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1

Electromagnetic acceleration of permanent magnets

S. N. Dolya

Joint Institute for Nuclear Research, Joliot - Curie 6, Dubna, Russia, 141980

Abstract

We consider the acceleration of the permanent magnets, consisting of NdFeB by means of

the running magnetic field gradient. It is shown that the specific magnetic moment per

nucleon in NdFeB is determined by the remained magnetization of the substance and for

NdFeB it is equal to the m

NdFeB

= 1.57 * 10

-10

eV / Gs. The maximum accessable gradient of

the magnetic field accelerating the permanent magnets is determined by the coercive force

H

c

~ 30 kGs. For the NdFeB magnets this gradient is equal to: dB

z

/ dz = 20 kGs / cm. The

finite velocity of the magnets V

fin

= 6 km / s, the length of acceleration is L

acc

= 637 m.

1. Introduction

The acceleration of the magnetic dipoles can be carried out in the Gauss gun

[1], wherein the iron core is magnetized by the field coil with a current and

further it is involved into the coil. When passing of the iron core through the

coil centre, the current in the coil abruptly stops and the iron core is flying to

the next coil where the same process takes place.

The Gauss gun has a number of drawbacks. First of all, while driving of the

core through the Gauss gun the core will deviate exponentially from the

selected synchronous phase since the phase motion in the Gauss gun in unstable

[2]. This means that if we calculate the movement of the core for some position

of the core relatively the position of the current pulse in the coil, which is called

the phase, then we find that the core while its moving along the coils with the

current deviates exponentially from the phase of the current in the coil.

Indeed, let the core be a little bit behind the phase in its movement and it has

arrived later to the coil where the current had already shut down. This means

that the core attains less energy during its passage through this coil and its being

late will grow with every following coil.

On the contrary, if the core comes ahead to some coil, it gets more energy and

to the following coil it will come with a growing advance.

Physically this means that the region of stable phases is located in the front

slope of the pulse, and steadily to accelerate the body it is possible only by

2

pushing the body with the current pulse. It is not possible steadily to accelerate

the core if the pulse pulls the core [3]. This is easier to understand trying to

speed up the iron core by means of a little magnet. This little magnet should be

taken into the hand and it will imitate the current pulse running on the coils.

The iron core is magnetized and attracted to the little magnet in the hand.

The hand with this magnet must move with acceleration imitating the current

pulse running through the coils of the Gauss gun. Then it will be clear that it is

impossible to reach the stable acceleration. If the hand moves too fast, the iron

core will get into the less and less field gradient of the little magnet and the core

will be behind the hand. If you start to move the hand a little bit slower, the iron

core, approaching the hand with the little magnet will get into the increasing

gradient of the magnetic field. It will accelerate faster and, finally, will catch up

with the hand holding the little magnet. Thus, it is not possible to obtain the

stable acceleration in this case.

The situation is different if the hand with a little magnet pushes the other little

magnet. When approaching each other the repulsive force between these little

magnets increases and the little magnet kept in the hand faster and faster

accelerates the other little magnet. If the little magnet goes too far from the one

in the hand it will get into the region of the less gradients of the magnetic field.

The little magnet in the hand is moving with acceleration and that is why it will

catch up with the other little magnet which is accelerated. In this case we will

observe the process of sustainable acceleration.

As we have mentioned above, this acceleration picture is explained by the

following: the region of stable phases of acceleration is located on the front

slope of the accelerating current pulse and it is necessary to push the little

magnet being accelerated but not pull it.

Another disadvantage of the Gauss gun is the low rate of acceleration of the

magnetic dipole in it.

The force of the magnetic dipole interaction F

z

with the gradient of the

magnetic field can be written as follows:

F

z

= m*dB

z

/dz, (1)

where m - the magnetic moment per mass unit, dB

z

/dz - the magnetic field

gradient.

3

The accelerator technology chooses the nucleon mass as the mass unit which

is equal to mc

2

≈ 1 GeV in energy units.

