IET Electric Power Applications

Research Article

Inductance and force calculations of circular

coils with parallel axes shielded by a cuboid

of high permeability

ISSN 1751-8660

Received on 6th October 2017

Revised 29th January 2018

Accepted on 1st February 2018

E-First on 4th April 2018

doi: 10.1049/iet-epa.2017.0646

www.ietdl.org

Yao Luo

1

, Yuanzhe Zhu

1

, Yue Yu

2

, Lei Zhang

3,4

1

School of Electrical Engineering, Wuhan University, Wuhan, People's Republic of China

2

Department of Electrical Engineering, Tsinghua University, Beijing, People's Republic of China

3

State Key Laboratory of Water Resources and Hydropower Engineering Science, Wuhan University, Wuhan, People's Republic of China

4

Department of Biological and Agricultural Engineering, Texas A&M University, College Station, Texas, USA

E-mail: sturmjungling@gmail.com

Abstract: A detailed analysis is presented for a boundary value problem of circular coils with parallel axes shielded by a cuboid

of high permeability. Field solutions are given by establishing a suitable ansatz of the magnetic scalar potential, which can

satisfy the boundary conditions on six surfaces of the cuboid without difficulty. Analytic expressions are also given for the self

and mutual inductance of shielded circular coils with rectangular cross section. By differentiating the self- and mutual magnetic

energy with respect to the centre coordinates of the shielded coils, the total forces exerted on them are further obtained, which

consist of self and mutual force components. Finally, the numerical results of the proposed method are compared with those of

the finite-element method simulations, and the proposed method proves to be accurate and efficient enough for practical

applications.

1Introduction

In many cases of electric power applications, a static or slowly

varying magnetic field should be confined to a certain region by

magnetic covers [1–3]. Under such circumstances, the field and

inductance of shielded coils will be changed owing to the existence

of shielding material, which cannot be taken into account by the

Neumann formula for inductance of coils in the free air [4–11].

Specific methods should be applied to this type of boundary value

problem (BVP) to take the reflection effect of boundary into

consideration. Normally, to evaluate the effects of shielding forms

such as a pair of parallel plates, circular hollow cylinder or hollow

sphere, we can refer to the relevant monographs [1–3]. However,

for a shielding form of a cuboid, no results have been found as yet

apart from some formulae published for the special case of this

problem [12]. The absence of this shielding form in the literature is

not due to its insignificance. In fact, the cuboidal shielding is

commonly used, and it is more convenient to manufacture and

install in practice. Moreover, it is not enough to simply employ the

finite-element method (FEM) to calculate the effects of cuboidal

shielding. For instance, using FEM to analyse an ‘inverse problem’

may lead to a rather cumbersome task [13] (the self and mutual

inductance of a coupled coil system must satisfy some equations,

so a direct solution with FEM will be infeasible). However,

analytical solutions, by contrast, are competent at this complicated

situation without the limitations of FEM.

In this work, analytical solutions will be given to the BVP

mentioned in the title as a generalisation of the results for coaxial

coils shielded by cuboid of high permeability [12]. Contrary to the

intuition, this seemingly simple problem cannot be tackled easily

with the routine method of magnetic vector potential (MVP) due to

the presence of all components of the magnetic field. In addition,

the solving process will become very tedious by starting with the

Bessel-integral formulas of MVP of the circular rings [14, 15],

since a complicated coefficient matrix of infinite dimensions will

arise from the boundary conditions on the surfaces of the cuboid.

Fortunately, these difficulties can be surmounted by using the

magnetic scalar potential (MSP). In view of the physical meaning,

a magnetic double layer with constant moment density will be

created by a ring carrying the current I [16], as a consequence, the

MSP will experience a jump quantity I while passing through the

area enclosed by the ring, and retain the continuity in the remaining

area. Accordingly, the coefficients of the ansatz can be determined

with the jump characteristics and orthogonality of the

eigenfunctions [17]. With the MSP, the field, inductance, and

forces between the shielded coils are easy to find. The validity of

the results will be confirmed by the numerical calculations

compared with those given by FEM simulations.

