Coaxial semi-infinite solenoid and superconducting disc

[Hak]

Homework Statement

A semi-infinite solenoid with a coil radius $r$ and a winding density $n$ (the number of coils per unit length) is located coaxially with a circular superconducting coil of radius $R$ so that its base is in the plane of the coil. It is known that $r \ll R$ . Initially, there was no current in the coil. The inductance of the coil is equal to $L$. The current strength in the solenoid is slowly increased from zero to $I$ and then maintained constant. The wires supplying current to the solenoid are arranged in such a way that their magnetic field and their interaction with other elements can be neglected. Let's direct the $x$ axis as shown in the figure.

1. Points $A$ and $C$ are located in the plane of the coil at distances $\frac{r}{3}$ and $3r$, respectively, from the axis of symmetry of the system. Find the projections of induction $B_{A_x}$ and $B_{C_x}$ of the magnetic field generated by the solenoid at points $A$ and $C$, respectively.

2. Find current strength in the coil. Which direction is it directed?

3. Find the magnitude and direction of the magnetic interaction force acting on the solenoid from the side of the coil.

Figure: https://ibb.co/R393bBJ

Relevant Equations

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I have no idea how to deal with the problem. Do you have any hints, please?

 

[kuruman]

I have no idea how to deal with the problem. Do you have any hints, please?

I have no idea how to deal with this problem either. Someone might have something for you if you posted the figure that is referenced in the statement of the problem. In any case forum rules require you to post your own thoughts and attempts at a solution before receiving help.

Please read, understand and follow our homework guidelines, especially item 4, here
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

 

[Hak]

I have no idea how to deal with this problem either. Someone might have something for you if you posted the figure that is referenced in the statement of the problem. In any case forum rules require you to post your own thoughts and attempts at a solution before receiving help.

Please read, understand and follow our homework guidelines, especially item 4, here
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

You're right, but in a previous message I transcribed my approach to a problem, without receiving any response. Of course, I have attempted my own solution, but it does not reach any conclusive point. In the hope of receiving help, however, I transcribe it.

 

[Hak]

The magnetic field generated by the solenoid at points $A$ and $C$ could be found by using the formula for the magnetic field inside an infinite solenoid: $$B = \mu_0 n I$$, where $\mu_0$ is the magnetic permeability of free space, $n$ is the winding density of the solenoid, and $I$ is the current in the solenoid. Since the solenoid is semi-infinite, we need to multiply this formula by a factor of $1/2$ to account for the missing half. Therefore, the magnetic field inside the solenoid is $$B = \frac{1}{2} \mu_0 n I$$ I don't know how to continue.

The current in the coil is induced by the changing magnetic flux due to the increasing current in the solenoid. The magnetic flux through the coil is given by $$\Phi_B = BA\cos{\theta}$$, where $B$ is the magnetic field due to the solenoid, $A$ is the area of the coil, and $\theta$ is the angle between the normal to the coil and the magnetic field. Since the coil is in the plane of the base of the solenoid, $\theta = 0$, and we have $$\Phi_B = BA = \frac{1}{2} \mu_0 n I \pi R^2$$, where $R$ is the radius of the coil. The induced emf in the coil is given by Faraday’s law: $$\mathcal{E} = -\frac{d\Phi_B}{dt}$$. Assuming that the current in the solenoid increases linearly from zero to $I$ in a time interval $\Delta t$, we have $$\mathcal{E} = -\frac{d}{dt}\left(\frac{1}{2} \mu_0 n I \pi R^2\right) = -\frac{1}{2} \mu_0 n \pi R^2 \frac{dI}{dt}$$. The current in the coil is related to the induced emf by Ohm’s law: $$\mathcal{E} = IR$$, where $R$ is the resistance of the coil. Therefore, we have $$I_c = -\frac{\mathcal{E}}{R} = \frac{1}{2R} \mu_0 n \pi R^2 \frac{dI}{dt}$$. This current is directed opposite to the current in the solenoid, according to Lenz’s law, which states that an induced current opposes the change that causes it.