In iron the magnetic moment per atom is [4] p. 524, m = 2.22 n

b

,

where n

b

= 9.27 * 10

-21

erg / Gs, the Bohr magneton, [4], p. 31. Taking into

account that the atomic weight of iron A is approximately equal to A = 56 and

going from erg units to the eV units (1 erg = 6.24 * 10

11

eV) it can be found that

the magnetic moment per nucleon in iron is equal to the following:

m

Fe

= 2.3*10

-10

eV/Gs*nucleon.

We assume that it is possible to create a running gradient of the magnetic field

equal to the value of dB

z

/ dz = 20 kGs / cm = 2 * 10

6

Gs / m. The finite velocity

of the core V

fin

= 6 km / s, expressed in the units of the velocity of light in

vacuum, will be equal to: β

fin

= V

fin

/ c = 2 * 10

-5

, where c = 3*10

10

cm / s – is

the velocity of light in vacuum. The finite energy per nucleon in the iron core

will be equal to:

W

fin

= mc

2

β

2fin

/2 = 10

9

*4*10

-10

/2 = 0.2 eV/nucleon.

The acceleration rate is ∆W / ∆z = F

z

= m * dB

z

/ dz = 2.3 * 10

-10

* 2*10

6

=

= 4.6 * 10

-4

eV / m * nucleon and the core will reach the finite energy

W

fin

= 0.2 eV / nucleon at the length of the accelerator equal to the following:

L

acc Fe

= W

fin

/(∆W/∆z) = 0.2/4.6*10

-4

= 435 m.

Note that it is not possible to sustainably accelerate the iron core in the Gauss

gun.

To increase the magnetic moment per nucleon in comparison with the iron is

possible if to use the superconducting coil with the current as the body being

accelerated [3]. For the coil with a diameter of d

turn

= 6 cm it is possible to

increase the magnetic moment per nucleon by 17 times.

2. The interaction of the running magnetic field gradient with a constant

magnet based on NdFeB

It is needed to preserve the superconductivity inthe current coil while

accelerating as well as during the time of flight of the accelerated magnetic

4

dipole to the target. It is not convenient. We consider a possibility of

accelerating the permanent magnet. The most powerful permanent magnet is

based on the composition of the magnets made of NdFeB.

We assume that the magnets are made of 30% of neodymium with atomic

mass A

Nd

= 144 and 70% of the iron having the atomic mass A

Fe

= 56. Find the

average atomic mass A

NdFeB

= 144 * 0.3 + 0.7* 56 = 82. The density of the

composition of NdFeB is taken to be equal ρ

NdFeB

= 7.4 g / cm

3

. Then, from the

relation of Avogadro we find the number of atoms per cubic centimeter:

6*10

23

………… 82 g

x …………… 7.4 g.

From the above we find: n

NdFeB

= 5 * 10

22

atoms in a cubic centimeter. The

role of the magnetic field of saturation will now be played by B

r

- the residual

magnetization of the sample consisting of NdFeB. Dividing the residual

induction by 4π and the density of atoms n

NdFeB

= 5 * 10

22

, we get the magnetic

moment per atom of the substance. Let us take a residual induction B

r

to be

equal to B

r

= 13.5 kGs.

Then we find that the magnetic moment per atom in the substance NdFeB is

roughly the same as in iron:

n

NdFeB

= 1.35*10

4

/(12.56*5*10

22

) =2.21n

b

.

However, due to the fact that the atomic weight NdFeB is greater than that of

the iron and, therefore, it contains a larger number of nucleons, the magnetic

dipole moment per nucleon in NdFeB, turns out to be smaller than that of the

iron and is equal to:

m

NdFeB

=1.57*10

-10

eV/Gs.

This means that the rate of accumulation of energy in the running gradient of

the magnetic field dB

z

/ dz = 20 kGs / cm in this case will be equal to:

∆W / ∆z = F

z

= m * dB

z

/ dz = 1.57 * 10

-10

* 2 * 10

6

=

=3.14 * 10

-4

eV / (m * nucleon). The finite energy W

fin

= 0.2 eV / nucleon will

be obtained by the NdFeB permanent magnet at the following length of

acceleration:

L

acc

= W

fin

/(∆W/∆z) = 0.2/3.14*10

-4

= 637 m.

