We previously Electrostatics found expression for electrostatics now lets look at magnetostatics

• Conditions for magnetostatics (no time dependence)

$$ \frac{\partial \vec j}{\partial t}=0 \qquad \frac{\partial \rho}{\partial t}=0 $$

💼 Case: Same as previous setup except this time we are looking at the magnetic field due to a wire

• 🧠 Reminder: notation

• We define the magnetic vector potential as

  $$ \boxed{\vec A(\vec r)=\frac{\mu_0}{4\pi} \int \frac{\vec j (\vec r')}{R}\,\text d \tau '} $$

  where $\vec j(\vec r')$ is the current density

🦚 Properties:

1. $\vec A(\vec r)$ is in the same direction as $\vec j (\vec r)$

2. The equation can be decomposed into 3 separate equations for each of the 3D coordinates $A_i$

3. We see the similarities between the equations for $V$ and $A_i$

   $$ A_i(\vec r)=\frac{\mu_0}{4\pi} \int \frac{j_i (\vec r')}{R}\,\text d\tau ' \qquad \qquad V(\vec r)=\frac{1}{4\pi \epsilon_0}\int\frac{\rho (\vec r')}{R} \,\text d \tau ' $$

   🗒️ Note: especially since even the constants are related ie $c=(\mu_0 \epsilon_0)^{-\frac 12}$ so $\mu_0 \propto \frac{1}{\epsilon_0}$

4. They can both be expressed as $4$-vectors and they are related in $4$-space

   $$ \left ( \frac{V}{c} ,\vec A \right ) \qquad \left ( \rho c,\vec j \right ) $$

• Since $\vec j (\vec r')=\rho ( \vec r')\vec v (\vec r')$ where $\vec v (\vec r')$ is the velocity we can write

  $$ \vec A(\vec r)=\frac{\mu_0}{4\pi} \int \frac{\rho(\vec r')\vec v(\vec r')}{R}\,\text d \tau ' $$

  • 🗒️ Note: We can make the equations $A_i$ and $V$ look even closer using this notation

Finding the magnetic field $\vec B$

$$ \vec B =\vec \nabla \times \vec A =\frac{\mu_0}{4\pi} \int \vec \nabla \times \left ( \frac{\vec j ( \vec r')}{R} \right )\,\text d \tau ' $$

• We use the vector identity $\vec \nabla \times (f\vec A)=f(\vec \nabla \times \vec A)-\vec A \times (\vec \nabla f)$ (expand for details)

$$ \vec B =\frac{\mu_0}{4\pi} \int \vec j(\vec r')\times\frac{ \hat R}{R^2}\,\text d \tau ' $$

• Replacing the current density (3D) by the current (1D) ($\vec j (\vec r') \,\text d \tau ' = I\,\text d \vec l'$) we get Biot-Savart law

  $$ \boxed{\vec B =\vec \nabla \times \vec A = \frac{\mu_0 I}{4\pi} \int \,\text dl' \times \frac{\hat R}{R^2}} $$

• Using the vector identity $\vec \nabla \cdot (\vec \nabla \times \vec A)=0$ we get

  $$ \vec \nabla \cdot \vec B = \vec \nabla \cdot ( \vec \nabla \times \vec A )=0 $$

Applying the Biot-Savart law

Infinite wire

For details on calculation go to B-field from infinitely long wire

• The $B$-field at a distance $R$ from an infinite wire carrying current $I$ is

  $$ B=\frac{\mu_0 I}{2\pi R} $$