Chapter 7: The size of a driving force of the electromagnetic engine and electric power consumption

Since it is “F=BIL”, to increase the size of a driving force by one electromagnetic engine,

there are three ways.

1.      To make the strength of the magnetic field of the superconductive magnet stronger.

2.      To make the strength of the ripple current stronger.

3.      To make the size of the ring of the superconductive magnet and the normal-conductive magnet bigger.

Of course, the driving force as a whole can be increased by combining and using several electromagnetic engines. Among these, "1. To make the strength of the magnetic field of the superconductive magnet stronger." can strengthen the electromagnetic force without raising electric power consumption but there is a limit in it. Also, using " 3. To make the size of the ring of the superconductive magnet and the normal-conductive magnet bigger." there is a danger that the weight and the size become bigger too much. I think that "2. To make the strength of the ripple current stronger. " is most effective. Then, the size of the driving force can be controlled by controlling the strength of the ripple current. In " Research on Momentum Order ", it is using modest numerical values but the driving force can be made very strong, too.

The strength of the ripple current can be strengthened to more than tens of thousands of amperes.

But, the electric power consumption becomes big. Therefore, I thought of making an electric resistance smaller by making the cross section area of the normal-conductive magnet through which the ripple current flows bigger. However, this time, by making the cross section area bigger, the weight of the equipment becomes heavier. However, I think that the increase of the strength of the electromagnetic force by making an electric current stronger can exceed the increase of the weight of the equipment sufficiently. But, electric power consumption with some degree will be necessary.

The specific example computation

☆ About the normal-conductive magnet

The magnetic field which the superconductive magnet gives is 5 Teslas.

It uses iron for the normal-conductive body as the normal-conductive magnet.

The length of the round of the normal-conductive magnet is 1.6 meters.

The cross section area of the normal-conductive body is 0.02 square meters.

The maximum strength of the ripple current which flows through the normal-conductive body is 48,000 amperes.

The electric resistivity of iron ( steel ) is 20×10⁻⁸ ohms・meters .

The density of iron is 7.87× 10³ kilograms per meter³.

Presupposing above figures, computations will be done.

An effective value will be calculated to use for the calculation of electric power consumption of the ripple current.

The effective value of an alternate current is calculated by multiply the maximum value of the alternate current by 1/√2 . The ripple current flows intermittently but since the alternate current is necessary before rectifying, the effective value is used as it is.

48,000 × 1/√2≒33950 amperes

An average value is used for the calculation of the strength of the electromagnetic force by the ripple current. It is because the average value of the electromagnetic force becomes an effective value of a driving force.

The average value of an alternate current is calculated by the maximum value of the alternate current by 2/π. Moreover, since the ripple current flows intermittently, it is divided by 2.

48,000 × 2/π÷2≒15280 amperes

Since an electric resistance of a normal-conductive body is calculated by " electric resistivity × length ÷ cross section area ", the electric resistance of this normal-conductive body is,

20×10⁻⁸×1.6÷0.02＝1.6×10⁻⁵  ohms

Since electric power consumption is calculated by " electric current ² × electric resistance ", using the effective value of the electric current, the electric power consumption of the normal-conductive body in this case is,

（3.395×10⁴）²×1.6×10⁻⁵≒18442 watts

By the way, since voltage is calculated by " electric current × resistance",

the voltage is,

3.395×10⁴×1.6×10⁻⁵＝0.5432 volt

Since the mass of material is calculated by " density × cross section area × length "

The weight of the normal-conductive body is,

7.87×10³×0.02×1.6＝251.84 kilograms

☆ About the copper cables to use for the wiring

The length of the copper cables is 1 meter.

It becomes this value in bringing a ripple current power supply or a transformer near the normal-conductive body.

The cross section area of the copper cables is 0.012 square meters.

The electric resistivity of copper is 2× 10⁻⁸ ohms・meters.

The density of copper is 8.96 × 10³ kilograms per meters³.

Presupposing above figures, computations will be done.

The electric resistance of the copper cables is,

2×10⁻⁸×1÷0.012＝1/6×10⁻⁵ ohms

Using the effective value of the electric current, the electric power consumption of the copper cables in this case is,

（3.395×10⁴）²×1/6×10⁻⁵≒1921 watts

By the way, voltage is calculated by " electric current × resistance ",

3.395×10⁴×1/6×10⁻⁵≒0.056584 volt

The weight of the copper cables is,

8.96×10³×0.012×1＝107.52 kilograms

The total electric power consumption of the normal-conductive body and the copper cables is,

18442＋1921＝20363 watts

This 20.4 kilowatts multiplied by time becomes necessary electric energy when making the electric current strongest. A power generator which can supply this electric energy is needed.

The total weight of the normal-conductive body and the copper cables is,

251.84＋107.52＝359.36 kilograms

☆ The force which the ripple current receives will be calculated.

It is calculated according to the formula of F=BIL.

Since seeking the strength of an effective force, the average value is used as the strength of the ripple current.

5×1.528×10⁴×1.6＝122240 Newtons

122240÷9.8≒12473.4 kilograms

That is, it is a force of about 12.47 tons.

12470−359.36＝12110.64 kilograms、

That is a leeway of about 12.1 tons.