5

3. Acceleration of the permanent magnets in a spiral waveguide

The Gauss gun, in fact, is an artificial line with definite parameters, i.e., in this

line there are individual capacitors which discharge onto separate inductors. If

you transfer from the line with the definite parameters to the line with

distributed parameters which is a spiral waveguide, it is possible to significantly

increase the average rate of the energy gain by the permanent magnet. This will

be obtained due to the fact that the gaps between the coils where the magnet

moves without acceleration will be excluded.

At the same time, as usually for linear particle accelerators we have to choose

a synchronous phase on the front slope of the current pulse and carry out a

detailed calculation of the dynamics of the magnet being accelerated. This

calculation for the magnet containing an iron core and a superconducting coil

was performed in [5].

Below we consider the acceleration of the permanent magnet made of NdFeB,

having a length of l

NdFeB

= 1 cm. We assume that the acceleration of the

permanent magnet is performed by the running gradient of the magnetic field.

This gradient is generated by the running current pulse via the coils of the

spiral.

As in [5] we obtain the relation between the value of the gradient and the

value of the magnetic field - dB

z

/ dz B

z

:

dB

z

/dz = k

3

B

z

=(2π/λ

s

)B

z

, (2)

where k

3

(the wave vector in the z-direction) is equal to: k

3

= 2π / λ

s

,

λ

s

– the slowed down wave length propagating in the spiral waveguide

is λ

s

= λ

0

* β

ph

; λ

0

= c / f - the wavelength in vacuum; c = 3*10

10

cm / s is the

velocity of light in vacuum; f is the frequency of the wave. The wave frequency

f is related to the duration of the current pulse propagating along the coils of the

spiral with τ ratio: f ≈ 1 / 2τ.

The phase velocity of the wave β

ph

, expressed in units of the velocity of light,

must coincide with the velocity of the accelerated permanent magnet. At the

end of acceleration this velocity must be equal to β

ph fin

= β

fin

= 2 * 10

-5

, that

corresponds to the normal velocity of the permanent magnet equal

V

fin

= β

fin

* c = 6 km / s.

6

Inside the spiral waveguide the components of the longitudinal electrical and

magnetic fields can be written as:

E

z

= E

0

I

0

(k

1

r),

B

z

= -i(k

1

/k)tgΨI

0

(k

1

r

0

)E

0

I

0

(k

1

r)/I

1

(k

1

r

0

), (3)

where I

0

, I

1

- a modified Bessel function, r

0

– the radius of the spiral,

k

1

= k (1 / β

2ph

- 1)

1/2

– the radial wave vector inside the spiral waveguide. The

total wave vector k = ω / (cε

1 / 2

), ε – is the dielectric constant of the medium,

which fills the spiral waveguide, ω = 2πf –is the angular frequency,

tgΨ = h / 2πr

0

– is the tangent of the spiral winding angle, h - the winding pitch

of the spiral. The value of E

0

is the amplitude of the electric field on the axis of

the spiral.

For the case of β

ph

<< 1 it can be considered k

1

/ k = 1 / β

ph

. We also assume

that the whole accelerator is divided into individual sections, where in each

section the following condition is fulfilled: x = k

1

r

0

= 2πr

0

/ (β

ph

λ

0

) = 1. Bearing

in mind that I

0

(1) / I

1

(1) = 2 and I

0

(0) = 1, we get:

B

z

/E

z

= 2tgΨ/ β

ph

. (4)

Since for the dense winding of the spiral and dielectric filling of the area

between the coil and the screen there is the following condition [5]:

β

ph

= √2 tgΨ/ε

1/2

, (5)

we, finally, obtain:

B

z

/E

z

= (2ε)

1/2

. (6)

Here we must remember the following. When accelerating of the iron cores

in the Gauss gun the iron core is magnetized by the current pulse accelerating

this core in such a way that the core is pulled into the coil. Similarly, the core is

pulled into the coil in a home electrical bell. At the acceleration of the

permanent magnets by the current pulse running via the spiral coils these

magnets will be pushed out from the coil by the magnetic field gradient, and

they will be demagnetized by the magnetic field of the current pulse.

