~ (R_1 \leq r \leq R_2) \end{align} $$

Let's note: The magnetic field of the insulation (or dielectric) of the coaxial cable falls off reciprocally with the radial distance.

Magnetic field inside the outer wire

To determine the magnetic field inside the outer wire, an ampere loop with radius \(r\) is placed inside the outer wire: \(R_2 \leq r \leq R_3\). The included current \(I\) is on the one hand the current \(I_{\text i}\) of the inner wire, on the other hand a variable current \(I(r)\), which of course depends on the radius \(r\) of the ampere loop:

Ampere law for the outer wire

Formula anchor$$ \begin{align} B \, 2\pi \, r ~=~ \mu_0 \, (I_{\text i} + I(r)) \end{align} $$

The current \(I(r)\) can be written with the constant current density \(j_{\text e}\) of the outer wire and the area \(A(r)\) enclosed by the ampere loop. This enclosed area is the area \(\pi \, r^2 \) of the ampere loop minus the area \(\pi \, {R_2}^2 \) (see Illustration 1):

The current density \(j_{\text e}\) is not known, so it is expressed by the given current \(I_{\text e}\) of the outer wire. The current density \(j_{\text e}\) is the current \(I_{\text e}\) per cross-sectional area through which this current flows. The cross-sectional area is the area \(\pi \, {R_3}^2 \) minus the area \(\pi \, {R_2}^2 \):

Enclosed current in the outer wire expressed with given quantities

If the current in the outer wire is the same current as in the inner wire, but flowing in the opposite direction (\(I := I_{\text i} = - I_{\text e}\)), then Eq. 13 becomes:

Magnetic field in the outer wire with opposing currents

To determine the magnetic field outside the coaxial cable, an ampere loop with radius \(r\) is placed outside the coaxial cable: \(r \geq R_3\). The enclosed current is the sum of the inner wire current and the outer wire current. Thus Eq. 4 becomes:

Ampere law for the outside of a coaxial cable

Formula anchor$$ \begin{align} B \, 2\pi \, r ~=~ \mu_0 \, (I_{\text i} + I_{\text e}) \end{align} $$

Rearrange eq. 15 with respect to the magnetic field \(B\):

So we can state: The magnetic field \(B(r)\) outside the coaxial cable falls off reciprocally with the radial distance \(r\).

If the current in the outer wire is the same current as in the inner wire, but flowing in the opposite direction (\(I := I_{\text i} = - I_{\text e}\)), then Eq. 16 becomes:

Magnetic field outside a coaxial cable with opposite currents

Formula anchor$$ \begin{align} B ~=~ 0 ~~~~ (r \geq R_3) \end{align} $$

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