Magnetic field of a circular loop with current

In this article, are derived the formulae for the magnetic flux produced by a circular current loop. The derivation is done via integrating the Biot-Savart-Laplace law along the contour of the loop.

We start by considering a circular loop with current $I$. Let us associate a polar coordinate system with the loop with its origin at the center of the circle (see fig.1). Here we define $\vec{r}$ as a vector from origin to a point belonging to the contour of the loop, and $\,\mathrm{d}\vec{r}$ as an ininitely small increment of that vector. Provided the contour of the loop is a planar circular curve of radius $a$, vector $\vec{r}$ can be indexed by the angle $\varphi$. We also define the observation point as the vector $\vec{r}_0$.

$$\mathrm{d}\vec{B}(\vec{r}_0)=\frac{\mu_0 I}{4\pi} \cdot \frac{[\,\mathrm{d}\vec{r} \times (\vec{r}_0-\vec{r})]}{|\vec{r}_0-\vec{r}|^3}$$

Biot-Savar-Laplace Law

Fig.1 Circular loop with current. Illustration of Biot-Savart-Laplace Law

 

Biot-Savart-Laplace Law gives the elementary contribution to the magnetic field $\,\mathrm{d}\vec{B}$ produced by an element $\,\mathrm{d}\vec{r}$ of the current loop. The expressions that constitute the above formula, are expanded explicitly as follows below.

$$\vec{r}_0 = (\rho\cos{\varphi}, \rho\sin{\varphi}, z) \overset{\varphi = 0}{\rightarrow} (\rho, 0, z)$$

$$\vec{r} = (a\cos{\varphi}, a\sin{\varphi}, 0)$$

$$\mathrm{d}\vec{r} = (-a\sin{\varphi}, a\cos{\varphi}, 0)\,\mathrm{d}\varphi$$

$$\vec{r}_0-\vec{r} = (\rho - a\cos{\varphi}, -a\sin{\varphi}, z)$$

$$[\mathrm{d}\vec{r} \times (\vec{r}_0-\vec{r})] = \begin{vmatrix}
 \vec{e}_x&  \vec{e}_y& \vec{e}_z\\ 
 -a\sin{\varphi}\,\mathrm{d}\varphi&  a\cos{\varphi}\,\mathrm{d}\varphi& 0\\ 
 \rho - a\cos{\varphi}& -a\sin{\varphi}& z 
\end{vmatrix} = (az\cos{\varphi}, az\sin{\varphi}, a^2 - a\rho\cos{\varphi})\,\mathrm{d}\varphi$$

$$|\vec{r}_0-\vec{r}|^3 = \left(\rho^2 + a^2 + z^2 -2\rho a\cos{\varphi}\right)^{\frac{3}{2}}$$

Now that we have all the neccessary expressions, we can obtain the answer for the magnetic field by direct integration of the Biot-Savart-Laplace law.

$$\vec{B}(\vec{r}_0) = \int_C{\,\mathrm{d}\vec{B}(\vec{r}_0)}$$

Thus, after all calculations, we obtain the expression for the axial component of the magnetic field at any given point of space. Note that due to the axial symmetry of the problem, the answer does not depend on the polar angle $\varphi$.

$$B_z(\rho, z) = \frac{\mu_0 I}{4\pi} \int_0^{2\pi}{\frac{\left( a^2 - \rho a\cos{\varphi}\right)\,\mathrm{d}\varphi}{\left(\rho^2 + a^2 + z^2 -2\rho a\cos{\varphi}\right)^{\frac{3}{2}}}}$$

This result has a few remarkable things about it:

  1. The magnitude of the magnetic field is inversely proportional to $a$, i.e. $B\propto\frac{1}{a}$
  2. The magnitude of the magnetic field is inversely proportional to $\rho^2$, i.e. $B\propto\frac{1}{\rho^2}$
  3. The magnitude of the magnetic field is inversely proportional to $z^3$, i.e. $B\propto\frac{1}{z^3}$
  4. These relations will conserve for large $a$, $\rho$ and $z$ regardless of the shape of the loop

Similarly, we can obtain the radial component of the magnetic field

$$B_r(\rho, z) = \frac{\mu_0 I}{4\pi} \int_0^{2\pi}{\frac{a\,z\,\,\mathrm{d}\varphi}{\left(\rho^2 + a^2 + z^2 -2\rho a\cos{\varphi}\right)^{\frac{3}{2}}}}$$

If we consider a loop of radius $a = 0.1$ m and carrying electric current of $I = 1$ A, then we can plot the following three graphs for the the magnetic field (fig.2, 3 and 4).

Axial magnetic field across the radial axis

Fig.2 Magnitude of the axial magnetic field component versus the distance away from the Z-axis

 

Radial magnetic field across

Fig.3 Magnitude of the radial magnetic field component versus the distance away from the Z-axis

 

We can also find the absolute magnitude of the magnetic field as $B = \sqrt{B_r^2 + B_z^2}$

Magnitude of the magnetic field

Fig.4 Magnitude of the magnetic field versus the distance from Z-axis