The configuration of this BVP is shown in Fig. 1 with the

notations of geometric parameters. Two circular coils with parallel

axes and centres (x

1

, y

1

, z

1

), (x

2

, y

2

, z

2

), are shielded by a cuboid

with length, width, and height of 2a, 2b, 2l, respectively. The axes

of coils are parallel with the side 2l of the cuboid. The analysis will

begin with the field of a single shielded circular ring, and the

results are developed further to the circular coils with rectangular

cross section.

2Ansatz for MSP of a circular ring shielded by a

cuboid of high permeability

We take central point of the cuboid as the origin of the Cartesian

coordinate system (x, y, z) and let x, y, z-axes parallel to the sides

of 2a, 2b and 2l, respectively. The MSP V(x, y, z) must satisfy the

Laplace equation ΔV = 0 and the magnetic field strength H will be

given by negative gradient of V. Supposing that a circular ring

(Ring 1) with radius r

1

and centre (x

1

, y

1

) is located in the plane z

= ζ

1

and carrying the current I

1

, then by separation of variables an

ansatz can be established as

V x, y, z = I

1

∑

m, n = 1

∞

A

mn

f

1

λ

m

, x

× f

2

λ

n

, y f

3

+

λ

mn

, z, ζ

1

for z > ζ

1

(1a)

and

V x, y, z = I

1

∑

m, n = 1

∞

A

mn

f

1

λ

m

, x

× f

2

λ

n

, y f

3

−

λ

mn

, z, ζ

1

for z < ζ

1

,

(1b)

IET Electr. Power Appl., 2018, Vol. 12 Iss. 5, pp. 717-727

© The Institution of Engineering and Technology 2018

717

where A

mn

are undetermined coefficients,

m

,

n

and

mn

are

eigenvalues such that

λ

mn

= λ

m

2

+ λ

n

2

.

(2)

In (1a) and (1 b), the axial functions f

3

±

λ

mn

, z, ζ

1

are prepared for

later consideration of the jump feature. Furthermore, the

eigenfunctions f

1

(

m

, x) and f

2

(

n

, y) should be the linear

combination of trigonometric functions, namely

f

1

λ

m

, x = A

1

cos λ

m

x + B

1

sin λ

m

x ,

f

2

λ

n

, y = A

2

cos λ

n

y + B

2

sin λ

n

y .

(3)

The permeability of the cuboid can be regarded as infinity to

simplify the analysis. It follows that the tangential components of

H must vanish on the surfaces, namely

H

y

±a, y, z = 0,

H

x

x, ± b, z = 0,

(4a)

and

H

z

±a, y, z = 0,

H

z

x, ± b, z = 0.

(4b)

Since H = − ∇V, it follows that

V ±a, y, z = 0,

V

x, ± b, z = 0.

(4c)

Substituting (3) into (4c) yields the following linear systems for A

1

,

B

1

and A

2

, B

2

, respectively

cos λ

m

a A

1

+ sin λ

m

a B

1

= 0,

cos

λ

m

a A

1

− sin λ

m

a B

1

= 0,

(5a)

and

cos λ

n

b A

2

+ sin λ

n

b B

2

= 0,

cos

λ

n

b A

2

− sin λ

n

b B

2

= 0.

(5b)

Hence, the existence of non-trivial solutions of (5a) demands

cos λ

m

a sin λ

m

a

cos λ

m

a −sin λ

m

a

= 0.

(6)

Thus the eigenvalues are given immediately

λ

m

=

mπ

2a

, m = 1, 2, 3, …

Consequently, the corresponding eigenfunction f

1

can be found

according to the Sturm–Liouville theory [18]

f

1

λ

m

, x = sin λ

m

a ⋅ cos λ

m

x

−cos λ

m

a ⋅ sin λ

m

x = sin λ

m

a − x .

