Coaxial semi-infinite solenoid and superconducting disc

[Hak]

Sounds good.

This is what the person who proposed the problem said:

"Your answer for $B_{C_x}$ is correct. Unfortunately, your answer for $B_{A_x}$ is incorrect. The exact value of $B_{A_x}$ is extremely close to $B_{A_x} = \mu_0 n I$. For this part, try thinking of adding another similar solenoid to our solenoid and then try to look for symmetry in this arrangement.
For part 2, your answer is incorrect. I think you are correct about your initial thoughts, since the coil is superconducting, the magnetic flux passing through it must remain constant and equal to zero.

For part 3, there are 4 ways that I know of to solve this problem. You can try thinking that the force of interaction between the coil and the solenoid is directed along their common axis."

Picture: https://ibb.co/R393bBJ

How do you recommend we proceed?

 

[Hak]

According to Lenz's Law, the current induced in the loop generates a counter-flux $\Phi_L$ that opposes the external flux $\Phi_{ext}$, which attempts to thread the loop. The net flux through the loop is equal to the difference of the external flux attempting to thread the loop and the counter-flux due to the induced current: $$\Phi_{net} = \Phi_{ext} - \Phi_L$$. In the case of a superconducting loop, its resistance is zero, then the opposition of the counter-flux to the external flux is complete and $\Phi_{net}$ remains constant perpetually. So, since the total flux through the superconducting loop remains constant at $0$ (the reason for this is that any change in flux requires a nonzero emf around the loop, which requires in infinite current, so magnetic flux through the loop cannot change), the flux from the self-inductance $L$ of the loop must be equal and opposite to the external flux. So:

$$\Phi_{net} = \Phi_{ext} - \Phi_L = 0 \Rightarrow \Phi_{ext} = \Phi_L$$, where $\Phi_{L} = LI_{ring}$.

We conclude that:
$$\Phi_{ext} = LI_{ring}.$$

$$I_{ring} = - \frac {\mu_0 \pi n I R^2}{2L}$$ is the correct solution. For point 1, $$B_{A_x} = \frac{\mu_0 n I}{2}$$ is the correct solution. How to approach point 3 with these results?

 

[kuruman]

Unfortunately, your answer for $B_{A_x}$ is incorrect. The exact value of $B_{A_x}$ is extremely close to $B_{A_x} = \mu_0 n I$.

I am not sure I buy this statement. In the plane of the last solenoid loop the direction of $\mathbf{B}$ changes from parallel to $x$ at the center to perpendicular to $x$ at the circumference. How close is "extremely close" when the radial distance from the axis is $\frac{1}{3}r$?

For part 3 you should proceed by thinking of the 4 ways to approach the force. I already gave you one way involving the radial component of $\mathbf{B}$ at the circumference. Another way I can think of is to consider that the force is the negative gradient of the potential energy. In this case $U=\frac{1}{2}LI_{ring}^2$ and $F=-dU/dx$. So you have to figure out $dI_{ring}/dx$. Alternatively, you can write $U=\frac{1}{2}\Phi^2/L$ in which case you have to find an expression for $d\Phi/dx.$

I hope that you experience with this problem has satisfied your curiosity about Physics Olympiad problems. I wish you good luck in Italy.

 

[Hak]

I followed the first way calculating the radial field $B_R$. My final result was $$F = - \frac{1}{RL}\bigg(\frac{\mu_0 \pi n I r^2}{2}\bigg)^2$$ and it was correct. Thank you very much.

 

[kuruman]

You are welcome. What expression did you use for $B_R$?

 

[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.