# Energy of a solenoid with a partially removed core

## Homework Statement

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

Neglecting hysteresis, what is the total magnetic energy of the core. When $f$ is changed by amount $\Delta f$ show the amount of work done by the power supply to keep $I$ 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 $H_1$ and $H_2$ 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 $B$ is $0$, the magnetic field must be continuous. This means the magnetic field is ($H=B/\mu$)
$$B=\frac{nI\mu\mu_0}{\mu_0+f(\mu-\mu_0)}$$

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 $\mu_r>1$ that the above expression is negative and so core gets pushed out.

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The ** are the sections I am not sure on

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