(7a)

From (5b), we have the following result in a similar manner:

λ

n

=

nπ

2b

, n = 1, 2, 3, …

and

f

2

λ

n

, y = sin λ

n

b ⋅ cos λ

n

y

−cos λ

n

b ⋅ sin λ

n

y = sin λ

n

b − y .

(7b)

The axial functions f

3

±

λ

mn

, z, ζ

1

of the ansatz (1a) and (1b) should

be treated in a slightly different manner. Considering that the total

field should be separated into the primary and secondary parts, and

the source can be regarded as a magnetic double layer, f

3

±

λ

mn

, z, ζ

1

should be expressed as

f

3

+

λ

mn

, z, ζ

1

= A

3

e

λ

mn

z

+ B

3

e

−λ

mn

z

+ e

−λ

mn

z − ζ

1

for z > ζ

1

,

(8a)

f

3

−

λ

mn

, z, ζ

1

= A

3

e

λ

mn

z

+ B

3

e

−λ

mn

z

− e

−λ

mn

ζ

1

− z

for z < ζ

1

.

(8b)

Moreover, for z = ± l the field components

H

x

= − ∂V /∂x, H

y

= − ∂V /∂y must vanish, that is

A

3

e

lλ

mn

+ B

3

e

−lλ

mn

+ e

−

l − ζ

1

λ

mn

= 0,

A

3

e

−lλ

mn

+ B

3

e

lλ

mn

− e

− l + ζ

1

λ

mn

= 0.

(9)

[For simplicity, the square brackets [] of cosh[

mn

(l + ζ

1

)],

sinh[

mn

(l + z)], sin[

m

(a − x)], cos[

n

(b − y)], etc., will be omitted

henceforth. Accordingly, the factor inside the round bracket always

belongs to the argument of the preceding function. Unless

otherwise stated, this rule will apply to all of the following

formulae in this work.] Solving (9) gives

A

3

= −

cosh λ

mn

l + ζ

1

e

−lλ

mn

sinh

2lλ

mn

,

B

3

=

cosh λ

mn

l − ζ

1

e

−lλ

mn

sinh

2lλ

mn

.

(10)

We, therefore, have the result of axial functions

f

3

+

λ

mn

, z, ζ

1

=

2sinh λ

mn

l − z cosh λ

mn

l + ζ

1

sinh 2lλ

mn

for z > ζ

1

,

(11a)

Fig. 1 Side and plan views of the circular coils with parallel axes shielded

by a cuboid of high permeability

718 IET Electr. Power Appl., 2018, Vol. 12 Iss. 5, pp. 717-727

© The Institution of Engineering and Technology 2018

f

3

−

λ

mn

, z, ζ

1

= −

2sinh λ

mn

l + z cosh λ

mn

l − ζ

1

sinh 2lλ

mn

for z < ζ

1

.

(11b)

Let F

R

denotes the area encircled by the circular ring, it follows

that V(x, y, z) will obtain a spring quantity I

1

when passing through

F

R

and maintain the continuity on the remaining area F of plane z

= ζ

1

(see Fig. 2). This statement can be written in the following

form:

V x, y, ζ

1

+ − V x, y, ζ

1

− =

I

1

on F

R

,

0 on F,

(12)

where V(x, y, ζ

1

+) and V(x, y, ζ

1

−) are obtained by letting z = ζ

1

in

(1a) and (1b), respectively. With (1a) and (1 b) and

f

3

+

λ

mn

, ζ

1

, ζ

1

− f

3

−

λ

mn

, ζ

1

, ζ

1

= 2,

(13)

it can be deduced from (12)

∑

m, n = 1

∞

A

mn

sin λ

m

a − x sin λ

n

b − y =

1/2 on F

R

,

0 on F .

