Coaxial semi-infinite solenoid and superconducting disc

[Hak]

According to Gauss's Theorem for the magnetic field, we have that the divergence of $$\displaystyle \textbf{B}$$ is equal to $$0$$: the total magnetic field flux through a closed surface non-secanting the solenoid is zero. Applying Gauss's Theorem to a spherical collinear distribution to the solenoid with radius $$R$$ and centered on the center of the base of the multiple-winding coil, the total magnetic field across that surface is given by the difference between the flux $$\displaystyle \Phi_0 = \Phi_{inside}$$ of the magnetic field $$\displaystyle B_0 = B_{inside} = \mu_0 nI$$ at the centre of the sphere-solenoid and the flux $$\Phi_R$$ of the radial magnetic field $$B_R$$ through the coil.
The flux $$\Phi_0$$ is given by: $$\displaystyle \Phi_0 = B_0 S_{circle}$$, where $$\displaystyle S_{circle} = \pi r^2$$ is the surface of the superconducting loop. So: $$\displaystyle \Phi_0 = \mu_0 n \pi I r^2$$.
The flux $$\Phi_R$$ is given by: $$\displaystyle \Phi_R = B_R S_{sphere}$$, where $$\displaystyle S_{sphere} = 4 \pi R^2$$ is the surface of the sphere-solenoid. So: $$\displaystyle \Phi_R = B_R \cdot 4 \pi R^2$$.

So, we obtain:
$$\displaystyle \nabla \cdot \textbf{B}= 0$$ $$\Rightarrow$$ $$\displaystyle \Phi_{tot} = 0$$ $$\Rightarrow$$ $$\displaystyle \Phi_0 - \Phi_R = 0$$ $$\Rightarrow$$ $$\displaystyle \Phi_0 = \Phi_R$$ $$\Rightarrow$$ $$\displaystyle \mu_0 n \pi I r^2 = B_R \cdot 4 \pi R^2$$ $$\Rightarrow$$ $$\displaystyle \boxed{B_R = \frac{\mu_0 n I r^2}{4 R^2}}$$.

 

[kuruman]

This equation for the flux $$\displaystyle \Phi_R = B_R 4\pi R^2$$ is valid only if $B_R$ constant everywhere on the surface of the sphere. That is clearly not the case here.

the total magnetic field flux through a closed surface non-secanting the solenoid is zero

The magnetic flux through any closed surface is zero regardless of whether you have a solenoid or where the surface is relative to it. That comes from the divergence of B being equal to zero. The flux that goes in must come out. The situation is different if you have an open surface, but you didn't specify one here. So $\Phi_R=0.$

I do not believe that your arguments are correct.

 

[Hak]

This equation for the flux $$\displaystyle \Phi_R = B_R 4\pi R^2$$ is valid only if $B_R$ constant everywhere on the surface of the sphere. That is clearly not the case here.

The magnetic flux through any closed surface is zero regardless of whether you have a solenoid or where the surface is relative to it. That comes the divergence of B being equal to zero. The flux the goes in must come out. The situation is different if you have an open surface, but you didn't specify one here. So $\Phi_R=0.$

I do not believe that your arguments are correct.

You are absolutely right, but the good thing is that this is also the official solution. This really is a dilemma.

 

[kuruman]

There is no dilemma. The official solution is incorrect. I suggest that you find another source of challenging problems.

 

[Hak]

Thank you, you have opened my eyes. It's not my fault, I thought Russian Olympiads were a good source of intriguing problems. If the official solutions are wrong, I would say it is very, very bad. Thank you for everything.