The iron core can be accelerated by any value of the magnetic field gradient

and by the value of the magnetic field B

acc

associated with this gradient. When

7

the induction reaches saturation the iron core simply ceases further

magnetization.

If the value of the field B

acc

exceeds the coercive force of the permanent

magnet B

acc

> H

c

, the permanent magnet is simply demagnetized. In this case the

dipole magnetic moment will become zero and the permanent magnet ceases

the acceleration by the running gradient of the magnetic field.

The best permanent magnets made of NdFeB have the coercive force

H

c

> 30 kGs. As it will be shown below, for the short spatial running current

pulses the gradient of the magnetic field is achievable: dB

z

/ dz> 20 kGs / cm.

4. Selection of main parameters of the spiral waveguide

We choose the radius of the spiral waveguide winding r

0

= 1 cm. This means

that we have chosen the slowed down wavelength equal to: λ

s

= 2πr

0

= 6 cm and

this wavelength will remain unchanged throughout the accelerator. Take the

initial velocity of acceleration of the permanent magnets to be equal to the

following: V

in

= 600 m / s, [5].

This velocity is assumed to achieve by the preliminary gas-dynamic

acceleration of permanent magnets. Thus, the electromagnetic acceleration is

performed from the initial velocity, expressed in terms of the velocity of light

β

in

= 2 * 10

-6

to the final velocity β

fin

= 2 * 10

-5

.

We find the spatial length of the current pulse accelerating the permanent

magnets. The wavelength λ

0

= λ

s

/ β

ph

and it is equal to λ

0

in

= 3 * 10

6

cm to start

acceleration and λ

0 fin

= 3 * 10

5

cm - for the end of acceleration. To support the

optimal condition between the power involved into the spiral waveguide and the

amplitude E

0

of the electrical field on the acceleration axis [6], it is necessary to

alter the length of the accelerating current pulse τ from section to section in the

process of acceleration.

For the start of acceleration the wave length λ

0 in

= c * 2τ

in

, from where

τ

in

= λ

0 in

/ 2c = 50 µs. At the end of acceleration the duration of the current

pulse accelerating the permanent magnet must be 10 times shorter:

τ

fin

= λ

0 fin

/ 2c = 5 µs. This pulse duration corresponds to the frequency of the

acceleration at the start of acceleration f

in

= 1 / 2τ

in

=10

4

Hz, and at the end of

acceleration it corresponds to f

fin

= 1 / 2τ

fin

= 10

5

Hz.

8

From (2) we find the magnetic field value corresponding to the magnetic field

gradient dB

z

/ dz = 20 kGs / cm. For λ

s

/ 2π = 1 cm the magnetic field of the

running pulse is: B

z

= 20 kGs.

Knowing the magnetic field of the pulse from the formula (6) we can find the

electric field of the pulse on the system axis E

z

= B

z

/ (2ε)

1/2

. Choose the value

of ε for the medium, which fills the space between the coil and the outer screen

ε = 1.28 * 10

3

. This value of ε is contained, for example, in barium titanate.

From the electro-technical point of view this type of filling the space between

the coil and the outer screen means that we have increased the capacity between

the coils of the spiral and the capacitance between the coils of the spiral and the

screen by ε times.

Introduction of the medium with a high value of ε allows one additionally to

slow down the wave accelerating the permanent magnets.

Substituting the numbers into the formula (6) we find that E

z

= 120 kV / cm.

This value is smaller than the amplitude value E

z

of the electric field on the axis

E

0

. It is smaller because we have to use a non-maximum value of the electrical

field. [2]. Now we choose sinφ

s

= 0.7, that means that we have chosen the

synchronous phase φ

s

= 45

0

.

The amplitude value of the field E

0

in this case turns out to be equal to:

Е

0

= E

z

/sinφ

s

= 170 kV/cm.