(14)

Multiplying both sides of (14) by the eigenfunctions, we have

∑

p, q = 1

∞

A

pq

sin λ

p

a − x sin λ

q

b − y sin λ

m

a − x

× sin λ

n

b − y =

sin λ

m

a − x sin λ

n

b − y /2 on F

R

,

0 on F,

(15)

where

λ

p

=

pπ

2a

, p = 1, 2, 3, …

λ

q

=

qπ

2b

, q = 1, 2, 3, …

(16)

Next, we integrate the left-hand side of (15) with respect to (x, y)

over the whole rectangle 2a × 2b, and the right-hand side (RHS)

with respect to the local cylindrical coordinates (r, φ) over F

R

, that

is

∑

p, q = 1

∞

A

pq

∫

−a

a

sin λ

p

a − x sin λ

m

a − x dx

×

∫

−b

b

sin λ

q

b − y sin λ

n

b − y dy

=

1

2

∫

0

2π

∫

0

r

1

sin λ

m

a − x sin λ

n

b − y rdrdφ,

(17)

where

x = x

1

+ rcos φ,

y = y

1

+ rsin φ .

(18)

By means of the orthogonal relation

∫

−a

a

sin λ

p

a − x sin λ

m

a − x dx = aδ

pm

,

∫

−b

b

sin λ

q

b − y sin λ

n

b − y dy = bδ

qn

,

(19)

where δ

pm

, δ

qn

are the Kronecker delta, it follows from (17)

A

mn

=

1

2ab

∫

0

2π

∫

0

r

1

sin λ

m

a − x

1

− rcos φ

× sin λ

n

b − y

1

− rsin φ rdrdφ .

(20)

Applying the integrals

∫

0

2π

cos λ

m

rcos φ cos λ

n

rsin φ dφ = 2πJ

0

λ

mn

r

(21)

and

∫

0

2π

cos λ

m

rcos φ sin λ

n

rsin φ dφ

=

∫

0

2π

sin

λ

m

rcos φ cos λ

n

rsin φ dφ

=

∫

0

2π

sin

λ

m

rcos φ sin λ

n

rsin φ dφ = 0

(22)

to RHS of (20), where J

n

(x) is the Bessel function of the first kind

of order n, we thus get the coefficient

A

mn

=

πr

1

abλ

mn

J

1

λ

mn

r

1

sin λ

m

a − x

1

sin λ

n

b − y

1

.

(23)

Eventually, combining (1a), (1b), (7a), (7b), (11a), (11b), and (23)

gives the solution of MSP

V x, y, z = ±

2πr

1

I

1

ab

∑

m, n = 1

∞

J

1

λ

mn

r

1

sin λ

m

a − x

1

× sin λ

n

b − y

1

sin λ

m

a − x sin λ

n

b − y

×

sinh λ

mn

l ∓ z cosh λ

mn

l ± ζ

1

λ

mn

sinh 2lλ

mn

,

z > ζ

1

,

z < ζ

1

.

(24)

Moreover, by the definition

H = − ∇V,

(25)

we find the expressions of magnetic field immediately

Fig. 2 Global and local coordinates of a circular ring shielded by a

cuboid of high permeability

IET Electr. Power Appl., 2018, Vol. 12 Iss. 5, pp. 717-727

© The Institution of Engineering and Technology 2018

719

H

x

x, y, z = ±

2πr

1

I

1

ab

∑

m, n = 1

∞

J

1

λ

mn

r

1

sin λ

m

a − x

1

× sin λ

n

b − y

1

cos λ

m

a − x sin λ

n

b − y

×

λ

m

sinh λ

mn

l ∓ z cosh λ

mn

l ± ζ

1

λ

mn

sinh 2lλ

mn

,

H

y

x, y, z = ±

2πr

1

I

1

ab

∑

m, n = 1

∞

J

1

λ

mn

r

1

sin λ

m

a − x

1

× sin λ

n

b − y

1

sin λ

m

a − x cos λ

n

b − y

×

λ

n

sinh λ

mn

l ∓ z cosh λ

mn

l ± ζ

1

λ

mn

sinh 2lλ

mn

,

H

z

x, y, z =

2πr

1

I

1

ab

∑

m, n = 1

∞

J

1

λ

mn

r

1

sin λ

m

a − x

1

× sin λ

n

b − y

1

sin λ

m

a − x sin λ

n

b − y

×

cosh λ

mn

l ∓ z cosh λ

mn

l ± ζ

1

sinh 2lλ

mn

.

z > ζ

1

z < ζ

1

.

