# Net force between two Electromagnets considering Back EMF

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1. May 23, 2015

### Student149

First this Q might be trivial thus apologies.

Consider two bar electromagnets that can simulate bar magnets M1 and M2, placed end to end at distance D apart (North Pole of M1 facing South Pole of M2). Magnet M1 is 'fixed' to a base. Assuming both electromagnets have the Current I1 and Voltage V1.

The magnets will attract each other and M2 will travel the distance D in some time T1.

How will the the Net effective force of attraction (considering the effects of Back EMF) thus time taken, change with change in:

1. Current
2. Voltage
3. Important: We have to keep in mind the 'Back EMF' generated due to motion, that negates this attractive motion and how it changes with Current and Voltage (the part which I am unsure of).
Q1. Is there a simple formula for the Net Force (and time taken) w.r.t. changes in the above 3.?
Q2. In other words how does changing the parameters, affects the time M2 will take to travel distance D?

2. May 23, 2015

### Hesch

The question is not trivial, it is complicated. A lot of information is needed, such as:

- magnetic induction ( B-field ).
- cross section area of bars.
- length of bars
- mass of M2
- magnetic permeability in bars
- characteristic of power supply ( current source / voltage source ).
- impedance in coils

I think that a numerical calculation by computer is necessary, that will do the job within some minutes.

3. May 23, 2015

### Student149

Thank you for the reply. I do understand that various parameters are involved. But we can assume the ones like:
- cross section area of bars.
- length of bars
- mass of M2
- magnetic permeability in bars
- characteristic of power supply ( current source / voltage source ).
- impedance in coils

as some random feasible values. The only thing of interest is that how the time taken to travel changes with change in Current and/or voltage. That too, to a reasonable approximation. Something of a graph[Current/Voltage vs Time taken to travel distance D] or a formula to generate it. I have no clue of both, thus needed some help.

4. May 23, 2015

### Hesch

B = μ * H = μ0 * μr * H. Say that μr (as for the core) is constant ≈1000. If you multiply the current by a factor k, both B and H will be multiplied by k.

The magnetic energy density, Emagn = ½ * B * H [ J / m3 ], so Emagn will be multiplied by k2.

F = ΔE / ΔV = ΔE / ( A * Δd ), ( E is total magnetic energy, V is the (small) volume of airgap, A is cross section area of airgap, d is distance at any time ), thus also the force will be multiplied by k2.

Now, F = m * a, and d = ½*a*t2 → t = √( 2d / a ), which leads to that time is inverse proportional to k.

Last edited: May 23, 2015
5. May 23, 2015

### Student149

Thank you for such a clear answer. Much grateful.

6. May 25, 2015

### Student149

Another doubt in the Q. I think we missed the effect of change in Back Emf when increasing the Current by a factor of k (and thus on the change in time taken to travel distance D). This was an important part. Any comments?

7. May 26, 2015

### Hesch

If you use a voltage-source as power supply ( constant voltage output ), then: Yes.

But if you use a current-source as power supply, it automatically compensate this back-emf, and the current will be constant no matter that some back-emf arises.

By a "normal" powersupply you can make it a current-source by:

1) Set the current-limit to some level.
2) Set voltage-output to maximum.

The supply will automatically control the output voltage so that the current will be: Iout = Ilimit.