We can use the following formula to estimate the value of the pulse voltage

which will run in the spiral and create an appropriate electric field E

0

:

U = Е

0

* λ

s

/2π

=170 kV.

From the relation between the power of the energy flow and the field strength

on the axis of the spiral wave guide we can find the required pulse power [5]. It

is equal to the following:

P =(c/8)E

02

r

02

β

ph

{(1+I

0

K

1

/I

1

K

0

)(I

12

– I

0

I

2

) +

ε(I

0

/K

0

)

2

(1+I

1

K

0

/I

0

K

1

)(K

0

K

2

– K

12

)}. (7)

The argument of the modified Bessel functions of the first and second kind

shown in the curly brackets is the value of x = 2πr

0

/ λ

s

, which we have chosen

9

to be equal to: x = 1. Then, for this argument, the second term in brackets is

much larger than the first one, and the bracket itself is: {} = 3.77 * ε, [6].

Substituting the numbers into the formula (7), we find:

P (W) = 3*10

10

*1.7*1.7*10

10

*2*10

-5

*3.77*1.28*10

3

/(8*300*300*10

7

) =

=12.3 MW.

The issues of the pulse attenuation while its propagating in the spiral

waveguide will not be considered in detail. We will not discuss the following

issues: the capture of permanent magnets into the acceleration mode, the

accuracy of the synchronization of start of the gas-dynamic acceleration and of

the phase of the current pulse propagating along the spiral. All these issues as

well as requirements to the accuracy of maintaining the parameters of

acceleration, release of the permanent magnets into the atmosphere of the Earth,

etc, are discussed it in detail in [5-8].

Let us concentrate only on the issues of focusing the permanent magnets, i.e.,

how to hold the permanent magnet near the axis of acceleration. You can

continue the analogy with the acceleration of the little permanent magnet by

using a little magnet in the hand. You can see that at the repulsion of the little

magnet being accelerated from the magnet clutched in hand, the little

accelerated magnet tends to deviate from the axis of the acceleration in the

transverse direction. It also tends to turn by180 degrees to be pulled to the little

magnet clutched in the hand.

Deviation from the acceleration axis is explained by the fact that in the

azimuthally symmetric acceleration field it is impossible to fulfill

simultaneously the conditions of the radial and phase stability [2]. The radial

instability corresponds to the area of phase stability, which we have chosen

because we are pushing the permanent magnet. This means that the accelerating

field pushes out the magnet being accelerated from the acceleration axis. Note

that in the Gauss gun where the running current pulls the core, this current coil

pulls the core into the center, i.e., it puts it onto the acceleration axis.

For the magnetic and electric dipole there is the force which turns the dipole

by 180 degrees. There is no such force while accelerating the charged particles.

To keep the magnetic dipoles near the axis of the acceleration, it is necessary

to introduce additional external fields or establish the conditions under which

the dipoles will be kept near the axis of the acceleration.

10

In [5] this holding of the magnetic dipoles near the axis of the acceleration is

performed by means of using the magnetic field which alters its sign in the

space. In [3, 8] it is proposed to introduce acceleration in the narrow channel to

keep the accelerated body near the acceleration axis. There is no opportunity for

the body under acceleration to turn in this narrow channel by 180 degrees and

this channel will also hold the body near the axis of acceleration.

5. Structure of the permanent magnet being accelerated

We will assume that the permanent magnet made of NdFeB consists of one

hundred individual thread magnets each of them has a diameter of 100 µ. We

consider that these magnets are set in a pack so that the transverse dimension of

this pack is about 1 mm. The length of each of the magnet, as we have

mentioned above is l

NdFeB

= 1 cm.

We assume that each segment of the thread is equipped with a sharp cone in

the head part, θ

cone

= 3 * 10

-2

, so that the drag coefficient of the permanent

magnet C

x

= θ

2cone

= 10

-3

, [9]. Since the length of the acceleration till the finite

velocity is large: L

acc

= 637 m, this kind of the accelerator can be positioned

only horizontally. For the magnetic dipoles could cross the atmosphere, each of

them must be given an azimuthally small asymmetry in the area of the cone.

