318. Semi-infinite solenoid and super conducting disc

Tarapìa Tapioco · Messaggio da **Tarapìa Tapioco** » 6 set 2023, 14:00

Following this way, I would get $F = \frac{\mu_0^2 n \pi r R^2 I^2}{2L (R-r)}$ . Let me know, sorry for bothering you so much.

Physicsguy51 · Messaggio da **Physicsguy51** » 6 set 2023, 15:26

Tarapìa Tapioco ha scritto: ↑6 set 2023, 14:00 Following this way, I would get $F = \frac{\mu_0^2 n \pi r R^2 I^2}{2L (R-r)}$ . Let me know, sorry for bothering you so much.

It is incorrect unfortunately, and you are not bothering me friend. In fact, I am very happy to have this discussion with you!
Give it a few more tries, I am attaching the link to the solutions. Once you give up, take a look at them (they're in russian).
https://pho.rs/p/2880/s

Tarapìa Tapioco · Messaggio da **Tarapìa Tapioco** » 6 set 2023, 15:29

You do not need to attach the solution. In my opinion, you can do it after assigning the relay, that is, when one of the users has answered all the questions correctly. However, thank you.

Tarapìa Tapioco · Messaggio da **Tarapìa Tapioco** » 6 set 2023, 15:56

$3.$ If we have the supercurrent $I_{ring}$ and the radial field $B_r$ at the loop, we can find the force as:

$F = I_{ring} l B_r \Rightarrow F = 2 \pi R I_{ring} B_r$ .

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

I am very convinced from the beginning that this is the correct answer. As already mentioned, I have to find a way to calculate $B_r$ .

Could you provide some very little information about the 4 methods you are talking about to solve this point? Of course, you don't need to say what they are, I just need to know what they are roughly hinged on.

Tarapìa Tapioco · Messaggio da **Tarapìa Tapioco** » 6 set 2023, 17:08

Excuse me. Is the result $F = \frac{\mu_0^2 \pi^2 n^2 I^2 r^4}{4RL}$ ? If this is not it, I really don't know which other way I should go...

Higgs · Messaggio da **Higgs** » 6 set 2023, 18:39

2. It is well known that the flow of B across a coil of induction L and with a current i is $\Phi= Li$ . This flow must be opposite to that generated by solenoid whose B we have seen to be $\frac {\mu_0 In}{2}$ . Hence the flow caming from solenoid is $\Phi_s= \frac{\mu_0 I n}{2}.\pi r^2$ Taking $\Phi = -\Phi_s$ we obtain for the current in super conductig coil $i=\frac{\mu_0\pi r^2nI }{2L}$ To morrow I'll try 3.

Tarapìa Tapioco · Messaggio da **Tarapìa Tapioco** » 6 set 2023, 19:44

Physicsguy51 ha scritto: ↑3 set 2023, 21:00 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 $r / 3$ and $3 r$ , 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.

Picture: https://ibb.co/R393bBJ

$1.$ Consider a generical point $P$ at the end face of the semi-infinite solenoid. If the magnetic field strength at point P is denoted by $B_1$ , then, using superposition, a symmetrically placed second coil, as shown in Figure below, would also produce a field of strength $B_2 = B_1 = B_{\text{end}}$ .

Immagine

Since the magnetic field at the centre of the solenoid, resultant magnitude of the superimposed fields, is clearly $\displaystyle B_{\text{tot}} = B_{\text{inside}} = B_0 = \mu_0 n I$ , it follows that:

$\displaystyle B_{\text{tot}} = B_1 + B_2$ $\displaystyle \Rightarrow$ $B_0 = B_1 + B_1$ $\Rightarrow$ $\displaystyle B_0 = 2 B_1$ $\Rightarrow$ $\displaystyle B_1 = \frac{B_0}{2} = \frac{\mu_0 n I}{2}$ $\Rightarrow$ $\displaystyle B_{\text{end}} = \frac{B_{\text{inside}}}{2} = \frac{\mu_0 n I}{2}$ .