(26)

Equations (26) lay the foundation for following derivations of

inductance and forces of circular coils.

3Self- and mutual inductance of circular coils

with rectangular cross section shielded by a

cuboid of high permeability

Expressions for the inductance of circular coils are deducible

immediately from the magnetic field given by (26). Supposing that

another circular ring (Ring 2) is placed in the aforesaid field of

Ring 1, with the centre (x

2

, y

2

) and radius r

2

and lying in the plane

z = ζ

2

, therefore the mutual inductance between the Rings 1 and 2

can be found by calculating the magnetic flux passing through the

area encircled by the Ring 2, namely

M =

μ

0

I

1

∫

0

2π

∫

0

r

2

H

z

x

2

+ rcos φ, y

2

+ rsin φ, ζ

2

rdrdφ

=

4μ

0

π

2

r

1

r

2

ab

∑

m, n = 1

∞

J

1

λ

mn

r

1

J

1

λ

mn

r

2

× sin λ

m

a − x

1

sin λ

n

b − y

1

sin λ

m

a − x

2

×

sin λ

n

b − y

2

cosh λ

mn

l ± ζ

1

cosh λ

mn

l ∓ ζ

2

λ

mn

sinh 2lλ

mn

,

ζ

2

> ζ

1

ζ

2

< ζ

1

.

(27)

When x

1

, y

1

, x

2

, y

2

are all equal to zero, (27) will degenerate into

(19a) and (19b) of [12]. For a circular ring of the infinitesimal

cross section, the self-inductance is divergent [19].

To calculate the self- and mutual inductance for the shielded

circular coils with rectangular cross section from (27), we consider

a pair of circular coils, one has the centre (x

1

, y

1

), inner and outer

radii R

1

, R

2

, axial length 2h

1

and N

1

turns (Coil 1), another has the

corresponding parameters of (x

2

, y

2

), R

3

, R

4

, 2h

2

and N

2

(Coil 2),

and the middle planes of them are set in z = z

1

and z = z

2

,

respectively (see Fig. 1). By evaluating the axial integral with

respect to the arguments ζ

1

, ζ

2

of (27) and using the integral

formula [20]

w

λ

mn

, R

1

, R

2

=

∫

R

1

R

2

rJ

1

λ

mn

r dr

=

π

2λ

mn

R

2

J

1

λ

mn

R

2

H

0

λ

mn

R

2

− J

0

λ

mn

R

2

H

1

λ

mn

R

2

−R

1

J

1

λ

mn

R

1

H

0

λ

mn

R

1

− J

0

λ

mn

R

1

H

1

λ

mn

R

1

,

(28)

where H

n

(x) is the Struve function of order n, we have the

following results:

i. For z

1

+ h

1

≤ z

2

− h

2

M = κ

12

∑

m, n = 1

∞

w

λ

mn

, R

1

, R

2

w λ

mn

, R

3

, R

4

× sin λ

m

a − x

1

sin λ

n

b − y

1

sin λ

m

a − x

2

×

sin λ

n

b − y

2

v

1

λ

mn

, z

1

, z

2

λ

mn

sinh 2lλ

mn

,

(29a)

where

v

1

λ

mn

, z

1

, z

2

= 4sinh λ

mn

h

1

sinh λ

mn

h

2

× cosh λ

mn

l + z

1

cosh λ

mn

l − z

2

/λ

mn

2

and

κ

12

=

μ

0

π

2

N

1

N

2

abh

1

h

2

R

2

− R

1

R

4

− R

3

is a coefficient relative to the number of turns and geometric

parameters of both coils and cuboid.

