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Energy of a solenoid with a partially removed core

  • #1
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Homework Statement


A solenoid of volume [itex]V[/itex], current [itex]I[/itex] and [itex]n[/itex] turns per unit length has an LIH core, relative permitivity is [itex]\mu_r[/itex]. This core is then slid out so that a fraction [itex]f[/itex] of the solenoid's length is filled with air/vacuum (and [itex]1-f[/itex] is filled with the core).

Neglecting hysteresis, what is the total magnetic energy of the core. When [itex]f[/itex] is changed by amount [itex]\Delta f[/itex] show the amount of work done by the power supply to keep [itex]I[/itex] constant is
$$
\Delta W = n^2I^2V\mu_0(1-\mu_r)\Delta f
$$

If the core is allowed to slide, which way does it move?

Solution
If [itex]H_1[/itex] and [itex]H_2[/itex] are the auxillary fields in the core and air respectively we have, from Amperes law:
$$
H_1(1-f)+H_2f=nI
$$

**Now, as the divergence of [itex]B[/itex] is [itex]0[/itex], the magnetic field must be continuous. This means the magnetic field is ([itex]H=B/\mu[/itex])
$$
B=\frac{nI\mu\mu_0}{\mu_0+f(\mu-\mu_0)}
$$

The total magnetic energy is the sum of the magnetic energies in the 2 parts:

$$
W=Vf \frac{B^2}{2\mu_0}+V(1-f) \frac{B^2}{2\mu}=\frac{Vn^2I^2}{2(\mu_0+f(\mu-\mu_0))}
$$

**Now I am not sure how to get from here to their expression.

**And I assume that as most [itex]\mu_r>1[/itex] that the above expression is negative and so core gets pushed out.

---------
The ** are the sections I am not sure on
 

Answers and Replies

  • #2
Charles Link
Homework Helper
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They seem to be working with the premise that the magnetic field ## B ## with the core is ## B=\mu_o \mu_r n I ##, and ## B=\mu_o n I ## in the part that doesn't have the iron core. In decreasing the fraction that has the core, perhaps it could be looked at as the radius is narrowed rather than the length being decreased. And perhaps a similar equation works when the length is decreased, at least in regards to energy, but I would need to study it further. ## \\ ## One additional item: Since ## L=\frac{\Phi}{I} ## and ## W=(\frac{1}{2})L I^2 ##, I get a (1/2) in the final answer. I'm not sure the ## H ## is nearly as uniform in this problem, as your assumption using a computation involving ampere's law equation assumes. ## \\ ## [Additional note that you might find of interest: This same ampere's law is routinely employed in a transformer problem that involves a thin air gap. See e.g. https://www.physicsforums.com/threads/mean-magnetic-path-length-of-the-magnetic-core-of-a-transformer.930760/#post-5876211 . The difference for that case is the ## H ## from the poles is assumed to cause an ## H ## in the gap that differs from the ## H ## in the material, and the ## H ## from the two opposite but nearby poles will essentially cancel in the material. ## \\ ## (Note: The ## H ## from the poles is computed for uniform ## M ## using ## \sigma_m=M \cdot \hat{n} ## along with the inverse square law. The ## H ## from the poles is added to the ## H ## from the current in the solenoid. The ## H ## from the poles in this solenoid problem with a partial core will cause the ## H ## for this problem to be non-uniform). ## \\ ## Because the poles come as an opposite pair, and because the lines of flux for ## B ## are assumed continuous, the ## H's ## in this air gap problem can be assumed to be uniform over the different regions. That is a different problem than you have here, where the pole faces of the material are individual. ] ## \\ ## Editing: For determining the dynamics of whether the core gets pulled into or gets pushed out of the solenoid when the field is applied, I think another equation is in order. This equation suggests there is a radially outward force/stress on the material because removing some of the material and decreasing the radius will lower the energy. I do think it is necessary to consider the energy term (energy density) of the magnetization in the magnetic field that is of the form ## U=-M \cdot B ##. The sign of this term indicates the state of lower energy is when the magnetization is at its maximum value, and aligned with the magnetic field. To just get the book's answer, the first paragraph above will do that, but I'm not sure the book's result is accurate. See also: https://reviseomatic.org/help/e-components/Solenoids.php ## \\ ## There is one additional thing that came up in a google in studying this, and that is the presence of eddy currents that can arise in the core material as the current in the solenoid is increased. Oftentimes, magnetic cores in solenoids are laminated to minimize these eddy currents that result from the Faraday EMF that occurs as the magnetic field is introduced. These currents go in a circular direction through the conductive material. See also: https://ucscphysicsdemo.wordpress.com/physics-5c6c-demos/lenzs-law-jumping-ring/ Note: This got more complicated than I think was intended. I think the original problem was supposed to be simple, but it's actually a topic that can get somewhat complicated.
 
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  • #3
Charles Link
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For some simpler info regarding the above, the core can be looked at as a material that enhances the magnetic field ## B=\mu_o H ## from the solenoid by a factor of ## \mu_r ##. The result will affect the inductance so that with a core the inductance of the solenoid is ## L= \mu_r \mu_o n^2 V ##. (This can be shown by computing ## B ## for a long solenoid, and then computing the flux ## \Phi ##). ## \mu_r ## can take on a wide range of values, but a ## \mu_r \approx 500 ## or greater is not uncommon. Since ## L=\frac{\Phi}{I} ## and the EMF ## \mathcal{E}=-\frac{d \Phi}{dt} ##, the voltage of the inductor is ## \mathcal{E}=V=-L \frac{dI}{dt} ##. Also power output ## P=IV ##, and energy output ## W=\int P \, dt ##. (Perhaps these simpler results will be found to be useful).
 
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