So, this argument shows that the axial component of the field, at any point on the end face is exactly $\displaystyle \frac{B_0}{2}$ , where $B$ is the (uniform) field throughout the inside the corresponding infinite solenoid. This is true because adding another semi-infinite solenoid simply doubles the axial field at any point on the end face, and cancels the radial field, resulting in a purely axial field.

Similar reasoning shows that the horizontal component of the magnetic field vector through $P$ is $\frac{B_0}{2}$ ; this would be true for any $P$ whose distance from the axis is less than the radius $R$ of the solenoid (see Figure below).

Immagine

The drawing below shows the plane of the coil as a dotted line, and point $A$ and $C$ approximately drawn to scale.

Immagine

The axial component (parallel to the axis of the solenoid) is non-zero at $A$ because it's inside the solenoid and off the axis. Moreover, it can be noticed that the point of interest $A$ is on the plane of the end of the solenoid at $x = 0$ , so we expect that:

$\displaystyle \boxed{B_{A_x} = \frac{\mu_0 n I}{2}}$ .

To conferm it analitically, it can be considered how the magnetic field vector tips from zero to $\displaystyle \frac{\pi}{2}$ as the axial distance changes from $0$ to $r$ on the red dotted line. So, the angle $\theta$ between the vector $\overrightarrow{B_A}$ and the perpendicular to the axis of symmetry is, keeping the ratio constant:

$\displaystyle \frac{r}{\frac{\pi}{2}} = \frac{\frac{r}{3}}{\theta}$ $\Rightarrow$ $\displaystyle \theta = \frac{\frac{\pi}{2}}{r}\frac{r}{3} = \frac{\pi}{6}$ .

Since point $A$ is inside the solenoid, we have:

$\displaystyle B_{A_x} = B_0 \sin \theta$ $\Rightarrow$ $\displaystyle B_{A_x} = {\mu_0 n I} \sin \bigg(\frac{\pi}{6}\bigg)$ $\Rightarrow$ $\displaystyle \color{Red} \boxed{\color{Black}B_{A_x} = \frac{\mu_0 n I}{2}}$ .

Now, it can be shown that in a semi-infinite solenoid (such as the one shown in Figure below), a field line crossing the endmost winding is a straight line perpendicular to the axis of solenoid: it can be demonstrated that for distances greater than the radius $R$ of the solenoid, the field lines are perpendicular to the axis of the solenoid, which means that the field has zero axial component at $C$ , which is outside the solenoid and off the axis, as we can see from the drawing above.

Immagine

Consider the figures above and below. The field line $FGH$ , which passes through the very end of the winding, is a straight line from $G$ out to infinity. Let’s assume (in search of a contradiction) that there exists a field line that crosses the line $GH$ with a vertical component, as shown in Figure (a) below. Imagine flipping the solenoid upside down to obtain the situation in Figure (b), and then reversing the direction of the current (so that it now has the same direction as in Figure (a)) to obtain the situation in Figure (c). Note that the field at the given point on the line $GH$ has a downward component in both figures (a) and (c) (or upward in both, if we had initially drawn it upward). Now join the two semi-infinite solenoids in figures (a) and (c) end to end, thereby creating an infinite solenoid. By superposition, the fields simply add, so we end up with a downward component at the given point along $GH$ . But this is a contradiction, because we know that the field of an infinite solenoid is zero outside the solenoid. We conclude that the field due to the semi-infinite solenoid at the given point must have had zero vertical component $x$ . In other words, it was horizontal, as we wanted to show. So, $\displaystyle \color{Red} \boxed{\color{Black}B_{C_x} = 0}$ .

Immagine

The arguments used in point $1.$ lead to more general statements about the field of a semi-infinite solenoid. Consider two points $P$ and $P'$ , symmetrically located with respect to the end plane, as shown in Figure below. The fields $\textbf{B}$ and $\textbf{B'}$ are related as follows:

Immagine

The radial components of $\textbf{B}$ and $\textbf{B'}$ are equal. The sum of the axial components of $\textbf{B}$ and $\textbf{B'}$ is equal to $B_0$ if $P$ lies inside the solenoid, or to zero if $P$ lies outside the solenoid (that is, above the top “edge” of the solenoid in the figure). The conclusions of item $1.$ follow in the special case where $P$ and $P'$ coincide.