This will result in appearing of a lifting force and of the lift coefficient C

y

. It is

clear that C

y

cannot be more than C

x

, we consider C

x

, C

y

= 10

-3

.

6. Ballistics

The equation of motion of a permanent magnet in the atmosphere can be

written as follows [9]:

V(t) =V

0

/[1+ ρC

x

V

0

*

S*t/2m], (8)

where V

0

is the initial velocity of the permanent magnet. ρ = ρ

0

exp [-y / H

0

] is

the barometric formula describing the decrease of the density of the atmosphere

with the growing altitude; ρ

0

= 1.3 * 10

-3

g / cm

3

is the air density at the Earth's

surface; H

0

= 7 km, S = πd

2

/ 4 ≈ 10

-4

cm

2

is the cross section of the permanent

magnet; the mass of the magnet m = ρ

NdFeB

* S * l

NdFeB

= 7 * 10

-4

g.

The change of the vertical velocity of the permanent magnet in dependence on

time is described with the following equation:

11

V

y

= C

y

ρV

x2

*S*t/2m. (9)

The table below shows the time dependence of the vertical velocity of the

magnetic dipole as well as of its height of lifting and horizontal velocity.

Table. Flight parameters for the case of C

x

, C

y

= 10

-3

.

t, s V

x

, km/s V

y

, km /s

y, km

ρ

air

, g/cm

3

0 6 0 0 1.3*10

-3

3 5.13 1 0.5 1.3*10

-3

13 3.5 3.4 22 7*10

-5

We consider that after reaching the height y = 22 km by the permanent

magnet the air resistance can be neglected and the further motion will be

performed in vacuum. We find the time of lifting the magnet till the highest

possible point: τ

fly

= V

y

/ g = 340 s, where g = 10

-2

km / s

2

is the acceleration of

gravity. The maximum height of the magnet being lifted is:

H

max

= V

2y

/ 2g = 570 km. The flight range of the magnet is equal to the

following: S

max

= V

x

* 2τ

fly

= 2300 km.

7. Splitting of permanent magnets in the space

We use the coefficients C

x

and C

y

which correspond to a separately moving

magnet with a diameter of 100 µ. When the magnets are assembled into the

pack mentioned above with a diameter of 1 mm, these formulae are not

justified.

By means of a micro explosion we give transverse angles of the

order of θ

┴

= β

┴

/ β

z

= 5 * 10

-4

to all the magnets. Having these angles the

permanent magnets will diverge from one another at a distance of l

b

= θ

┴

* L

dr

.

This value for the distance L

dr

= 20 m will be equal to

l

b

= 5 * 10

-4

* 20m = 1 cm.

When colliding with the sharp cone the air molecules will get the transverse

velocity:V

┴

= V

0

* sinθ

cone

= 6 km / s * (1/30) = 200 m / s. If we split 100

permanent magnets on the area of 1 cm

2

, then the average distance between

them in this case is 0.1 cm. The residual gas molecules, which obtain the

transverse velocity of V

┴

= 200 m / s after colliding with the sharp cone of the

permanent magnet, will fly the distance to a neighboring magnet equal

12

to 0.1 cm during τ

1

= 0.1 / 2 * 10

4

= 5µs.

The magnet moving with the velocity of V

0

= 6 km / s will fly the distance

equal to its length l

NdFeB

= 1 cm, during the time equal to τ

2

= 1/6 * 10

5

= 1.7 µs.

It is seen that τ

2

<τ

1

and it means that the residual gas molecules will reach the

neighboring region later than this point has already been crossed by the

neighboring magnet. Thus, the influence of all the magnets on each other may

be neglected, i.e., we can assume that each magnet is moving independently of

the movement of other magnets.

The pack of the magnetic dipoles having a cross section of about 1 cm

2

can be

injected into the atmosphere.

