Inductance of a Solenoid with a Core inside

In summary, The energy of the field in the part with the core is ##\frac{1}{2}\frac{B^2}{\mu}## where ##\mu## is the magnetic permittivity of that region of space.
  • #1
titansarus
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Homework Statement
We have an Ideal Solenoid with current ##I## and turn per length ##n## and a core with magnetic permittivity of ##\mu## is inside it like the figure. the length ##x_0## of solenoid has the core and the rest is empty. Find the Inductance of a Solenoid.
Relevant Equations
##\int B ds = \mu I##, ##L_T = L_1 + L_2##

The question said the ##\mu## in question is the ##\mu## in the above equation so no need to worry about scale factor.
For finding magnetic field ##B##, We see this question like two Solenoids. for the first one, we have ##\int B ds = \mu I## so ##B x_0 = \mu I n x_0 ## so ##B = \mu n I##. For the second one we have ##B = \mu_0 n I##. For the Inductance we have ##L = \mu l n^2 A## so we have ##L_1 = \mu x_0 n^2 A## and for the second one we have ##L_2 = \mu_0 (L-x_0) n^2 A##. And the ##L_{total} = L_1 + L_2##

Is my reasoning right? I think I didn't considered mutual Inductance but the question is from an exam and it was orally said you don't need to consider mutual inductance.

inductance.png
 
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  • #2
Ignoring mutual inductance makes no sense. The inductance for any solenoid is approximately calculated assuming full coupling between each winding. That's where all those n^2 and N^2 terms come from. Why should it be different for the same solenoid except part of it has a slug inside?

I would approach this from an energy viewpoint. Very straightforward that way.
 
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  • #3
rude man said:
Ignoring mutual inductance makes no sense. The inductance for any solenoid is approximately calculated assuming full coupling between each winding. That's where all those n^2 and N^2 terms come from. Why should it be different for the same solenoid except part of it has a slug inside?

I would approach this from an energy viewpoint. Very straightforward that way.
I don't know how to solve this with energy? How is that? I just know energy is ##\frac{1}{2} L i^2##. But I don't know how to use this the calculate inductance in this question.
 
  • #4
titansarus said:
I don't know how to solve this with energy? How is that? I just know energy is ##\frac{1}{2} L i^2##. But I don't know how to use this the calculate inductance in this question.
You're off to a good start. Now, what are the B fields in both parts of the solenoid, and then the energies of the B fields in both parts of the solenoid?

BTW were you given the cross-sectional area of the solenoid? You need that parameter.
 
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  • #5
rude man said:
You're off to a good start. Now, what are the B fields in both parts of the solenoid, and then the energies of the B fields in both parts of the solenoid?

BTW were you given the cross-sectional area of the solenoid? You need that parameter.

I will get ##B## as ##B_{with~ core} = \mu n I ## and ##B_{without~core} = \mu_0 n I##, Right? And the energy of ##B## field is ##\frac{1}{2}\frac{B^2}{\mu}## where ##\mu## is the magnetic permittivity of that region of space. is this Right? Should I say that sum of these two energies of ##B## fields equals ##\frac{1}{2} L i^2## where ##L## is the equivalent induction?

For the cross-section area, The area of Core and Solenoid are both equal to ##A##. (The core completely fits into the solenoid)
 
  • #6
Is 1/2 BH = 1/2 B^2/μ energy or energy density?
BTW you can help me too. How did you manage to write B^2 the right way, i.e. B superscript 2?
 
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  • #7
rude man said:
Is 1/2 BH = 1/2 B^2/μ energy or energy density?
BTW you can help me too. How did you manage to write B^2 the right way, i.e. B superscript 2?
I think i was wrong. ##\frac{1}{2} \frac{B^2}{ \mu}## is energy density. I think I must multiply it by ##A * x## for the part that has the core and by ##A * (L-x)## for the part that hasn't. Beside this, Is my way correct.

For the subscript and superscript and actually any formula, you must write the formula between two #. Like this picture:

##\int_0^\infty \mu_1^2 \frac{1}{2} dx##
latex2.png
 
  • #8
titansarus said:
I think i was wrong. ##\frac{1}{2} \frac{B^2}{ \mu}## is energy density. I think I must multiply it by ##A * x## for the part that has the core and by ##A * (L-x)## for the part that hasn't. Beside this, Is my way correct.

For the subscript and superscript and actually any formula, you must write the formula between two #. Like this picture:

##\int_0^\infty \mu_1^2 \frac{1}{2} dx##View attachment 245286
Right! You're there!

I'm familiar (sort of) with LaTex so OK now I know. Before the latest console revision you could write subscripts and superscripts without resorting to LaTex. Thx.
 
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1. What is inductance?

Inductance is a property of an electrical circuit that describes its ability to generate an electromotive force (EMF) in response to a changing current. It is measured in units of Henry (H).

2. What is a solenoid with a core inside?

A solenoid is a coil of wire that is used to generate a magnetic field when an electric current is passed through it. A solenoid with a core inside refers to a solenoid that contains a ferromagnetic material, such as iron, inside the coil to increase the strength of the magnetic field.

3. How does a core inside a solenoid affect its inductance?

The presence of a core inside a solenoid increases its inductance by increasing the magnetic field strength within the coil. This is because the core becomes magnetized in the presence of the magnetic field, which in turn creates a stronger magnetic field within the coil.

4. How is the inductance of a solenoid with a core inside calculated?

The inductance of a solenoid with a core inside can be calculated using the formula L = μ₀μᵣN²A/l, where μ₀ is the permeability of free space, μᵣ is the relative permeability of the core material, N is the number of turns in the coil, A is the cross-sectional area of the coil, and l is the length of the coil.

5. What are some real-world applications of a solenoid with a core inside?

Solenoids with cores inside are commonly used in various electrical and electronic devices, such as relays, transformers, and motors. They are also used in industrial equipment for tasks such as controlling valves and switches, and in medical equipment for tasks such as generating magnetic fields for MRI machines.

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