The magnetic interaction force between two coils can be found by using Ampère’s force law: $$dF = \frac{\mu_0}{4\pi}\frac{I_1 I_2}{r^2}\left(dl_1\times dl_2\right)$$, where $I_1$ and $I_2$ are the currents in each coil, $r$ is the distance between two differential elements of length $dl_1$ and $dl_2$. To find the total force, we need to integrate this expression over both coils. However, this integration is very complicated and does not have a simple analytical solution. Therefore, we will use an approximation based on two assumptions: (1) the radius of the solenoid $r$ is much smaller than the radius of the coil $R$, and (2) the distance between the coils $d$ is much smaller than both radii. Under these assumptions, we can approximate both coils as circular loops with currents $I_1$ and $I_2$, and radii $r$ and $R$, respectively. The distance between their centers is $d$. The force between two circular loops can be found by using a formula derived by Biot and Savart: $$F = -\frac{\mu_0 I_1 I_2 r^2 R^2}{4(d^2 + R^2)^{3/2}}$$. This force is attractive and directed along the axis of symmetry of the system.

 

[kuruman]

There is a problem with this equation
$$I_c = -\frac{\mathcal{E}}{R} = \frac{1}{2R} \mu_0 n \pi R^2 \frac{dI}{dt}$$
You are using the same symbol $R$ for the resistance of the coil and for the radius of the coil. Even if you changed one of the symbols, what is the resistance of a superconducting loop?

 

[Hak]

You are right, caught up in my haste I did not check well before posting. Obviously, the resistance of a superconducting loop is equal to $0$. But the current will not be infinite though - for a given inductance $L$, the current increases according to $$V=−L\frac{dI}{dt}$$. So?

 

[kuruman]

So you need to answer question 1 that has nothing to do with the presence of the coil. How do you propose to proceed?

This doesn't work

Since the solenoid is semi-infinite, we need to multiply this formula by a factor of 1/2 to account for the missing half.

because the equation $B=\mu_0~n~i$ works for points far away from the ends. The point of interest is at one end. It looks like you need to do an integral.

 

[Hak]

So you need to answer question 1 that has nothing to do with the presence of the coil. How do you propose to proceed?

This doesn't work

because the equation $B=\mu_0~n~i$ works for points far away from the ends. The point of interest is at one end. It looks like you need to do an integral.

So, we need to know the magnetic field near the edge of the solenoid. For locations along the axis of the coil, we can consider $\overrightarrow{dB}$ due to one ring of current and integrate over all coils to find $\overrightarrow{B}$. In essence, I apply the principle of superposition. A coil of wire, radius $R$, carries a current $I$. It creates a magnetic field at generical point $P$. The magnetic field due to a tiny segment of the coil, up at the top of loop, is $$\overrightarrow{dB} = \frac{\mu_0 I}{4 \pi}\frac{d\overrightarrow{S} \times \hat{r}}{r^2}$$. A field of the same strength will be created by a segment at bottom of loop. The vertical component cancel out, leaving: $$dB_{horiz} = \frac{\mu_0 I}{4 \pi}\frac{dS}{D^2 +R^2}\frac{R}{\sqrt{(D^2 +R^2)}}$$. If we integrate over every tiny segment in the wire, we find the total magnetic field: $$\overrightarrow{B} = \int_{}^{}d\overrightarrow{B} = \frac{\mu_0 I}{4 \pi}\frac{R}{(D^2 +R^2)^\frac{3}{2}}\int_{}^{}dS = \frac{\mu_0 I}{4 \pi}\frac{R}{(D^2 +R^2)^\frac{3}{2}} \cdot 2 \pi R \Rightarrow \overrightarrow{B} = \frac{\mu_0 I}{2}\frac{R^2}{(D^2 +R^2)^\frac{3}{2}}$$. That gives us the magnetic field along the axis of a single loop. To calculate $\overrightarrow{B}$ along axis of a solenoid, we must integrate over all the loops. Consider a section of the solenoid of length $dx$. The total current winding around the solenoid in that section is $$ I_{tot} = I(n \cdot dx)$$. This section is located a distance $$z = D+x$$ away from point $P$. So it contributes $$B = \frac{\mu_0 I n dx}{2}\frac{R^2}{((D+x)^2 +R^2)^\frac{3}{2}}$$. And so the total magnetic field at a point $P$ which is $D$ away from the edge of the solenoid is $$B_{tot} = \int_{0}^{L}\frac{\mu_0 I n dx}{2}\frac{R^2}{((D+x)^2 +R^2)^\frac{3}{2}} = \frac{\mu_0 I n R^2}{2}\int_{0}^{L}\frac{dx}{((D+x)^2 +R^2)^\frac{3}{2}}$$. It simplifies a bit to substitute $z = D+x$. So: $$B_{tot}= \frac{\mu_0 I n R^2}{2}\int_{D}^{D+L}\frac{dz}{(z^2 +R^2)^\frac{3}{2}}$$. The integral can be done: $$\int_{}^{}\frac{dz}{(z^2 +R^2)^\frac{3}{2}} = \frac{z}{R^2(z^2 +R^2)^\frac{1}{2}}$$. So:
$$B_{tot}= \frac{\mu_0 I n R^2}{2 R^2}\bigg[\frac{z}{R^2(z^2 +R^2)^\frac{1}{2}}\bigg]_{D}^{D+L}$$ $$\Rightarrow$$ $$B_{tot}= \frac{\mu_0 I n}{2}\bigg[\frac{D+L}{\sqrt{(D+L)^2+R^2}} - \frac{D}{\sqrt{D^2+R^2}}\bigg]$$.