$2.$ The current strength in the coil is a manifestation of a Meissner effect situation. The supercurrent provides a counter-flux that excludes all the flux that comes out of the solenoid. As seen in point $1.$ , 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. According to Lenz's Law, the current induced in the loop generates a counter-flux $\Phi_{\text{ring}}$ from the self-inductance $L$ of the loop that opposes the external flux $\Phi_{\text{ext}}$ , which attempts to thread the loop. The net flux through the loop is equal to the sum of the external flux attempting to thread the loop and the counter-flux due to the induced current: $\displaystyle \Phi_{\text{net}} = \Phi_{\text{ext}} + \Phi_{\text{ring}}$ . 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 $\Phi_{\text{net}}$ remains constant perpetually; if it could change, the induced voltage would produce an infinitely large current. 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 counter-flux of the loop must be equal and opposite to the external flux. So:

$\displaystyle \Phi_{\text{net}} = \Phi_{\text{ext}} + \Phi_{\text{ring}} = 0 \Rightarrow \Phi_{\text{ext}} = - \Phi_{\text{ring}}$ , with

$\displaystyle \Phi_{\text{ext}} = B_{\text{end}} \cdot S$ , where $\displaystyle B_{\text{end}} = \frac{\mu_0 n I}{2}$ and $\displaystyle S = \pi r^2$ . So: $\displaystyle \Phi_{\text{ext}} = \frac {1}{2} \mu_0 n I \pi r^2$ .

Counter-flux, that is flux due to self-inductance $L$ of the loop, is: $\displaystyle \Phi_{\text{ring}} = LI_{\text{ring}}$ .

By equating the two expressions:

$\displaystyle \Phi_{\text{ext}} = \frac {1}{2} \mu_0 n I \pi r^2 = - LI_{\text{ring}}$ $\Rightarrow$ $\displaystyle I_{\text{ring}} = - \frac{\frac {1}{2} \mu_0 n I \pi r^2}{L}$ $\Rightarrow$ $\displaystyle \color{Red} \boxed {\color{Black} I_{\text{ring}} = - \frac{\mu_0 n I \pi r^2}{2L}}$

The induced electromotive force with different polarities induces a current whose magnetic field opposes the change in magnetic flux through the loop in order to ensure that the original flux is maintained through the loop when current flows in it. That's only how the negative sign in the equation came into being. So, the current in the coil is directed in such a way that it opposes the change in magnetic flux through the coil due to the increasing current in the solenoid. This is according to Lenz’s law, which states that an induced current always flows in a direction that creates a magnetic field that opposes the change in magnetic flux that caused it. In this case, since the magnetic field of the solenoid is pointing along the positive x-axis, the induced current in the coil must create a magnetic field that points along the negative x-axis. Therefore, using the right-hand rule, we can conclude that the current in the coil is directed clockwise when viewed from above.

$3.$ Consider the small circular current loop of radius $r$ placed in the magnetic field of the semi-infinite solenoid of radius $R$ . We want to find the force on the solenoid caused by the current ring. Because of the radial component of the field, $B_R$ , there is a net force on the solenoid as a whole. It arises because the field $B$ has an outward component $B_R$ everywhere around the ring, whereas the vertical component $B_x$ produces a force in the horizontal plane that simply stretches or compresses the solenoid – negligibly, assuming it is fairly rigid. Therefore, if the current flows in the direction indicated, each element of the loop, $dl$ , must be experiencing a downward force of magnitude $\displaystyle I B_R dl$ . If $B_R$ has the same magnitude at all points on the ring, as it must in the symmetrically spreading field assumed, the total downward force will have the magnitude:

$\displaystyle \boxed{F = - l I_{\text{ring}} B_R}$ , where $\displaystyle l = 2 \pi R$ is the length of the solenoid, taken as a circumference, and $B_R$ is the radial field. This expression has a negative sign because the force is repulsive and directed along the negative x-axis.
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 $\displaystyle \Phi_0 = \Phi_{\text{inside}}$ of the magnetic field $\displaystyle B_0 = B_{\text{inside}} = \mu_0 nI$ 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: $\displaystyle \Phi_0 = B_0 S_{\text{circle}}$ , where $\displaystyle S_{\text{circle}} = \pi r^2$ 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: $\displaystyle \Phi_R = B_R S_{\text{sphere}}$ , where $\displaystyle S_{\text{sphere}} = 4 \pi R^2$ is the surface of the sphere-solenoid. So: $\displaystyle \Phi_R = B_R \cdot 4 \pi R^2$ .

