Vector Potential, Magnetic Field, Mutual Inductance, Magnetic Force, Torque and Stiffness Calculation between Current-Carrying Arc Segments with Inclined Axes in Air

: In this paper, the improved and the new analytical and semi-analytical expressions for calculating the magnetic vector potential, magnetic field, magnetic force, mutual inductance, torque, and stiffness between two inclined current-carrying arc segments in air are given. The expressions are obtained either in the analytical form over the incomplete elliptic integrals of the first and the second kind or by the single numerical integration of some elliptical integrals of the first and the second kind. The validity of the presented formulas is proved from the particular cases when the inclined circular loops are addressed. We mention that all formulas are obtained by the integral approach, except the stiffness, which is found by the derivative of the magnetic force. The novelty of this paper is the treatment of the inclined circular carting-current arc segments for which the calculations of the previously mentioned electromagnetic quantities are given.

integrals whose kernel functions are also the incomplete elliptic integrals of the first and second kind. We mention that some formulas are improved such as the calculation of the magnetic vector potential and magnetic field. The new formulas are given for the mutual inductance, the magnetic force, the torque, and the stiffness. In these formulas the angles of current-carrying arcs are arbitrary. The validity of all formulas is verified with the corresponding calculations for the inclined circular loops. For the convenience of a reader, all derived formulas including were programmed by using the Mathematica. The Mathematica files with the implemented formulas are available from the author as supplementary materials to this paper.

Basic expressions
Let us take into consideration two current-carrying arc segments as showed in Fig.1, where the center of the larger segment (primary coil) of the radius RP is placed at the plane XOY whose center is O (0,0,0). The smaller circular segment (secondary coil) of the radius RS is placed in an inclined plane whose general equation is, where a, b and c are the components of the normal ⃗ ⃗ on the inclined plane in the center of the secondary circular segment C (xC, yC, zC). The segments are with the currents RP RS, respectively. For circular segments (see Figure 1) we define, [22][23][24]: 1) The primary circular segment of radius RP is placed in the plane XOY (Z = 0) with the center at O (0, 0, 0). An arbitrary point P (xP, yP, zP) of this segment has parametric coordinates, = cos( ) , = sin( ), = 0, ∈ ( 1 , 2 ) (2) 2) The differential of the primary circular segment is given by, = {− sin( ) , cos( ), 0} , ∈ ( 1 , 2 ) 3) The primary circular segment of radius RS is placed in the inclined plane (1) Let us introduce the following substitution − = − 2 .
(10), (11) and (12) become, The final solutions for (8), (9) and (10) can be obtained analytically in the form of the incomplete elliptic integrals of the first and the second kind and the simple elementary func- Finally, where, ( , ) and ( , ) , [40][41] are the incomplete elliptic integrals of the first and the second kind.

Special cases
where ( ) and ( ), [40][41] are the complete integrals of the first and the second kind.
Expressions (19), (20) and (21) are valid for = 0. This is the singular case. The point S is between 1 and 2 on the cercle.
The point S is between 2 and 1 + 2 on the cercle.
Thus, all results are obtained in the closed form over the incomplete elliptic integrals of the first and the second kind as well as over some elementary functions.

Magnetic field calculation at the point S (xS ,yS, zS)
The magnetic field ⃗ ( ) produced by the primary circular segment of the radius RP carrying the current IP, can be calculated in an arbitrary point S (xS yS, zS) by, where , and are previously given.
Let us introduce the following substitution − = − 2 .
The final solutions for (32), (33) and (34)  This is the singular case, (see Figure 2) where the point S is between 1 and 2 .
This is well-known expressions [11] obtained in the form of the complete elliptic integrals of the first and second kind K(k) and E(k).

Magnetic force calculation between two inclined current-carrying arc segments
The magnetic force between two inclined arc segments carrying currents IP and IS can be calculated by, where ⃗ ( ) is the magnetic field produced by the primary current IP in the first arc segment, acting at the point S of the second arc segment.

Magnetic torque calculation between two inclined current-carrying arc segments
Torque is defined as the cross product of a displacement and a force. The displacement is from the center for taking torque, which is arbitrarily defined, to the point S of application of the force to the body experiencing the torque. Previously, we calculated the magnetic force between two arc current0carrying segments where we used the analytical expressions of the magnetic field at the at the point S of the second arc segment. The magnet field is produced by the current in the primary arc segment. We use the same reasoning for the torque and from (55) we have, 4 3 Using (7), (35), (36), (37) and developing the double cross product in (57) we obtain the final components of the torque.

