Axial Electromagnetic Forces in Multi-Turn Coils

Hello! Welcome back to our course on designing high-field magnets.

In our previous lesson, we focused on withstanding the immense outward radial forces (hoop stress) in a pulsed magnet. We saw how shrink-fitting a strong reinforcement jacket creates a compressive pre-stress, giving the coil a crucial mechanical advantage against the explosive Lorentz forces.

However, radial forces are only part of the story. The $J \times B$ force is a vector, and it also has a significant axial component. These axial forces act along the length of the coil, trying to crush it towards its center while simultaneously pulling its ends apart. Neglecting them would lead to catastrophic failure just as surely as neglecting hoop stress.

Today's lesson directly addresses this challenge. Our learning outcome is to calculate the axial electromagnetic forces between different sections of a multi-turn coil. We will build up the physics from a simple two-loop interaction to a full, thick solenoid, giving you a powerful analytical tool for magnet design.

1. The Origin of Axial Forces

In an ideal, infinitely long solenoid, the magnetic field is purely axial ( $B = B_{z} z^$ ) inside and zero outside. Since the current flows azimuthally ( $J = J_{ϕ} ϕ^$ ), the Lorentz force $F = J \times B$ is purely radial ( $F \propto ϕ^\times z^= - r^$ ).

However, any real solenoid is finite. Near the ends, the field lines curve outwards, creating a significant radial component of the magnetic field, $B_{r}$ . It is this radial field that generates the axial force:

$F_{a x ia l} \propto J_{ϕ} ϕ^\times B_{r} r^= J_{ϕ} B_{r} (ϕ^\times r^) = - J_{ϕ} B_{r} z^$

This force is directed towards the mid-plane ( $z = 0$ ) where $B_{r}$ changes sign.

For a turn in the top half of the coil ( $z > 0$ ), $B_{r} > 0$ , so the force is in the $- z^$ direction (compressive).
For a turn in the bottom half ( $z < 0$ ), $B_{r} < 0$ , so the force is in the $+ z^$ direction (also compressive).

The net effect is that the two halves of the coil are powerfully forced together. The total compressive force at the mid-plane can be enormous—many tonnes in a high-field magnet. Furthermore, this compression implies that the turns at the very ends of the coil are experiencing a strong tension, being pulled inwards by the rest of the coil.

Electromagnetic Forces in Multi-Coil Systems — This image illustrates the force components in a two-coil system. While it shows separate HV and LV coils, the physics is directly analogous to the forces between different layers or sections of a single, multi-turn magnet. The interaction of the current in one section with the magnetic field from another generates both radial forces ($F_r$) that push the sections apart and axial forces ($F_a$) that compress or stretch the winding.

2. The Building Block: Force Between Two Current Loops

To analyze a complex solenoid, we first need a precise formula for the axial force between two simple, coaxial current loops. You could derive this by starting with the Biot-Savart law to find the off-axis $B_{r}$ field of one loop and then integrating the Lorentz force on the second loop, but this is cumbersome.

A more elegant approach, which aligns with your preference for detailed derivations, uses the concept of magnetic vector potential. The following paper provides a concise summary of this method and its result.

Analytical and FEM Based Calculation of Electromagnetic Forces ...

The paper 'Analytical and FEM Based Calculation of Electromagnetic Forces...' by Shiri and Moghadam provides the analytical foundation we need. We will focus on their derivation of the force between two current-carrying rings.

Please read Section II, 'MAGNETIC FORCE BETWEEN TWO CONCENTRIC CIRCULAR ELEMENTS' (pages 866-867). Pay attention to the overall method: the force is derived from the magnetic vector potential. You don't need to replicate the derivation, but focus on understanding the final expression, Equation (4). Notice that the force depends on the radii (a, b), the axial separation (z), the currents (I_1, I_2), and functions K(k) and E(k), which are complete elliptic integrals.

The key takeaway is that there is an exact, albeit complex, analytical expression $F_{z} (I_{1}, I_{2}, r_{1}, r_{2}, z)$ for the axial force between two current loops. The elliptic integrals arise because of the cylindrical geometry of the problem. While their form is intimidating, they are standard mathematical functions that can be computed numerically with high precision.

3. Scaling Up: The Mesh-Matrix Method for Solenoids

Now, how do we apply this two-loop formula to a real solenoid, which has a thick, rectangular cross-section containing thousands of turns? We can approximate the continuous current distribution by discretizing it into a grid of discrete current filaments. This powerful and intuitive technique is often called the mesh-matrix method or filament method.

The procedure is as follows:

Discretize the Coil: Imagine the coil's cross-section (defined by its height and radial thickness) is divided into a fine grid, for example, $n_{a}$ segments axially and $n_{r}$ segments radially. Each small rectangle in this grid is a "filament".
Assign Current to Filaments: The total current of the coil, $N I$ (number of turns $\times$ current), is distributed among these filaments. Assuming a uniform current density, the current $i_{f i l}$ in each filament is:
$i_{f i l} = \frac{N I}{n _{a} \times n _{r}}$
Sum Pairwise Forces: Each filament is now treated as a simple current loop located at its geometric center. To find the total axial force on a specific filament j, we simply apply the principle of superposition: we calculate the axial force exerted by every other filament l on filament j using the two-loop formula from the previous section, and sum them all up.
$F_{j, a x ia l} = \sum_{l \neq = j} F_{z} (i_{f i l}, i_{f i l}, r_{j}, r_{l}, z_{j l})$
where $r_{j}, r_{l}$ are the radii of the filaments and $z_{j l}$ is their axial separation.

