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# A rectangular core has fixed permeability $\mu_{r}>>1,$ a square cross section of dimensions $a \times a,$ and has centerline dimensions around its perimeter of b and d. Coils 1 and 2, having turn numbers $\(N_{1}\) and \(N_{2}\),$ are wound on the core. Consider a selected core cross-sectional plane as lying within the xy plane, such that the surface is defined by 0 < x < a, 0 < y < a. (a) With current $I_1$ in coil 1, use Ampere’s circuital law to find the magnetic flux density as a function of position over the core cross-section. (b) Integrate your result of part (a) to determine the total magnetic flux within the core. (c) Find the self-inductance of coil 1. (d) Find the mutual inductance between coils 1 and 2.

Solution

Verifieda)

Since $\mu_r >> 1$ the magnetic scattering is negligible.

Apply the Ampere's circuit law for the red conture C shown in figure below.

$\begin{align*} \oint_C H\cdot dL &=N_1I_1 \\ \\ \text{Use that } \ B&=\mu H =\mu_r \mu_0 H \\ \\ \oint_C \frac{B}{\mu_r \mu_0} \cdot dL &=N_1I_1 \end{align*}$

$B$ is a constant along the contour of $C$ and $\mu_r$ and $\mu_0$ are also constants so

$\begin{equation*} \frac{B}{\mu_r \mu_0} \oint_C dL =N_1I_1 \\ \end{equation*}$

$\oint_C dL$ is a contour $C$ circumference.

$\begin{align*} \oint_C dL&=2\cdot (b-\frac{a}{2}+x)+2\cdot (d-\frac{a}{2}+x) \\ \oint_C dL &=2x+2b+2d-a \\ N_1I_1&=\frac{B}{\mu_r \mu_0}\cdot (2x+2b+2d-a) \\ B &=\boxed{\frac{N_1I_1}{\mu_r \mu_0 (2x+2b+2d-a)}} \end{align*}$

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