Please provide some theoretical procedure which equates Biot-Savart law with the Maxwell's 4th equation for steady current, which is Ampere's law $$\quad \nabla\times{\bf B} = \mu_0{\bf J}.$$


1 Answer 1


Let me answer the second query.

We know $$\mathbf{ B}=\text{curl}\;\mathbf{ A} $$ where $\bf B$ is the magnetic field & $\bf A$ is vector-potential.

Now, using Maxwell's equation, we get $$\text{curl}\;(\text{curl}\; \mathbf A)= \mu_0 \mathbf J . $$ By cracking a bit-algebra, we get, $$-\frac{\partial^2 A_x}{\partial x^2}-\frac{\partial^2 A_x}{\partial y^2}- \frac{\partial^2 A_x}{\partial z^2} +\frac{\partial}{\partial x}\left(\frac{\partial A_x}{\partial x}+ \frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)= \mu_0 J_x.$$

Now, for convenience, we take $\text{div}\; \mathbf A= 0\; .$

This makes the above relation looks like $$-\frac{\partial^2 A_x}{\partial x^2}-\frac{\partial^2 A_x}{\partial y^2}- \frac{\partial^2 A_x}{\partial z^2} = \mu_0 J_x\;.$$ The vector potential at $(x_1,y_1,z_1)$ is then given by $$\mathbf A(x_1,y_1,z_1)= \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(x_2,y_2,z_2)\;\mathrm dv_2}{r_{12}}\;.$$

Now, consider a loop of wire carrying current $I$.

Now, $\mathrm{d} v_2= a\; \mathrm{d} l$ where $\mathrm dl$ is an infinitesimal section of the wire; $\mathbf{J} \; \mathrm{d} v_2= I\; \mathrm{d}\mathbf{l} .$

Therefore our vector-potential for the thin-wire carrying steady current is given by $$\mathbf A= \frac{\mu_0 I}{4\pi}\int \frac{\mathrm d\mathbf l}{r_{12}}.$$

Let us focus on that section of wire which happens to let the current in the $\hat{\mathbf x}$ direction & is located at the origin of our frame.

Then at a certain point $(x,y)$ in $xy$ plane, contribution to the vector-potential from the infinitesimal wire-section at the origin is given by $$\mathrm{d}\mathbf A= \hat{\mathbf x} \;\frac{\mu_0 I}{4\pi} \frac{\mathrm{d}\mathbf l}{\sqrt{x^2 + y^2}} \;.$$

Since, $\mathbf A$ is in $xy$ plane, its curl must point in the $\hat{\mathbf z}$ direction. therefore, \begin{align}\mathrm{d}\mathbf B &= \text{curl}\;\mathrm{d}\mathbf A\\& =\hat{\mathbf z}\left(-\frac{\partial A_x}{\partial y}\right)\\ &= \hat{\mathbf z}\;\frac{\mu_0 I}{4\pi} \frac{ \mathrm{d} l\; y}{(x^2 + y^2)^{3/2}}\\ &= \hat{\mathbf z}\;\frac{\mu_0 I}{4\pi} \frac{\mathrm{d}l\; \sin\varphi}{r^2}\;. \end{align}

Now, that being deduced, we can say that this would be valid for any general coordinate system; all that matters is the relative orientation of the element $d\mathbf l$, radius vector $\bf r$ from the element to the concerned point.

The contribution to the magnetic field from any element of wire can be taken to be a vector perpendicular to the plane containing $\mathrm{d} \mathbf l$ & $\bf r$ & the angle between them namely $\varphi\; .$

This can be compactly written as $${\mathrm{d}\mathbf B= \frac{\mu_0}{4\pi} \frac{I\; \mathrm{d}\mathbf l \times \hat{ \mathbf r}}{r^2}}\; .$$ And this is Biot-Savart Law, which you wanted.

  • $\begingroup$ Can Mr. Downvoter please tell me how can I please him? For I'm not an AI that can decipher the reason for the discontentment. $\endgroup$
    – user36790
    Mar 3, 2016 at 12:35

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