This process does not sit well with me. Where do I go wrong? How could I apply it to the case of a semi-infinite solenoid?

 

[kuruman]

The vertical component cancel out, leaving $~\dots$

They don't because you are asked to find the field at points A and C which are off-axis. Below is the appropriate figure where $ds_1$ and $ds_2$ are diametrically opposed current elements. The magnetic field contributions $dB_1$ and $dB_2$ don't have equal lengths because the distances to the current elements are not equal. Furthermore, the angle between $dB_1$ and $dB_2$ is not bisected by the horizontal. So the symmetry argument does not apply.

At this point, I must ask you what level course are you in that requires you to solve this problem? Direct integration involves elliptic integrals which cannot be solved analytically which means that some approximation techniques might be necessary.

 

[Hak]

They don't because you are asked to find the field at points A and C which are off-axis. Below is the appropriate figure where $ds_1$ and $ds_2$ are diametrically opposed current elements. The magnetic field contributions $dB_1$ and $dB_2$ don't have equal lengths because the distances to the current elements are not equal. Furthermore, the angle between $dB_1$ and $dB_2$ is not bisected by the horizontal. So the symmetry argument does not apply.

View attachment 331481

At this point, I must ask you what level course are you in that requires you to solve this problem? Direct integration involves elliptic integrals which cannot be solved analytically which means that some approximation techniques might be necessary.

This year I will attend a University in Italy, Faculty of Physics. This problem is from some Physics Olympiads, so I tried to tackle it out of personal curiosity, even though I have never dealt with these topics. How do you recommend I proceed? Aren't there any tricks or special conditions that can make me avoid using the elliptic integral? Usually, procedures in the Physics Olympiads do not involve such advanced concepts as elliptic integrals...

 

[kuruman]

I found this on the web. I think you will find it useful. The answer for part 1 hinges on the observation that the flux through the end of the semi-infinite solenoid is half the flux in the middle.

 

[Hak]

I found this on the web. I think you will find it useful. The answer for part 1 hinges on the observation that the flux through the end of the semi-infinite solenoid is half the flux in the middle...

Thank you so much. However, I had already seen this useful video. I still don't understand how the observation that "the flux through the end of the semi-infinite solenoid is half the flux in the medium" can be determinative of point 1. I only found the following:

In terms of angle $\theta_1$ and $\theta_2$, the magnetic field at a point on the axis is given as $$B = \displaystyle\frac{\mu_0 nI}{2}(cos \theta_1 + cos \theta_2)$$. For a semi-infinite solenoid we use $\theta_1 = \theta$ and $\theta_2 = 0$, and we have: $$cos \theta = \frac{x}{(x^2+R^2)^\frac{1}{2}}$$. Thus, magnetic induction as a function of distance $x$ is given by: $$ B= \displaystyle\frac{\mu_0 nI}{2}\bigg[\frac{x}{(x^2+R^2)^\frac{1}{2}} +1 \bigg]$$.

