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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.

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The ** are the sections I am not sure on