ii. For z

1

− h

1

≤ z

2

− h

2

< z

1

+ h

1

≤ z

2

+ h

2

M = κ

12

∑

m, n = 1

∞

w λ

mn

, R

1

, R

2

w λ

mn

, R

3

, R

4

× sin λ

m

a − x

1

sin λ

n

b − y

1

sin λ

m

a − x

2

×

sin λ

n

b − y

2

v

2

λ

mn

, z

1

, z

2

λ

mn

sinh 2lλ

mn

,

where

v

2

λ

mn

, z

1

, z

2

=

1

λ

mn

2

h

1

+ h

2

+ z

1

− z

2

λ

mn

sinh 2lλ

mn

+sinh λ

mn

h

1

− l + z

1

sinh λ

mn

h

2

− l − z

2

+sinh λ

mn

h

1

+ l + z

1

sinh λ

mn

h

2

− l + z

2

+2sinh h

2

λ

mn

sinh λ

mn

h

1

− l − z

1

cosh λ

mn

l − z

2

.

(29b)

iii. For z

2

− h

2

≤ z

1

− h

1

< z

1

+ h

1

≤ z

2

+ h

2

(see (29c))

In a similar manner, we deduce that the self-inductance, for a

shielded circular coil with the centre (x

1

, y

1

, z

1

), inner and outer

radii R

1

, R

2

, axial length 2h

1

and N

1

turns, is

L =

μ

0

π

2

N

1

2

abh

1

2

R

2

− R

1

2

∑

m, n = 1

∞

w

2

λ

mn

, R

1

, R

2

×

sin

2

λ

m

a − x

1

sin

2

λ

n

b − y

1

v

4

λ

mn

, z

1

λ

mn

sinh 2lλ

mn

(30)

with the axial function given by

v

4

λ

mn

, z

1

=

1

λ

mn

2

cosh 2λ

mn

(h

1

− l) − cosh 2lλ

mn

+cosh 2h

1

λ

mn

cosh 2z

1

λ

mn

−cosh 2z

1

λ

mn

+ 2h

1

λ

mn

sinh 2lλ

mn

.

720 IET Electr. Power Appl., 2018, Vol. 12 Iss. 5, pp. 717-727

© The Institution of Engineering and Technology 2018

4Magnetic force exerted on circular coils

shielded by a cuboid of high permeability

The magnetic forces between the coils are of practical importance.

Before we commence the specific calculation, some preliminary

discussions should be given for the force calculation of coils with

the presence of magnetic materials. Unlike the force calculation of

coils in the free air, in which we only need to consider the

interaction between the coils, in this BVP the interaction between

coils and that between coil and cuboid should be considered

simultaneously. Hence, if we calculate the force from the magnetic

energy, both mutual and self-magnetic energy (mutual and self-

inductance) should be taken into account to obtain the total force.

To clarify the process of force calculation, we will begin with the

definition of magnetic force. With the aid of the obtained magnetic

field (26) of Ring 1, the mutual magnetic force F

2

m

exerted on

Ring 2 can be obtained immediately. According to the definition of

magnetic force, it follows that [21, 22]

F

2

m

=

∮

I

2

dl

2

× B

1

= I

2

r

2

∫

0

2π

B

z

x

2

+ r

2

cos φ, y

2

+ r

2

sin φ, z cos φe

x

+B

z

x

2

+ r

2

cos φ, y

2

+ r

2

sin φ, z sin φe

y

−

B

y

x

2

+ r

2

cos φ, y

2

+ r

2

sin φ, z sin φ

+B

x

x

2

+ r

2

cos φ, y

2

+ r

2

sin φ, z cos φ

e

z

dφ,

(31)

where ‘dl

2

’ is the infinitesimal element of Ring 2, and the local

polar coordinate (r, φ)