 

[kuruman]

Angles $\theta_1$ and $\theta_2$ are related to the ends of a solenoid that is finite at both ends. You have defined them as the complementary angles of the ones shown in the video. When you write $$B = \displaystyle\frac{\mu_0 nI}{2}(cos \theta_1 + cos \theta_2)$$ that gives the field at a point inside a solenoid that is finite at both ends. For an infinite solenoid, $\theta_1=\theta_2=0$ because the ends are infinitely far away from all points inside.

Then $$ B_{\text{inf}} =\frac{\mu_0 nI}{2}(1 +1)=\mu_0 n I $$ For a semi-infinite solenoid, only one end is infinitely far away from all inside points, say $\theta_2=0.$ Then for a point inside the semi-infinite solenoid $$B_{\text{semi-inf}} =\frac{\mu_0 nI}{2}(1 +\cos\theta_1).$$ At the open end of the semi-infinite solenoid, $\theta_1=\frac{\pi}{2}$ so that $$B_{\text{semi-inf}}^{\text{end}} =\frac{\mu_0 nI}{2}(1 +0)=\frac{1}{2}B_{\text{inf}}.$$

 

[Hak]

So? I can see that your approach coincides with mine, but how to apply it to the two points $A$ and $C$? Is it enough to substitute $x = \frac{r}{3}$ and $x = 3r$ in my last equation?

 

[kuruman]

So? I can see that your approach coincides with mine, but how to apply it to the two points $A$ and $C$? Is it enough to substitute $x = \frac{r}{3}$ and $x = 3r$ in my last equation?

That will not work. Coordinate $x$ is along the axis of the solenoid. Points $A$ and $C$ are both on the plane of the end of the solenoid at $x=0.$

 

[Hak]

So how can I arrive at calculating $B$ in $A$ and $C$ from "observation that the flux through the end of the semi-infinite solenoid is half the flux in the middle"? As you can see, I am really struggling, although I have attempted many approaches.

 

[Hak]

I don't know how to proceed for step 1, but, waiting for some hints, I tried to develop some ideas for step 2.

I calculated the mutual inductance $M$ considering the situation where the current $I$ is flowing through the solenoid and using magnetic field flux through the ring:

$$\Phi = MI \Rightarrow B_{semi-inf}S = MI \Rightarrow M = \frac{B_{semi-inf} S}{I}$$, where $B_{semi-inf} = \frac{1}{2}\mu_0 n I$ and $S = \pi R^2$. So: $$M = \frac{\mu_0 n I \pi R^2}{2I} \Rightarrow M = \frac{\mu_0 n \pi R^2}{2}$$.

From Faraday's Law, we have:

$$\mathcal{E_{solenoid}} = - M \frac{dI_{ring}}{dt} - L_{solenoid}\frac{dI}{dt}$$
$$\mathcal{E_{ring}} = - M \frac{dI_{solenoid}}{dt} - L\frac{dI_{ring}}{dt}$$, where $\mathcal{E_{solenoid}}= Ik_{solenoid}$ and $\mathcal{E_{ring}} = I_{ring} k_{ring} = 0$ because the resistance $k_{ring}$ of a superconducting loop is $0$ (I indicated the resistances with $k$ to avoid any confusion with the radius $R$).

From this point on, I am no longer able to continue, since the values of $L_{solenoid}$ and $k_{solenoid}$ are not known. Even if $L_{solenoid}$ were considered negligible, the integration would obviously not return a meaningful result.

 

[kuruman]

In part 1 you are looking for the axial component of the magnetic field at points in the plane of the last loop. If you follow the video, it is shown that for distances greater than the radius of the solenoid, the field lines are perpendicular to the axis of the solenoid which means that the field has zero axial component at $C.$ That's the easy part. Point $A$ is harder, but I suspect it involves a flux argument of the kind that used to prove item 4 in the video.

