Is Biot-Savart law valid to derive the magnetic field for a point charge moving at a velocity $v$ at any point?

Question

According to Biot-Savart Law, the magnetic field at a point by current of length $dl$ is given by: $$\vec dB = {\mu _0 I \vec{dl}×\vec r \over 4π r^3}\tag{01}\label{01}$$

Now for a point charge , we can write $dl$ as $v$ $dt$ and $I$ $dt$ is the charge that passes which is $q$ ,so the expression becomes: $$ \vec B = {\mu _0 q \vec{v}×\vec r \over 4π r^3}\tag{02}\label{02}$$

But I had read in David J. Griffiths - Electrodynamics that the Biot-Savart's law is valid only for steady currents and an isolated moving charge is not a steady current.

Later he explicitly states that this formula is wrong! But here someone has given a very fancy proof, which I don't really understand , that this formula is valid

This left me confused, wether the above way of solving for magnetic field is correct or not. Please help

I'm still in highschool , so don't know what the derivation of Biot-Savarts law is.

Answer 1

Magnetostatics (and electrostatics for that matter) still work then the individual charges are moving provided they are moving at speeds much less than the speed of light $c$. In particular, a current is a collection of individual moving charges, and a steady current would be a large number of charges moving with the same velocity in the same direction. The discrete nature of the charges is ignored, just like one ignores the discrete nature of water molecules when considering the steady flow of water through a pipe. Thus the expression in terms of $v$ is valid for individual charges, at least for $v/c\ll 1$.

When $v/c$ is not much less than $1$, the fields transforms in a certain way under Lorentz transformations: in your specific problem one would go to the rest frame of the particle, compute the $\vec E$ field there, and then boost to the moving frame to obtain the resulting electric and magnetic fields.

(Unfortunately the SI units are not very insightful but in CGS there would be a factor of $v/c$ in your expression for $\vec B$ and the condition $v/c\ll 1$ would be much more obvious.)

Answer 2

This is derived in the context of the quasi static regime where amperes law$ \int B \cdot dl = \mu_0 I$ still holds, taking the divergence of this gives$ \nabla \cdot J = 0$ the current density of a point charge doesn't follow this.

$\nabla × A = B$

$\nabla × (\nabla × A) = \mu_0 J$

$\nabla(\nabla \cdot A) - \nabla^2 A = \mu_0 J$

Set $\nabla \cdot A = 0 $ due to field invariance

$\nabla^2 A = -\mu_0 J$

$ A(r',t) =\iiint \frac{\mu_0}{4\pi} \frac{J(r',t)}{|r-r'|} d^3r'$

for a point charge

$J(r',t) = Q r_{s}'(t)\delta^3(r'-r_{s}(t))$

where $ r_{s}(t) $is the position vector of the charge

A =$\iiint \frac{Q\mu_0}{4\pi} \frac{ r_{s}'(t)\delta^3(r'-r_{s}(t))}{|r-r'|} d^3r'$

= $\frac{Q\mu_0}{4\pi} \frac{r_{s}'(t)}{|r-r_{s}(t)|}$

$B = \nabla × (\frac{Q\mu_0}{4\pi} \frac{r_{s}'(t)}{|r-r_{s}(t)|})$

Changing coordinates such that$ |r-r_{s}(t)|$ is just radial distance R from the charge,

$ \frac{Q\mu_0}{4\pi} \nabla × ( \frac{r_{s}'(t)}{R})$

=$ \frac{Q\mu_0}{4\pi} (\nabla(\frac{1}{R}) × r_{s}'(t) +\frac{1}{R}\nabla × r_{s}'(t) ) $

=$ \frac{Q\mu_0}{4\pi} \nabla(\frac{1}{R}) ×r_{s}'(t) $

=$ -\frac{Q\mu_0}{4\pi} \frac{\hat r ×r_{s}'(t)}{R^2} $

=$ -\frac{Q\mu_0}{4\pi} \frac{\hat r ×v}{R^2} $

=$ \frac{Q\mu_0}{4\pi} \frac{v×\hat r}{R^2} $

Answer 3

$$ \vec{dB} = \frac{\mu _0}{4π r^3}.dq \vec{v}×\vec r $$

thus, $$ \vec{B} = \int \frac{\mu _0}{4π r^3}.dq \vec{v}×\vec r $$

where the $dq$ defines with kind of charge you are dealing with, the distribution of charge is related the dq by $$\rho=\frac{dq}{dv},$$ $$\sigma =\frac{dq}{ds}$$ or $$ \lambda=\frac{dq}{dl}$$ Using these last definitions of charge density you calculate the integral magnetic field. Let's now discuss, the point charge case. The charge point is dimensionless and this means that the integral over qd gives q and finally one finds, $$ \vec{B} = \int \frac{\mu _0}{4π r^3}.dq \vec{v}×\vec r=\frac{\mu _0}{4π r^3}.dq \vec{v}×\vec r \int dq $$ It is obvuious that $\int dq=q$ (no spatial dependence) which leads to $$ \vec{B} =\frac{\mu _0}{4π r^3}.q \vec{v}×\vec r $$.

Answer 4

Definitely it works for steady currents. If charges accelerate(alternating current) there appears vortex magnetic field. The second equation is also for constantly moving charged particles.

$$ \vec dB = {\mu _0 I \vec{dl}×\vec r \over 4π r^3}={\mu _0 \frac{q}{dt} \vec{dl}×\vec r \over 4π r^3}={\mu _0 q \vec{v}×\vec r \over 4π r^3}$$

Current is created by only one charge.