318. Semi-infinite solenoid and super conducting disc

[Physicsguy51]

A semi-infinite solenoid with a coil radius and a winding density (the number of coils per unit length) is located coaxially with a circular superconducting coil of radius so that its base is in the plane of the coil. It is known that . Initially, there was no current in the coil. The inductance of the coil is equal to . The current strength in the solenoid is slowly increased from zero to 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 axis as shown in the figure.

1. Points and are located in the plane of the coil at distances and , respectively, from the axis of symmetry of the system. Find the projections of induction and of the magnetic field generated by the solenoid at points and , 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.


Picture: https://ibb.co/R393bBJ

[Tarapìa Tapioco]

Is the value of needed to calculate the current in item 2?

[Higgs]

1. 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 r/3 from the axis while C is an external point at distance 3r from the axis. Now it is clear that has an unic component of B wich is parallel to x. Hence I think that . 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 low we obtain or It is a negligible fraction of B or : Is it correct at this point to say ?

[Tarapìa Tapioco]

Sorry, but I do not agree with @Higgs in a total way. According to my theory, the second part is correct. In a semi-infinite solenoid, a field line crossing the endmost winding is a straight line perpendicular to the axis of solenoid: it can be demonstrated and 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 . So, . All the steps taken by @Higgs in this regard are correct. The drawing below shows the plane of the coil as a dotted line, and points and approximately drawn to scale. The axial component (parallel to the axis of the solenoid) is zero at and non-zero at .



Unfortunately, I disagree on the first part. 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 and the line element are perpendicular to each other and . So Ampere's law in this case gives , which is not very helpful. Moreover, the equation works for points far away from the ends, whereas the point of interest is on the plane of the end of the solenoid at .

I think it's a Meissner effect situation. The supercurrent provides a counter-flux that excludes all the flux that comes out of the solenoid. Half the flux in the solenoid goes out the sides, so we have to calculate how much current is needed to generate that amount of counter-flux. Since the radius of the solenoid is much larger than the radius of the coil, I think it is safe to calculate the counter-flux through the end of the solenoid using the expression for the field at the center of a ring of current. According to Lenz's Law, the current induced in the loop generates a counter-flux that opposes the external flux , 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: . 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 remains constant perpetually. So, since the total flux through the superconducting loop remains constant at (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 counter-flux of the loop must be equal and opposite to the external flux. So:

, with

I would say that the magnetic field at the center of the loop is , that produces flux through the end of the solenoid. Since the flux through the loop must be zero:



Though, the value of self-inductance provided by the text does not appear in this result, which is a problem.

If we have the supercurrent and the radial field at the loop, we can find the force as:

.

The key for this item is finding on the dotted red line (see figure above) as a function of . I don't know how to do that for large distances from the axis.

Could you post some hints about the observations I just put forward, so that I can begin to develop a real procedure?

[Physicsguy51]

1. 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 r/3 from the axis while C is an external point at distance 3r from the axis. Now it is clear that has an unic component of B wich is parallel to x. Hence I think that . 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 low we obtain or It is a negligible fraction of B or : Is it correct at this point to say ?

Your value of is extremely close to the correct value. Your answer of is correct!

[Physicsguy51]

Sorry, but I do not agree with @Higgs in a total way. According to my theory, the second part is correct. In a semi-infinite solenoid, a field line crossing the endmost winding is a straight line perpendicular to the axis of solenoid: it can be demonstrated and 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 . So, . All the steps taken by @Higgs in this regard are correct. The drawing below shows the plane of the coil as a dotted line, and points and approximately drawn to scale. The axial component (parallel to the axis of the solenoid) is zero at and non-zero at .



Unfortunately, I disagree on the first part. 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 and the line element are perpendicular to each other and . So Ampere's law in this case gives , which is not very helpful. Moreover, the equation works for points far away from the ends, whereas the point of interest is on the plane of the end of the solenoid at .

I think it's a Meissner effect situation. The supercurrent provides a counter-flux that excludes all the flux that comes out of the solenoid. Half the flux in the solenoid goes out the sides, so we have to calculate how much current is needed to generate that amount of counter-flux. Since the radius of the coil is much larger than the radius of the solenoid, I think it is safe to calculate the counter-flux through the end of the solenoid using the expression for the field at the center of a ring of current. According to Lenz's Law, the current induced in the loop generates a counter-flux that opposes the external flux , 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: . 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 remains constant perpetually. So, since the total flux through the superconducting loop remains constant at (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 counter-flux of the loop must be equal and opposite to the external flux. So:

, with

I would say that the magnetic field at the center of the loop is , that produces flux through the end of the solenoid. Since the flux through the loop must be zero:



Though, the value of self-inductance provided by the text does not appear in this result, which is a problem.

If we have the supercurrent and the radial field at the loop, we can find the force as:

.

The key for this item is finding on the dotted red line (see figure above) as a function of . I don't know how to do that for large distances from the axis.

Could you post some hints about the observations I just put forward, so that I can begin to develop a real procedure?

Your answer for is correct. Unfortunately, your answer for
is incorrect. 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. Good luck!

[Tarapìa Tapioco]

@Physicsguy51 I wanted to make it clear that mine are not answers, but only theories and observations. The value of cannot be incorrect because I have not yet given it!

[Tarapìa Tapioco]

Is ? And is ?

[Physicsguy51]

Is ? And is ?

Both are correct. The current is in opposite direction hence I_ring will have a -ve sign.

[Tarapìa Tapioco]

Ok, thank you very much for your diligent reply!