Is the magnetic field from this current rotationally symmetric?

[Trollfaz]

Imagine a very large disc with current converging to it's center as $$ I=- \frac{k}{r} \hat{r}$$
Obviously $\theta$ is absent in the equation so the magnetic field by this current should look the same no matter which angle one is observing. Then does this mean that the magnetic field is dependent on r and exists as circles (anticlockwise above the disc, clockwise below the disc) and weakens with r under Biot Savart Law

 

[Orodruin]

Your current density violates conservation of charge, which follows directly from Maxwell’s equations.

 

[Trollfaz]

Conservation of charge is only true if the system is isolated entirely but this can be maintained by external sources constantly inputting or withdrawing charges

 

[Orodruin]

Conservation of charge is only true if the system is isolated entirely but this can be maintained by external sources constantly inputting or withdrawing charges

When I say conservation of charge I refer to the continuity equation for charge. This is a local statement and your setup is violating it.

 

[Dale]

Conservation of charge is only true if the system is isolated entirely but this can be maintained by external sources constantly inputting or withdrawing charges

That inputting or withdrawing charge is a current.

The conservation of charge follows from Maxwell’s equations. If it is violated then Maxwell’s equations do not hold. Without them is there even a magnetic field at all? Certainly Biot Savart would not apply either.

 

[Vanadium 50]

This reminds me of "Can God create a stone so heavy he can't lift it?"

Asking what Maxwell's equations say for a system that violates them (and of course, cannot be physically realized) is pointless. Some would use other adjectives.

 

[vanhees71]

First of all a current is a scalar. What you seem to mean is a current density. Now you write
$$\vec{j}=-\frac{k}{r} \hat{r}.$$
This is obviously a stationary current density, and thus you should have $\vec{\nabla} \cdot \vec{j}=0$, but for your field
$$\vec{\nabla} \cdot \vec{j} = -\frac{k}{r^2} \neq 0,$$
i.e., there's no solution for the magnetostatics problem,
$$\vec{\nabla} \times \vec{B} = \mu_0 \vec{j}.$$
Maxwell's equations in general are only solvable, when the electromagnetic charge is conserved, i.e., if the continuity equation,
$$\partial_t \rho + \vec{\nabla} \cdot \vec{j}=0,$$
holds. This "integrability condition" follows from the gauge invariance of Maxwell's theory.

 

[Orodruin]

This is obviously a stationary current density, and thus you should have ∇→⋅j→=0, but for your field
∇→⋅j→=−kr2≠0,
i.e

This would be true in spherical coordinates. However, the OP is referring to a disc, which implies $r$ should probably be interpreted as the cylinder coordinate radius. If so, $\nabla\cdot\vec j=0$ everywhere … except in $r=0$, where it is not.

 

[vanhees71]

Well, ill-defined problems can lead easily to misunderstandings. So what's meant then seems to be
$$\vec{j}=-\frac{k}{R} \hat{R} \delta(z),$$
where $(R,\varphi,z)$ are cylinder coordinates. That also obviously violates $\vec{\nabla} \cdot \vec{j}=0$ ;-)).

 

[Orodruin]

Well, ill-defined problems can lead easily to misunderstandings. So what's meant then seems to be
$$\vec{j}=-\frac{k}{R} \hat{R} \delta(z),$$
where $(R,\varphi,z)$ are cylinder coordinates. That also obviously violates $\vec{\nabla} \cdot \vec{j}=0$ ;-)).

Not as obvious as you might think. The only violation occurs in R=0 where the coordinates are singular. Since the delta function depends only on z, it will not contribute to any divergence. The divergence becomes $[\partial_R(-k\delta(z))]/R = 0$ for $R>0$. It takes an additional argument to conclude the non-zero divergence at the origin so I’m not sure I would call it obvious.

 

[Trollfaz]

So the Biot Savart Law and Ampere's law only applies to steady current. Is the definition of a steady current one that results in no charge to the system

 

[Orodruin]

Time independent and $\nabla\cdot\vec j =0$, which leads to no charge accumulation. Somewhat simplified: charge accumulation would mean a changing charge, which would mean a changing E field, which would give additional contributions to the B field.

 

[vanhees71]

Not as obvious as you might think. The only violation occurs in R=0 where the coordinates are singular. Since the delta function depends only on z, it will not contribute to any divergence. The divergence becomes $[\partial_R(-k\delta(z))]/R = 0$ for $R>0$. It takes an additional argument to conclude the non-zero divergence at the origin so I’m not sure I would call it obvious.

To see what happens at $R=0$ let's integrate over an arbitrary collinear cylinder $Z$ of radius $a$. You formally get
$$\int_Z \mathrm{d}^3 x \vec{\nabla} \cdot \vec{j} = \int_{\partial Z} \mathrm{d}^2 \vec{f} \cdot \vec{j}.$$
The bottom and top circles don't contribute, and the mantle area element normal vector is $\mathrm{d}^2 \vec{f}=a \hat{R} \mathrm{d} \varphi \mathrm{d} z$:
$$\int_{M} \mathrm{d}^2 \vec{f} \cdot \vec{j}=\int_{0}^{2 \pi} \mathrm{d} \varphi \int_{-h/2}^{h/2} \mathrm{d} z a (-k/a) \delta(z)=-2 \pi k \neq 0.$$
So, as is also intuitively obvious, the singularity is such that there is a non-zero current, leading to an accumulation of charge at the center of the disk. I'm a bit lost at the task to find the electromagnetic fields. You have to assume some singular time-dependent charge density to get the charge conservation right...