Force/ enegry equations for maglev (electromangetism)

[richard cash]

i am having a few problems with a project I am doing for university. The basics of which are force/energy/field size needed to lift magnets.

Basically, i need to show some calculations for the repulsion/attraction forces etc and my knowledge on electromangetism is limited.

For the first system of electromangetic attraction (EMS maglev suspension) we have a mass(train) being suspended above a track and the vehicle curls round the track. Under the track the electromagnet on the train is attracted to a permanent magnet. I have tried to get some equations relating the field strength, current, turns in the wire etc but just can't seem to get any equations right.

also, there is a repulsive system (EDS) whereby an electromagnet (usually a superconductor) is above a track on a train and induces currents in the track that repel the train. I can't figure out and sort of equation for the field strength, energy used or force required to left the train here either.

Could someone please help me out.

 

[Crosson]

Hmm, if you assume the bottom of the train is flat, and that the track is flat, you may as well use this formula.

The pressure on the bottom of the train will be:

P = [tex] \frac{B ^2}{2\mu_0} [/tex]

Assuming equal currents in the train and track. The expression above, is true for pressure, and for energy density (joules per cubic meter) so if you multiply the pressure by the total volume in which B is significant, you get total energy.

To find the current required to produce this B, use:

[tex]B = \frac{\mu_0 i}{A} [/tex]

where i is current and A is cross sectional area.

 

[Graeme Robinson]

Hello there,

I can understand your difficulties with the project you are working on. Electromagnetism can be a complex topic, but with some basic equations and principles, we can solve the problems you are facing.

Firstly, let's start with the EMS maglev suspension system. As you mentioned, the train is being suspended above the track and the vehicle curls around the track. This system works on the principle of magnetic attraction between the permanent magnet on the track and the electromagnet on the train. The force of attraction between two magnets is given by the equation:

F = (μ0/4π) * (m1 * m2)/d^2

Where:
F is the force of attraction
μ0 is the permeability of free space (4π * 10^-7)
m1 and m2 are the magnetic moments of the two magnets
d is the distance between the two magnets

In this case, the force of attraction is counteracted by the weight of the train, which is given by the equation:

Fg = m * g

Where:
Fg is the weight of the train
m is the mass of the train
g is the acceleration due to gravity (9.8 m/s^2)

Now, for the EDS system, the train is being lifted by the repulsive force between the superconductor on the train and the induced currents in the track. The strength of the magnetic field produced by the superconductor is given by the equation:

B = μ0 * I/2πr

Where:
B is the magnetic field strength
μ0 is the permeability of free space
I is the current passing through the superconductor
r is the distance from the superconductor

The induced current in the track is given by Faraday's law of electromagnetic induction:

ε = -N * dΦ/dt

Where:
ε is the induced electromotive force (emf)
N is the number of turns in the track
Φ is the magnetic flux through the track
t is time

The repulsive force can be calculated using the Lorentz force equation:

F = q * v * B

Where:
F is the repulsive force
q is the charge of the train (assumed to be positive)
v is the velocity of the train
B is the magnetic field strength

To calculate the energy used, we can use the equation:

E = P * t