Mutual inductance calculation between two current-carrying arc segments with inclined axes
The mutual inductance between current-carrying arc segments with inclined axes in air can be calculated by, where , ⃗⃗ and are previously given.
From, (3), (7) and (61) the mutual inductance can by calculated by, We take the substitution − = − 2 that leads to final solution for the mutual induct- Thus, the calculation of the mutual inductance is obtained by the simple integration where the kernel functions are given in the analytical form over the incomplete elliptic integrals of the first and the second kind.

Special cases
This case is the singular case. The first arc segment lies in the plane z = 0 and the second in the plane y = constant. There are two possibilities for this case.
These vectors must be used in equation (63).
All previous electromagnetic quantities are obtained by using the integral approach.

Stiffness calculation between two inclined current-carrying arc segments
The stiffness is the extent to which an object resists deformation in response to an applied force. Knowing the magnetic force between two inclined current-carrying arc segments the corresponding stiffness between them can be calculated by the derivate of the corre- Even though it is the tedious work we give only the stiffness kzz from (66) which is the axial stiffness. This developed formula can serve the potential readers to make other stiffness, for example, by Mathematica or MATLAB programming, The calculation of other stiffness will be subject of our future work. In this paper we give the benchmark example for calculating the axial stiffness between two coaxial current circular loops.
The magnetic force between two coaxial circular loops is, where, and are the currents in the primary and secondary loop.

and
are the corresponding radii of loops.
Obviously, we can find analytically only the stiffness kzz because others are zero.
This stiffness is given by, where, From (71), (72) and (73) the axial stiffness kzz is, where, As mentioned before, this formula (74) will serve as the benchmark example to verify the validity of the general expression for the stiffness kzz. In Appendix D we give the complete expressions of this axial stiffness.
Here, we give only final expressions of kzz.
where, Let us begin with the circular loop for which is φ1 = 0 and φ2 = 2π.
From (16), (17) and (18) we have the components, and the total magnetic vector potential as follows. From [11] we obtained the same results for the total magnetic vector potential. Obviously, it is the well-known formulas for the current loop.
From [11] we obtained the same results for the magnetic field. Obviously, it is the wellknown formula for the current loop.
Now, let us apply these equations for the same problem but with the different positions of angles, for example, φ1 = π/6 and φ2 = 3π/4. We obtain,  We begin with two inclined circular loops. The last results for φ4 = 2π, are obtained in [23] and [24].
Using the presented method here we have, The last results are obtained in [23] and [24].
Thus, when segments lead to the circular loops, we can see the results that converge to those of the circular loops. The secondary coil is in the plane y = 2. Coils are with perpendicular axes. Calculate the magnetic force between coils. All currents are unit.
This case is the singular case because a = c = 0. Let us begin with two perpendicular current loops, [23] and [24] in which we found, Thus, we obtained for the case 5.1.3 the same results as in [23], [24] and [31]. For the case 5.1.2 we obtain the same results as in [23] and [24] but with opposite signs for each component, because in them we did not take into consideration these unit vectors.
Thus, this singular case where the angles are arbitrary, can be used as the benchmark example which in the limit where the current coils are the current loops as proved previously. Let us begin with two circular loops for which is φ1 = 0, φ2 = 2π, φ3 = 0 and φ4 =2π. Let us take φ1 = π/12, φ2 = π, φ3 = 0 and φ4 =2π.
The approach presented here gives, All results are in an excellent agreement.
Applying (63) the mutual inductance is, We find the same result in [22]. This is the singular case. Let us begin with two circular loops for which is φ1 = 0, φ2 = 2π, φ3 = 0 and φ4 =2π.
For y = − 20 cm the mutual inductance is, For y = 0 cm the mutual inductance is Let us begin with two circular loops for which is φ1 = 0, φ2 = 2π, φ3 = 0 and φ4 =2π.
Here, we taste tree cases 1) For all cases the mutual inductance [22] gives,

= 0
By this work we obtained the same value. It means that for any position of the secondary loop the mutual inductance is zero when the center of the second loop is positioned in the origin O. The same results are obtained in [37].

Conclusions
In this paper we give some ameliorated and new formulas for calculating important electromagnetic quantities such as the magnetic vector potential, the magnetic field, the magnetic force, the mutual inductance, and the stiffness between two inclined current-carrying arc segments in air. The angles of arc segments are arbitrary. All formulas are developed in the close form over the incomplete elliptic integrals of the first and the second kind (the magnetic vector potential and the magnetic field) and in the simple integral form whose kernel functions are also given in the close form over the incomplete elliptic integrals of the first and the second kind (the magnetic force, the magnetic torque, the mutual inductance, and the stiffness which has proven invaluable in validating the method presented here.

Conflicts of Interest:
The author declares no conflict of interest.