For part 2, I don't believe that flux goes through the coil. It's a Meissner effect situation. The supercurrent provides a counterflux that excludes all the flux that comes out of the solenoid. As you know from the video, half the flux in the solenoid goes out the sides. So you have to calculate how much current is needed to generate that amount of counterflux. Since the radius of the coil is much larger than the radius of the solenoid, I think it is safe to calculate the counterflux through the end of the solenoid using the expression for the field at the center of a ring of current.

 

[Hak]

If you follow the video, it is shown that for distances greater than the radius of the solenoid, the field lines are perpendicular to the axis of the solenoid which means that the field has zero axial component at $C$. That's the easy part.

Thank you, but I did not understand properly. It is absolutely true that "for distances greater than the radius of the solenoid, the field lines are perpendicular to the axis of the solenoid", but according to what is it possible to say that $3r > R$? If not, the field in $C$ may not be $0$, or is it? Could I have an explanation for this?

Point $A$ is harder, but I suspect it involves a flux argument of the kind that used to prove item 4 in the video.

I didn't quite understand that either. Item 4 in the video was about to prove that a line in the central region at a distance $r$ from axis exits from the end at a distance $\sqrt{2} r$ from the axis. Thus, the equation $$\frac {1}{2}B_0 \pi r'^2 = B_0 \pi r^2$$ was set up and it was found that $r' = \sqrt{2} r$. In this case, I set the equation $$\frac {1}{2}B_0 \pi r'^2 = B_0 \pi \bigg(\frac{r}{3}\bigg)^2$$ and I found that $$r' = \frac{\sqrt{2}}{3} r$$. From this point on, I don't know how to continue with the available data. I attempted to found the magnetic induction at a point at a very small distance $z = \frac{r}{3}$ from the axis by using the gaussian pill box method. Assuming the total effecting leaving flux equal to $0$, we have: $$(dB_x)\pi z^2 + B_r \cdot 2 \pi z dx = 0 \Rightarrow B_z = - \frac{z}{2} \frac {dB_x}{dx}$$. I supposed $z \ll x$, so that $$B_x = \frac {\mu_0 n I}{2} \bigg[1 + \frac{x}{\sqrt{x^2+R^2}}\bigg]$$. So: $$B_z = \frac {\mu_0 n I}{2} \bigg[\frac{z}{2} + \frac{R^2}{(x^2+R^2)^{\frac{3}{2}}}\bigg]$$. I don't know how to continue.

For part 2, I don't believe that flux goes through the coil. It's a Meissner effect situation. The supercurrent provides a counterflux that excludes all the flux that comes out of the solenoid. As you know from the video, half the flux in the solenoid goes out the sides. So you have to calculate how much current is needed to generate that amount of counterflux. Since the radius of the coil is much larger than the radius of the solenoid, I think it is safe to calculate the counterflux through the end of the solenoid using the expression for the field at the center of a ring of current.

You are absolutely right. 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 $\Phi_{L} = LI_{ring}$ and $\Phi_{ext} = BS$ with $B = \frac{\mu_0I}{2r}$ (the field at the center of a ring of current) and $S = \pi r^2$. We conclude that:
$$\Phi_{ext} = LI_{ring} \Rightarrow I_{ring} = \frac{\Phi_{ext}}{L} = \frac{\mu_0 \pi I r}{2L}$$. I feel that this solution is almost correct, but still something is missing to be totally correct.

About item 3, I had thought of the same procedure as in my second message, but the result evidently cannot be correct. Could you give me some advice?

I apologize for any annoying caused.

 

[kuruman]

$\dots~$ but according to what is it possible to say that 3r>R?

According to the statement of the problem

Homework Statement: $\dots$ It is known that $r \ll R$ .

It is not possible to say that $3r>R$ and I don't know where you got that idea, The drawing below shows the plane of the coil as a dotted line and points A and C approximately drawn to scale. The axial component (parallel to the axis of the solenoid) is zero at C and non-zero at A.