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

Substituting into the first expression for $F$ :

$\displaystyle F = - l I_{\text{ring}} B_R$ $\Rightarrow$ $\displaystyle F = - 2 \pi R \frac{\mu_0 n I \pi r^2}{2L} \frac{\mu_0 n I r^2}{4 R^2} = - \frac{\mu_0^2 n^2 \pi^2 I^2 r^4}{4RL}$ $\Rightarrow$ $\displaystyle \color {Red} \boxed {\color {Black} F = - \frac{1}{RL} \bigg(\frac{\mu_0 n \pi I r^2}{2}\bigg)^2}$ .

Physicsguy51 · Messaggio da **Physicsguy51** » 6 set 2023, 19:57

Tarapìa Tapioco ha scritto: ↑6 set 2023, 17:08 Excuse me. Is the result $F = \frac{\mu_0^2 \pi^2 n^2 I^2 r^4}{4RL}$ ? If this is not it, I really don't know which other way I should go...

Correct! Amazing

Physicsguy51 · Messaggio da **Physicsguy51** » 6 set 2023, 19:58

Tarapìa Tapioco ha scritto: ↑6 set 2023, 19:44 $1.$ Consider a generical point $P$ at the end face of the semi-infinite solenoid. If the magnetic field strength at point P is denoted by $B_1$ , then, using superposition, a symmetrically placed second coil, as shown in Figure below, would also produce a field of strength $B_2 = B_1 = B_{end}$ .

Since the magnetic field at the centre of the solenoid, resultant magnitude of the superimposed fields, is clearly $\displaystyle B_{tot} = B_{inside} = B_0 = \mu_0 n I$ , it follows that:

$\displaystyle B_{tot} = B_1 + B_2$ $\displaystyle \Rightarrow$ $B_0 = B_1 + B_1$ $\Rightarrow$ $\displaystyle B_0 = 2 B_1$ $\Rightarrow$ $\displaystyle B_1 = \frac{B_0}{2} = \frac{\mu_0 n I}{2}$ $\Rightarrow$ $\displaystyle B_{end} = \frac{B_{inside}}{2} = \frac{\mu_0 n I}{2}$ .

So, this argument shows that the axial component of the field, at any point on the end face is exactly $\displaystyle \frac{B_0}{2}$ , where $B$ is the (uniform) field throughout the inside the corresponding infinite solenoid. This is true because adding another semi-infinite solenoid simply doubles the axial field at any point on the end face, and cancels the radial field, resulting in a purely axial field.

Similar reasoning shows that the horizontal component of the magnetic field vector through $P$ is $\frac{B_0}{2}$ ; this would be true for any $P$ whose distance from the axis is less than the radius $R$ of the solenoid (see Figure below).

The drawing below shows the plane of the coil as a dotted line, and point $A$ and $C$ approximately drawn to scale.

The axial component (parallel to the axis of the solenoid) is non-zero at $A$ because it's inside the solenoid and off the axis. Moreover, it can be noticed that the point of interest $A$ is on the plane of the end of the solenoid at $x = 0$ , so we expect that:

$\displaystyle \boxed{B_{A_x} = \frac{\mu_0 n I}{2}}$ .

To conferm it analitically, it can be considered how the magnetic field vector tips from zero to \frac{\pi}{2} as the axial distance changes from $0$ to $r$ on the red dotted line. So, the angle \theta between the vector $\overrightarrow{B_A}$ and the perpendicular to the axis of symmetry is, keeping the ratio constant:

$\displaystyle \frac{r}{\frac{\pi}{2}} = \frac{\frac{r}{3}}{\theta}$ $\Rightarrow$ $\displaystyle \theta = \frac{\frac{\pi}{2}}{r}\frac{r}{3} = \frac{\pi}{6}$ .

