# Force and energy of the magnet

1. Jan 8, 2016

### cdummie

1. The problem statement, all variables and given/known data
If we have a system that is shown in the picture, the upper piece is magnet, and if we know the following data:
S = 1cm2 which is area of material, S0 = 1.2cm2 which is the area of fissure. To bring these tow pieces of this system magnetic induction in the fissure should be B0= 1.6 T. Dimensions of this magnetic circuit are: l1 = 15cm, l2 = 10cm, l0 = 1mm. Determine the force needed to bring together these two parts, and magnetic energy "wasted" to establish magnetic field right before bringing these two parts together.

2. Relevant equations
$w_m = \frac{1}{2}BH \\ F_m = - \frac{dW_m}{dx}$

3. The attempt at a solution
density of magnetic energy is $w_m = \frac{1}{2}BH$ volume of both pieces of material together v=(l1+l2)S density of magnetic energy in fissures $w_{m_0} = \frac{1}{2}B_0H_0$ ,
volume of fissures $v_0 = 2xS$ (x instead of l0 because it changes when there two parts are getting closer, and force has only x component, which is the vertical one in this case). when the lower part in moving volume of fissure, magnetic energy of fissure and magnetic energy of part with material change, only the volume of part with material remains the same. But, if we say that flux is constant then, the only thing that affects the change of energy is the volume of the fissure.

$dW_m = B_0H_0Sdx \\ F_m = - \frac{dW_m}{dx} \\ F_m = 2SB_0H_0$

Since i have B0 i can easily determine H0 which would mean that i solved this without using the magnetization characteristics diagram. Am i doing something wrong?

Picture of magnet

Magnetic characteristics of material diagram.

2. Jan 8, 2016

### Hesch

Yes, you can do that. But how come, that S ≠ S0 ?

Contrary, you cannot solve this part of the problem, without using the characteristics:

3. Jan 8, 2016

### cdummie

It's just a mistake in the text of the example, but i wanted it to remain the as it is, maybe it makes no sense, but it won't affect the procedure of solving.
Why i can't? I mean the only thing that affects the change of energy is change of volume of fissure. Which means that i don't need H or B, i need only H0 and B0 along with S and dx. I assume i made mistake somewhere, i'd like to know where.

4. Jan 8, 2016

### Hesch

No, when you change the energy in the fissure, you must change the energy in the core as well. Bcore ≈ Bfissure.

μcore is not constant, but = dB/dH, which can be determined from the magnetic characteristic.

Some magnetic energy is also "wasted" in the core.

5. Jan 8, 2016

### Hesch

There is no change of volume right before bringing these two parts together. The only wasted energy is due to magnetizing the airgap and the core.

6. Jan 9, 2016

### cdummie

What is dWm equal to then? is it: $dW_m = \frac{1}{2}BHS(l_1 + l_2) + B_0H_0Sdx$ ?

7. Jan 9, 2016

### Hesch

The magnetic energy density in the core

emag = ½BH = ( the area under the characteristic from H=0 to H=2000 ), because μ = dB/dH is varying in the interval.
Had μ been constant, the area would have been ½BH ( area of a triangle ).

Hope you can see it, I'm unable to explain it in english.

8. Jan 9, 2016

### cdummie

I don't quite understand this, does it means that i can use 1\2BH or not? What happens to the Wm(magnetic energy)? About that triangle you mentioned, i don't see any triangles when H is from 0 to 2000, i can see it when H goes from 0 to 500. But, how can i determine the value of H for which i'm finding density of energy. I mean if i use Amphere's law i have H(l1+l2) + Bl00 = NI but i don't have N and i don't have I either, what i am supposed to do?

9. Jan 9, 2016

### Hesch

See the attached.

As for the airgap, you calculate emag as the /////-area in the upper figure. emag = ½*B0*H0 [J/m3] (area of triangle)

As for the core, you calculate emag as the /////-area in the lower figure ( not a triangle ).

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10. Jan 9, 2016

### cdummie

I think i understand that part now. After this i actually need to calculate the energy in the core using the diagram, i will try that it should work. Thanks.