The angle given to the magnets - θ

┴

= β

┴

/ β

z

= 5 * 10

-4

, will lead to the fact

that after the flight over the distance S

max

= 2300 km the magnets will occupy

the area of 1 * 1 km

2

. At the gradual distribution of the 100 magnets over this

square, the distance between the neighboring magnets will be equal to 100 m.

8. The thermal

loading on the

cone

located

at the head of the permanent

magnets

At this longitudinal intersection of the atmosphere the permanent magnets

lose a significant part of their kinetic energy. Indeed, the initial energy of each

magnet having been released into the atmosphere is equal to the following:

E

kin in

= mV

20

/2 = 7*10

-4

*36*10

10

/(2*10

7

)=12.6 J.

After passing through the atmosphere, this energy is reduced till the following

value:

E

kin fin

= m(V

2x

+V

2y

)/2 = 8.4 J,

so that the energy losses of each magnet passing through the atmosphere are:

∆E

kin

= 4.2 J.

Let the cone of the permanent magnet head be made of graphite.

The length of this cone can be found from the relationship:

l

cone

= d/θ

cone

= 3 mm. The volume of the cone is V

cone

= (1/3) d

2

l

cone

= 10

-5

cm

3

.

The density of graphite is ρ

C

= 2.4 g / cm

3

, so that the mass of the cone is equal

to m

cone

= 2.4 * 10

-5

g. The thermal capacity of graphite is equal to [10], p. 201

C

C

= 200 J / g * degree, graphite remains solid till the temperature T = 4200 C

0

,

13

after it is sublimated. Multiplying the thermal capacity by the temperature of

4200C degrees, we find that the graphite cone is resistant till energy release in it

equal to 20 J.

The losses of kinetic energy of the permanent magnet are distributed as

follows: only 1-2% is allocated in the cone, the remained energy is transferred

to the molecules of the residual gas. Considering that 2% of lost energy will be

allocated in the cone, the energy release in it will be 0.1 J that is by 2 orders of

magnitude smaller than the energy of 20 J, which is necessary to break the

cone.

9. The optical observation of the flight trajectory of the permanent magnets.

We assume that the optical observation of the flight trajectories of the

permanent magnets will be performed from the distance of H

w

= 300 km. If we

have assumed that the permanent magnets occupy an area with the dimensions

of 1 * 1 km, then this area is visible at the angle of θ

w

= 1/300 = 3 * 10

-3

.

We provide each permanent magnet with several angle reflectors. It is clear

that the dimensions of the reflectors can not be greater than 100 µ. At the area

of the angle reflector equal to100 * 100 µ

2

, the ratio of the reflection area to the

area of observation is of the value equal to 1 km

2

is 10

-4

cm

2

/ 10

10

cm

2

= 10

-14

.

Obviously, the angle of the reflector light is equal to the angle θ

w

= 3 * 10

-3

of

the incident radiation. It means that in the radiation area the reflected radiation

will cover the same area - 1 km

2

. At the diameter of the telescope receiving

mirror of the order of 3 m its square will be equal to 10 m

2

and the relation of

the mirror to the area of the reflected radiation in this case is equal

to 10 m

2

/ 10

6

m

2

= 10

-5

. The total losses of the radiation intensity with respect

to the radiated power are, thus, of the order of 10

-19

.

Let the flight trajectories of the permanent magnets be irradiated with the

laser wavelength equal to λ

l

= 0.4 µ. This means that the energy of each

quantum of light emitted by the laser is equal to 3 eV. Since

1 J = 6.24 * 10

18

eV, the radiation of 100 J of energy per pulse will be equal to

the radiation of 2* 10

20

light quanta. From these emitted quanta

of light: 2* 10

20

, taking into account the losses of intensity equal to10

-19

, about

20 quanta will arrive to the photo detector array.

If this photo detector array is composed of micro pixel avalanche

14

photodiodes, whose description can be found in [11], it should be stressed that

the quantum efficiency of these photodiodes is of the order of 1. These

photodiodes operate at the room temperature, they have the single photoelectron

resolution, i.e., they can count individual quanta with efficiency close to 100%.