x = x

2

+ rcos φ

y = y

2

+ rsin φ

(32)

is introduced to assist the integration. Using the angular integrals

∫

0

2π

sin λ

m

r

2

cos φ cos λ

n

r

2

sin φ cos φdφ =

2πλ

m

J

1

λ

mn

r

2

λ

mn

,

(33a)

and

∫

0

2π

cos λ

m

r

2

cos φ cos λ

n

r

2

sin φ cos φdφ

=

∫

0

2π

cos

λ

m

r

2

cos φ sin λ

n

r

2

sin φ cos φdφ

=

∫

0

2π

sin

λ

m

r

2

cos φ sin λ

n

r

2

sin φ cos φdφ = 0

(33b)

it follows that

∫

0

2π

sin λ

m

a − x

2

− r

2

cos φ

× sin λ

n

b − y

2

− r

2

sin φ cos φdφ

= − 2πλ

m

J

1

λ

mn

r

2

cos λ

m

a − x

2

sin λ

n

b − y

2

/λ

mn

.

(34)

We thus obtain the x-component of F

2

m

F

2, x

m

= −

4μ

0

π

2

r

1

r

2

I

1

I

2

ab

∑

m, n = 1

∞

λ

m

J

1

λ

mn

r

1

J

1

λ

mn

r

2

× sin λ

m

a − x

1

sin λ

n

b − y

1

cos λ

m

a − x

2

×

sin λ

n

b − y

2

cosh λ

mn

l ± ζ

1

cosh λ

mn

l ∓ ζ

2

λ

mn

sinh 2lλ

mn

,

ζ

2

≥ ζ

1

ζ

2

< ζ

1

.

(35)

Comparing (27) with (35), it is observed that

F

2, x

m

=

∂W

m

∂x

2

= I

1

I

2

∂M

∂x

2

,

where W

m

is the mutual magnetic energy. In fact, this is a general

theorem which establishes the relationship between magnetic force

and mutual inductance between two coils [23–27]. Consequently,

the remaining force components F

2, y

m

, F

2, z

m

are also deducible from

differentiating (27) with respect to the corresponding coordinates

y

2

, ζ

2

, and we will not repeat them here due to the space

limitations.

Similarly, by differentiation with respect to the corresponding

centre coordinates of coils in (29a)–(29c), the mutual force F

2

m

exerted on Coil 2 is obtained with no difficulty

i. For z

1

+ h

1

≤ z

2

− h

2

F

2, x

m

= I

1

I

2

∂M

∂x

2

= − κ

12

I

1

I

2

∑

m, n = 1

∞

λ

m

w

λ

mn

, R

1

, R

2

w λ

mn

, R

3

, R

4

× sin λ

m

a − x

1

sin λ

n

b − y

1

cos λ

m

a − x

2

×

sin λ

n

b − y

2

v

1

λ

mn

, z

1

, z

2

λ

mn

sinh 2lλ

mn

,

(36a)

M = κ

12

∑

m, n = 1

∞

w

λ

mn

, R

1

, R

2

w λ

mn

, R

3

, R

4

× sin λ

m

a − x

1

sin λ

n

b − y

1

sin λ

m

a − x

2

×

sin λ

n

b − y

2

v

3

λ

mn

, z

1

, z

2

λ

mn

sinh 2lλ

mn

,

where

v

3

λ

mn

, z

1

, z

2

=

2

λ

mn

2

h

1

λ

mn

sinh 2lλ

mn

+ sinh λ

mn

h

1

⋅ cosh λ

mn

l − z

1

sinh λ

mn

(h

2

− l − z

2

)

+sinh

λ

mn

h

1

cosh λ

mn

l + z

1

sinh λ

mn

(h

2

− l + z

2

)

.

(29c)

IET Electr. Power Appl., 2018, Vol. 12 Iss. 5, pp. 717-727

© The Institution of Engineering and Technology 2018

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