📂Electrodynamics

Steady Current and Biot-Savart Law

Definition¹

A steady current refers to the flow of charge that continues without changing in amount or direction.

Description

Since the current does not change over time, the magnetic field created by the steady current also does not change over time. The ‘direction of progress’ mentioned here is a different concept from the direction of a vector we commonly think of. It means that as long as the flow continues in one direction, even if it flows through a curved conductor, the direction of progress does not change. Let’s denote the volume charge density as $\rho$, and the volume current density as $\mathbf{J}$. If the current generated by this is a steady current, then by definition, the following equation holds.

$$ \dfrac{\partial \rho}{\partial t} = 0 \quad \text{and} \quad \dfrac{\partial \mathbf{J}}{\partial t}=0 $$

Therefore, according to the continuity equation, the following equation applies.

$$ \nabla \cdot \mathbf{J} = 0 $$

Of course, steady current is theoretical, and it does not actually exist, so the discussion on steady current is entirely theoretical. However, in many areas of physics, such theory closely approximates reality.

Formula

The magnetic field created by a steady current can be calculated with the following formula, which is called the Biot-Savart Law.

$$ \mathbf{B}(\mathbf{r})=\dfrac{ \mu_{0}}{4\pi}\int \dfrac{\mathbf{I} \times \crH}{\cR ^2}dl^{\prime}=\dfrac{ \mu_{0}}{4\pi} I \int \dfrac{d \mathbf{l}^{\prime} \times \crH}{\cR ^2} $$

Here, $\bcR$ refers to the displacement vector, and the constant $\mu$ is the permeability. $\mu_{0}$ is the permeability in a vacuum. The Biot-Savart law for surface and volume currents is represented using surface current density and volume current density.

$$ \begin{align*} \mathbf{B}(\mathbf{r}) =&\ \dfrac{ \mu_{0}}{4\pi}\int \dfrac{\mathbf{K}(\mathbf{r}^{\prime}) \times \crH}{\cR ^2}da^{\prime} \\ \mathbf{B}(\mathbf{r}) =&\ \dfrac{ \mu_{0}}{4\pi}\int \dfrac{\mathbf{J}(\mathbf{r}^{\prime}) \times \crH}{\cR ^2}d\tau^{\prime} \end{align*} $$

Example

Calculate the magnetic field at a point located at a perpendicular distance $s$ from a wire carrying a steady current of $I$.

$$ \begin{align*} |d\mathbf{l}^{\prime} \times \crH | =&\ |d\mathbf{l}^{\prime}||\crH|\sin \alpha \\ =&\ dl^{\prime} \sin \alpha \\ =&\ dl^{\prime} \sin \left( \theta + \frac{\pi}{2} \right) \\ =&\ dl^{\prime} \cos \theta \end{align*} $$

Since $l^{\prime}=s\tan \theta$,

$$ dl^{\prime}=\dfrac{s}{\cos ^2 \theta}d\theta $$

Since $s=\cR \cos \theta$,

$$ \dfrac{1}{\cR ^2}=\dfrac{\cos ^2 \theta}{s^2} $$

Substituting into the Biot-Savart Law, and calculating the magnitude of $\mathbf{B}(\mathbf{r})$,

$$ \begin{align*} B =&\ \left| \dfrac{ \mu_{0}}{4\pi} I \int \dfrac{d \mathbf{l}^{\prime} \times \crH}{\cR ^2} \right| \\ =&\ \dfrac{ \mu_{0}}{4\pi} I \int \dfrac{ \left| d \mathbf{l}^{\prime} \times \crH \right| }{\cR ^2} \\ =&\ \dfrac{\mu_{0} I}{4\pi} \int \left( \dfrac{\cos ^2 \theta}{s^2} \right) \left( \dfrac{s}{\cos^2\theta} \right) \cos \theta d\theta \\ =&\ \dfrac{\mu_{0} I}{4\pi s} \int \cos \theta d\theta \end{align*} $$ If it had been a case for a segment of wire as shown in diagram$(2)$, then the integration limits would be from $\theta _{1}$ to $\theta_2$. Since the example is regarding an infinitely long wire, as in diagram $(2)$, situations are like $\theta_{1}=-\dfrac{\pi}{2}$, $\theta_2=\dfrac{\pi}{2}$. Therefore, the magnitude of the magnetic field is $$ \begin{align*} B =&\ \dfrac{\mu_{0} I}{4\pi s} \int_{-\frac{\pi}{2} }^{\frac{\pi}{2}} \cos \theta d\theta \\ =&\ \dfrac{\mu_{0} I}{4\pi s} \left(\sin {\textstyle \frac{\pi}{2}}- \sin {\textstyle \frac{-\pi}{2}} \right) \\ =&\ \dfrac{\mu_{0} I}{2\pi s} \end{align*} $$

Its direction is out of the paper according to the right-hand rule. If the right-hand side is set as $\hat{\mathbf{z}}$ in the cylindrical coordinate system,

$$ \mathbf{B}=\dfrac{\mu_{0} I}{2\pi s} \hat{\boldsymbol{\phi}} $$

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