This means that the flight trajectories of each of the 100 accelerated magnets

will be detected reliably.

If the laser emits one pulse per second – it will work with the frequency

of 1 Hz, then its average power is equal to 100 W.

10. The penetration depth of the permanent magnets in aluminum

The depth of penetration of the permanent magnets, moving at the velocity of

V

0

= (3.5

2

+ 3.4

2

)

1/2

= 4.9 km / s, in the aluminum it can be estimated from the

following considerations. Since these processes are fast, it is necessary to take

into account the interaction only with those atoms which are in the trajectory of

the permanent magnet.

We find the energy that will be needed to heat up one centimeter of aluminum

from the room temperature till the melting point. The cross section of this

aluminum is equal to the cross section of the permanent magnet,

i.e. S

tr

= 10

-4

cm

2

. This energy is necessary to be added with the energy of the

phase transition of the solid body to the liquid. The liquid must be heated to the

temperature of evaporation and take into account the energy of the phase

transition of liquid to vapor.

The total energy expenditures are equal to the following: ∆ E

1 cm

= 3.7 J / cm.

The kinetic energy of the permanent magnet having the velocity

V

0

= 4.9 km / s, is equal to: E

kin fin

= 8.4 J. From the analysis of the collision

processes of the meteorites with solids it can be concluded that at such

velocities about 40% of the meteorite kinetic energy is used for heating the

meteorite and evaporation of the target substance. We assume that from the

total kinetic energy which is equal to E

kin fin

= 8.4 J, the permanent magnet

spends E

kin fin1

= 3.3 J to evaporate aluminum.

Now, dividing E

kin fin1

by the value of specific energy losses ∆E

1 cm

, we find

the depth of penetration of the permanent magnet into aluminum:

L

Al

= E

kin fin1

/ ∆E

1 cm

≈ 1 cm.

15

11. Conclusion

When we develop the permanent magnets with residual induction greater

than, B

r

> 13.5 kGs it will be possible to achieve a more specific magnetic

moment per nucleon than in the NdFeB magnets:

m> m

NdFeB

= 1.57 * 10

-10

eV / Gs. If you can get a great coercive force of the

magnets than H

c

= 30 kGs, it will be possible to increase the magnetic field

gradient accelerating the magnetic dipoles by more than 20 kGs/cm.

References

1. http://en.wikipedia.org/wiki/Coilgun

2. http://cas.web.cern.ch/cas/brunnen/presentations/pdf/leduff.pdf

3. S. N. Dolya, Acceleration of magnetic dipoles by the sequence of current

turns, http://arxiv.org/ftp/arxiv/papers/1401/1401.4518.pdf

http://link.springer.com/article/10.1134%2FS1063784214110085#page-1

4. Table of Physical quantities, ed. by I. K. Kikoin, Moscow, Atomizdat,

1976

5. S. N. Dolya, Electromagnetic way of accelerating the magnetic dipoles,

http://arxiv.org/ftp/arxiv/papers/1312/1312.5046.pdf

6. S. N. Dolya, Artificial Micrometeorites,

http://arxiv.org/ftp/arxiv/papers/1407/1407.8542.pdf

7. S. N. Dolya, Electromagnetic acceleration of electrically charged bodies,

http://arxiv.org/ftp/arxiv/papers/1402/1402.3035.pdf

8. S. N. Dolya,

Electrodynamics acceleration of electrical dipoles,

http://arxiv.org/ftp/arxiv/papers/1312/1312.7393.pdf

9. S. N. Dolya, About the electrodynamics acceleration of

cylinder-shaped particles,

http://arxiv.org/ftp/arxiv/papers/1311/1311.5315.pdf

10. Tables of physical quantities, ed. I. S. Grigor’iev and

E. Z. Mei’likhov, Moscow, Energoatomizdat, 1991

16

11. S. N. Dolya, Registration of the radiation reflected from the water

surface at a wavelength λ = 375 nm,

https://www.researchgate.net/publication/275337938_Registration_of_the_radi

ation_reflected_from_the_water_surface_at_a_wavelength___375_nm