This method transforms a complex continuous problem into a large but straightforward summation. Given your experience with Python and Julia, you can see how this would be implemented with a nested loop over all pairs of filaments. The following reading explains this exact process.

Analytical and FEM Based Calculation of Electromagnetic Forces ...

Let's return to the Shiri and Moghadam paper to see how they implement this filament method.

Read Section III, 'CALCULATION OF AXIAL DISTRIBUTION OF THE FORCE' (page 867). This section formalizes the mesh-matrix method we just discussed. See how their Equation (8) is just Equation (4) from the previous section, but now applied to two filaments j and l.

4. Visualizing the Force Distribution

By performing this calculation for every filament in the grid, we can build a complete map of the axial force distribution throughout the coil's winding. This reveals the critical stress points.

The typical distribution has several key features:

Symmetry: The force distribution is symmetric about the coil's mid-plane ( $z = 0$ ).
Zero Force at the Mid-plane: A filament located exactly at the mid-plane experiences zero net axial force because the forces from the top and bottom halves of the coil perfectly cancel out.
Maximum Total Compression at the Mid-plane: While the local force is zero, the mid-plane is where the cumulative compressive force is at its maximum. It must withstand the force from the entire top half pushing down and the entire bottom half pushing up.
Maximum Local Stress at the Ends: The largest axial force on any single filament occurs at the ends of the coil (largest $∣ z ∣$ ). As seen in the paper, these forces also tend to be larger towards the outer radius of the winding. These end turns are under the most tension and are a primary failure concern.

The final reading assignment illustrates these results with clear plots.

Analytical and FEM Based Calculation of Electromagnetic Forces ...

Now let's examine the results of the calculation. This will give you a concrete understanding of where the stresses are highest.

Read Section IV.A, 'Analytical results' (pages 867-868). Study Figures 4 and 5 carefully. Figure 4(a) shows the axial force on the outermost layer as a function of axial position—note how it peaks at the ends. Figure 5 provides a 2D map of the force, confirming that the highest forces are concentrated at the corners of the coil's cross-section.

Test your understanding!

Imagine you have designed a multi-layer pulsed magnet. The calculations show that the axial compressive force at the mid-plane exceeds the strength of the epoxy and insulation. Your goal is to reduce this total compressive force.

Considering the physics we've discussed ( $F_{a x ia l} \propto J_{ϕ} B_{r}$ ), which of these two strategies is more likely to be effective, and why?

A) Make the coil axially shorter, while keeping the number of turns $N$ and current $I$ the same. This increases the linear turn density.
B) Make the coil axially longer, while keeping $N$ and $I$ the same. This decreases the linear turn density.

Show answer

Strategy (B) is the effective one.

Here's the reasoning: The axial force arises from the interaction of the current with the radial field component, $B_{r}$ . For a finite solenoid, $B_{r}$ is strongest near the ends. By making the coil axially longer (more "slender"), you are making it behave more like an ideal, infinite solenoid.

In a longer coil:

The field lines in the central region become more parallel to the axis, reducing $B_{r}$ for a large portion of the winding.
The "end effects," where $B_{r}$ is significant, are pushed further apart.
While the force on the end turns might not change drastically, the integrated compressive force from each half of the coil on the other is reduced because the contributions from the central turns (where $B_{r}$ is now smaller) are diminished.

Conversely, making the coil shorter (Strategy A) makes it a "squat" coil, closer to a Helmholtz pair configuration. This enhances the radial field component everywhere, dramatically increasing the axial compressive forces. This is a well-known trade-off in magnet design: shorter, more compact coils can generate higher central fields for a given $N I$ , but they suffer from much higher axial forces and poorer homogeneity.

Conclusion

Today, we've moved from the radial to the axial dimension of stress in a solenoid. You now have a complete framework for understanding and, crucially, calculating the axial forces that try to crush and tear a coil apart.

Key Takeaways:

Axial forces originate from the interaction between the azimuthal current ( $J_{ϕ}$ ) and the radial magnetic field component ( $B_{r}$ ) that exists near the ends of any finite coil.
These forces are compressive, squeezing the coil towards its mid-plane.
We can compute the detailed force distribution using the mesh-matrix method: discretize the coil into filaments and sum the analytically calculated pairwise forces between them.
The analysis reveals that while the total compressive load is highest at the mid-plane, the largest local tensile stresses are found on the conductors at the very ends of the coil.

Next Lesson Preview:
Knowing the force distribution is the diagnostic step. The next step is the engineering solution. Now that we can calculate these multi-tonne axial forces, how do we build a structure to contain them? In our next lesson, we will focus on how to design axial pre-compression systems and support structures to counteract these electromagnetic forces and prevent mechanical failure.

Recommended resources

1. The Origin of Axial Forces

2. The Building Block: Force Between Two Current Loops

3. Scaling Up: The Mesh-Matrix Method for Solenoids

4. Visualizing the Force Distribution

Conclusion