About part 3, if you have the super current $I_{sc}$ and the radial field $B_r$ at the loop, you can find the force as $F=2\pi R I_{sc}B_r$. The key for parts 2 and 3 is finding $B_r$ on the dotted red line as a function of $r$. I don't know how to do that for large distances from the axis.

 

[Hak]

Ok, thank you so much, but about part 2 you said that

it is safe to calculate the counterflux through the end of the solenoid using the expression for the field at the center of a ring of current.

So, is my process correct? Based on your last message, I didn't understand whether the method you recommended previously about item 2 is still valid or not. I'm having a hard time.

 

[Hak]

For example, is it not possible to use trigonometry to find $B_r$?

 

[kuruman]

So, is my process correct? Based on your last message, I didn't understand whether the method you recommended previously about item 2 is still valid or not. I'm having a hard time.

I would say that the magnetic field at the center of the loop is $$B_{sc}=\frac{m_0I_{sc}}{2R}.$$ That produces flux $$\Phi_{sc}=\frac{\mu_0I_{sc}}{2R}\pi r^2$$through the end of the solenoid. Since the flux through the loop must be zero, $$\frac{\mu_0I_{sc}}{2R}\pi r^2=\frac{1}{2}\mu_0~n~i~\pi r^2.$$Solve for $I_{sc}.$

For example, is it not possible to use trigonometry to find $B_r$?

I think one needs a method for finding how the magnetic field vector tips from zero to 90° as the axial distance changes from zero to $r$ on the red dotted line.

 

[Hak]

Ok, but why does the text provide the self-inductance $L$ as a known datum, if it is then not provided in the procedure?

 

[Hak]

A friend of mine stated the following:

"Points $A$ and $C$ are located on the plane of the last coil of solenoid. In particular $A$ is an internal point of solenoid at distance $\frac{r}{3}$ from the axis while $C$ is an external point at distance $3r$ from the axis. Now it is clear that $B_A$ has an unic component of $B$ which is parallel to $x$. Hence I think that $$B_{A_x} = B_A = \mu_0 n I$$. On the contrary $C$ as external point has a tangent component of $B$. If we consider an Amperian coil of radius $3r$ in the plane ortogonal to axis containing just the last coil of solenoid, by Ampere law we obtain $$B_t \times (2 \pi 3r) = \mu_0 I$$ or $$B_t = \frac{\mu_0 I}{6 \pi r}$$. It is a negligible fraction of $B$ or $$\frac{B_t}{B} = \frac{1}{6 \pi r n}$$. We can say that $$B_{C_x} = 0$$."
Is it correct? In my honest opinion, only the second part is correct.

 

[kuruman]

Your friend does not understand Ampere's law. To begin with, an Amperian loop

in the plane ortogonal to axis containing just the last coil of solenoid

has no current cutting through its plane so the enclosed current is zero. This is consistent with the fact that the component of the field at the circumference of the Amperian loop is radial which means that the field vector $\mathbf{B}$ and the line element $d\mathbf{s}$ are perpendicular to each other and $\oint \mathbf{B}\cdot d\mathbf{s}=0.$ So Ampere's law in this case gives $0=0$ which is not very helpful.

I agree that $B_{C_x} = 0$ and I said so in post #20:

The axial component (parallel to the axis of the solenoid) is zero at C and non-zero at A.

I have said all I can say about this problem and I am sorry I cannot help you more. If you come across a solution, let me know.

 

[Hak]

Thank you very much. I cannot understand why the text gives the self-inductance $L$, if then- as you have shown- it is not needed for current calculation.

 

[kuruman]

Thank you very much. I cannot understand why the text gives the self-inductance $L$, if then- as you have shown- it is not needed for current calculation.

I showed you how I would do it. That doesn't make my approach correct. I did not write the problem, so I don't have an opinion why its author gave the inductance.

 

[Hak]

So, let's do this. I'll just try to ask for some hints from the person who proposed the problem, surely he will know best what interpretation of the text the author intended. As soon as I know more, can I write again below to try to figure out the solution together?

 

[kuruman]

Sounds good.

 

[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.