Magnetic field of radially emitted alpha particles by a sphere of polonium A layer of polonium is deposited on the surface of a sphere of radius $R$. The metal emits alpha particles. Assume that these particles are emitted radially outward ($\vec e_r$), thus forming a current $\vec J(\vec r)=J(r)\vec e_r$. Will there be a magnetic field associated with this current?
 A: The Maxwell way
First I assume that the current is homogeneous. In that case the important point to realise is that if current goes away, the sphere becomes charged. If the total current going away is $J$ and the process starts $t=0$ then the charge on the sphere is $q=-J t$.
Maxwell tells us
$$ \vec \nabla \times \vec B= \mu_0 \vec j +\mu_0 \varepsilon_0 \partial_t \vec E $$
As 
$$\partial_t \vec E= \frac{1}{4 \pi \varepsilon_0} \partial_t \frac {-J t}{r^2}\frac{\vec r}{r}= -\frac{1}{4 \pi \varepsilon_0} \frac {J}{r^2}\frac{\vec r}{r}$$ 
and
$$\vec j= \frac{J}{4 \pi r^2} \frac{\vec r}{r}$$
the radial current density you finally get
$$ \vec \nabla \times \vec B= \mu_0 \frac{J}{4 \pi r^2} \frac{\vec r}{r} -\mu_0 \varepsilon_0 \frac{1}{4 \pi \varepsilon_0} \frac {J}{r^3}\vec r =0$$
So in the perfect case the answer is no, but you have to consider the accumulating charge on the sphere as well. On the other hand, decay is a statistical process such that there will be fluctuations in current density over time and solid angle. These fluctuations will results in some magnetic fields, but much smaller than what you would expect from a current $\vec J$.
The symmetry way
As the current is radial the whole system is fully rotational symmetric. As a consequence there is no specific direction in which the magnetic field could point other than in or out. If that would be the case, we would have created a magnetic monopole, which does not exist. The magnetic field, hence must be zero.
A little bit of both
If it was only about the current and we forget about the changing electric field we get something like
$$ \vec \nabla \times \vec B= \mu_0 \vec j$$
Symmetry requires the the current density is of the form (we know how it looks, but the general case is enough) $\mu_0 \vec j(\vec r)=\vec r f(r)/r$. We than can immediately say that there is an $F(r)$ with $\partial_\xi F'(\xi)=f(\xi)$ such that
$$\vec \nabla\cdot F(r)= \frac{\vec r}{r}f(r)$$
The current is a gradient field of a potential and an according source.
So the Maxwell equation without the electric field would require something like
$$ \vec \nabla \times \vec B= \vec \nabla\cdot F(r)$$
i.e.
$$ \vec \nabla\cdot (\vec \nabla \times \vec B)= \Delta F(r)$$
The lef hand side is zero, while we know that there is a source on the right, which is a contradiction.