Since point $A$ is inside the solenoid, we have:

$\displaystyle B_{A_x} = B_0 sin \theta$ $\Rightarrow$ $\displaystyle B_{A_x} = \frac{\mu_0 n I} sin (\frac{\pi}{6})$ $\Rightarrow$ $\displaystyle \color{Red} \boxed{\color{Black}B_{A_x} = \frac{\mu_0 n I}{2}}$ .

Now, it can be shown that in a semi-infinite solenoid (such as the one shown in Figure below), a field line crossing the endmost winding is a straight line perpendicular to the axis of solenoid: it can be demonstrated that for distances greater than the radius $R$ 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, which is outside the solenoid and off the axis, as we can see from the drawing above.

Consider the figures above and below. The field line FGH, which passes through the very end of the winding,is a straight line from G out to infinity. Let’s assume (in search of a contradiction) that there exists a field line that crosses the line GH with a vertical component, as shown in Figure (a) below. Imagine flipping the solenoid upside down to obtain the situation in Figure (b), and then reversing the direction of the current (so that it now has the same direction as in Figure (a)) to obtain the situation in Figure (c). Note that the field at the given point on the line GH has a downward component in both figures (a) and (c) (or upward in both, if we had initially drawn it upward). Now join the two semi-infinite solenoids in figures (a) and (c) end to end, thereby creating an infinite solenoid. By superposition, the fields simply add, so we end up with a downward component at the given point along GH. But this is a contradiction, because we know that the field of an infinite solenoid is zero outside the solenoid. We conclude that the field due to the semi-infinite solenoid at the given point must have had zero vertical component $x$ . In other words, it was horizontal, as we wanted to show. So, $\displaystyle \color{Red} \boxed{\color{Black}B_{C_x} = 0}$ .

The arguments used in point $1.$ lead to more general statements about the field of a semi-infinite solenoid. Consider two points $P$ and $P'$ , symmetrically located with respect to the end plane, as shown in Figure below. The fields $\textbf{B}$ and $\textbf{B'}$ are related as follows:

The radial components of $\textbf{B}$ and $\textbf{B'}$ are equal. The sum of the axial components of $\textbf{B}$ and $\textbf{B'}$ is equal to $B_0$ if $P$ lies inside the solenoid, or to zero if $P$ lies outside the solenoid (that is, above the top “edge” of the solenoid in the figure). The conclusions of item $1.$ follow in the special case where $P$ and $P'$ coincide.

2. The current strength in the coil is a manifestation of a Meissner effect situation. The supercurrent provides a counter-flux that excludes all the flux that comes out of the solenoid. As seen in point $1.$ , 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. According to Lenz's Law, the current induced in the loop generates a counter-flux $\Phi_{ring}$ from the self-inductance $L$ of the loop 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: $\displaystyle \Phi_{net} = \Phi_{ext} + \Phi_{ring}$ . 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; if it could change, the induced voltage would produce an infinitely large current. 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 counter-flux of the loop must be equal and opposite to the external flux. So:

$\displaystyle \Phi_{net} = \Phi_{ext} - \Phi_{ring} = 0 \Rightarrow \Phi_{ext} = \Phi_{ring}$ , with

$\displaystyle \Phi_{ext} = B_{end} \cdot S$ , where $\displaystyle B_{end} = frac{\mu_0 n I}{2}$ and $\displaystyle S = \pi r^2$ . So: $\displaystyle \Phi_{ext} = \frac {1}{2} \mu_0 n I \pi r^2$ .

Counter-flux, that is flux due to self-inductance $L$ of the loop, is: $\displaystyle \Phi_{ring} = LI_{ring}$ .

