# Magnetic Circuit Problem Involving a Solenoid and Plunger

1. Jan 21, 2010

### Bizkit

1. The problem statement, all variables and given/known data
The problem can be found http://whites.sdsmt.edu/classes/ee382/homework/382Homework1.pdf" [Broken] (the last one), along with a picture of the circuit.

2. Relevant equations
R = l/(µS)

mmf = NI = ΨR

B = Ψ/S

3. The attempt at a solution
I've never done a problem like this before. I've looked for information and example problems to help me do it in my book and online, but I haven't found anything that can help me. The part about it that really confuses me is the use of the plunger. Does it move? If it does, how do I know what the length of the gap is? If it doesn't, do I just ignore the part of the plunger that is below the rest of the circuit because it acts like an open circuit (I don't know if that's true or not)? I was hoping someone here could help me figure out what to do. Thanks.

Last edited by a moderator: May 4, 2017
2. Jan 22, 2010

### Maxim Zh

The basics of the magnetic-electric analogy are here:
http://en.wikipedia.org/wiki/Magnetic_circuit" [Broken]

The MMF is

$$\varepsilon = IN$$

The magnetic resistances are

$$R_1 = \frac{h - l_p - l_g}{\mu_c w^2};$$

$$R_2 = \frac{2h - l_s}{\mu_c w^2};$$

$$R_s = \frac{l_s}{\mu_0 w^2};$$

$$R_p = \frac{l_p}{\mu_p w^2};$$

$$R_g = \frac{l_g}{\mu_0 w^2}.$$

The total magnetic resistance is calculated like the corresponding electrical resistance:

$$R = R_1 + \frac{R_2 + R_s}{2} + R_p + R_g.$$

Then the magnetic flux is

$$\Phi = \frac{\varepsilon}{R}.$$

$$l_p$$ is the length of the upper part of the plunger. So there is no problem.

yes

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Last edited by a moderator: May 4, 2017
3. Jan 22, 2010

### Bizkit

Thanks for the reply. I'm pretty sure I understand what to do now.