By equating the two expressions:

$\displaystyle \Phi_{ext} = \frac {1}{2} \mu_0 n I \pi r^2 = LI_{ring}$ $\Rightarrow$ $\displaystyle I_{ring} = - \frac{\frac {1}{2} \mu_0 n I \pi r^2}{L}$ $\Rightarrow$ $\displaystyle \color{Red} \boxed {\color{Black} I_{ring} = \frac{\mu_0 n I \pi r^2}{2L}}$

The induced electromotive force with different polarities induces a current whose magnetic field opposes the change in magnetic flux through the loop in order to ensure that the original flux is maintained through the loop when current flows in it. That's only how the negative sign in the equation came into being. So, the current in the coil is directed in such a way that it opposes the change in magnetic flux through the coil due to the increasing current in the solenoid. This is according to Lenz’s law, which states that an induced current always flows in a direction that creates a magnetic field that opposes the change in magnetic flux that caused it. In this case, since the magnetic field of the solenoid is pointing along the positive x-axis, the induced current in the coil must create a magnetic field that points along the negative x-axis. Therefore, using the right-hand rule, we can conclude that the current in the coil is directed clockwise when viewed from above.

$3.$ Consider the small circular current loop of radius $r$ placed in the magnetic field of the semi-infinite solenoid of radius $R$ . We want to find the force on the solenoid caused by the current ring. Because of the radial component of the field, $B_R$ , there is a net force on the solenoid as a whole. It arises because the field $B$ has an outward component $B_R$ everywhere around the ring, whereas the vertical component $B_x$ produces a force in the horizontal plane that simply stretches or compresses the solenoid – negligibly, assuming it is fairly rigid. Therefore, if the current flows in the direction indicated, each element of the loop, $dl$ , must be experiencing a downward force of magnitude $\displaystyle I B_R dl$ . If $B_R$ has the same magnitude at all points on the ring, as it must in the symmetrically spreading field assumed, the total downward force will have the magnitude:

$\displaystyle \boxed{F = - l I_{ring} B_R}$ , where $\displaystyle l = 2 \pi R$ is the length of the solenoid, taken as a circumference, and $B_R$ is the radial field. This expression has a negative sign because the force is repulsive and directed along the negative x-axis.
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:
$\displaystyle \nabla \cdot \textbf{B}= 0$ $\Rightarrow$ $\displaystyle \Phi_{tot} = 0$ $\Rightarrow$ $\displaystyle \Phi_0 - \Phi_R = 0$ $\Rightarrow$ $\displaystyle \Phi_0 = \Phi_R$ $\Rightarrow$ $\displaystyle \mu_0 n \pi I r^2 = B_R \cdot 4 \pi R^2$ $\Rightarrow$ $\displaystyle \boxed{B_R = \frac{\mu_0 n I r^2}{4 R^2}}$ .

Substituting into the first expression for $F$ :

$\displaystyle F = - l I_{ring} B_R$ $\Rightarrow$ $\displaystyle F = - 2 \pi R \frac{\mu_0 n I \pi r^2}{2L} \frac{\mu_0 n I r^2}{4 R^2} = - \frac{\mu_0^2 n^2 \pi^2 I^2 r^4}{4RL}$ $\Rightarrow$ $\displaystyle \color {Red} \boxed {\color {Black} F = - \frac{1}{RL} \bigg(\frac{\mu_0 n \pi I r^2}{2}\bigg)^2}$ .

Brilliant! Relay goes to you

Physicsguy51 · Messaggio da **Physicsguy51** » 6 set 2023, 19:58

So, did you find the problem challenging? Also, let me know are you a student or a professor?

318. Semi-infinite solenoid and super conducting disc

Re: 318. Semi-infinite solenoid and super conducting disc

Re: 318. Semi-infinite solenoid and super conducting disc

Re: 318. Semi-infinite solenoid and super conducting disc

Re: 318. Semi-infinite solenoid and super conducting disc

Re: 318. Semi-infinite solenoid and super conducting disc

Re: 318. Semi-infinite solenoid and super conducting disc

Re: 318. Semi-infinite solenoid and super conducting disc

Re: 318. Semi-infinite solenoid and super conducting disc

Re: 318. Semi-infinite solenoid and super conducting disc

Re: 318. Semi-infinite solenoid and super conducting disc