Apply a field and move the wall over a grain boundary. Consider the coercive force.

<p>
</p>

Consider the Hamiltonian.

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</p>

What is D? Why is dH set to zero?

<center>

<br>

<math>H = \int_{-D}^

Unknown macro: {D}

\left [A_i \left ( \frac

Unknown macro: {d theta}
Unknown macro: {dx}

\right )^2 + K_i \sin^2 \theta - HM \cos \theta \right] dx</math>

<br>

<math> \delta H = 0 </math>

<br>

<math>\delta H = -2A_i \frac

Unknown macro: {d^2 theta}
Unknown macro: {d x^2}

+ K_i \sin 2 \theta + HM \sin \theta</math>

<br>

</center>

Three different expressions result from the last equation above. Multiply by the term below.

<p>
</p>

How does multiplying by this term produce equations below?

<center>

<br>

<math>\int \frac

Unknown macro: {d x}

dx</math>

<br>

<math>-A_i \left ( \frac

Unknown macro: {d theta_i}
Unknown macro: {dx}

\right )^2 + K_i \sin^2 \theta - HM \cos \theta = D_i</math>

<br>

<math>-A_1 \left ( \frac

Unknown macro: {d theta_1}

\right )^2 + K_1 \sin^2 \theta - HM \cos \theta = D_1</math>

<br>

<math>-A_2 \left ( \frac

Unknown macro: {d theta_2}
Unknown macro: {dx}

\right )^2 + K_2 \sin^2 \theta - HM \cos \theta = D_2</math>

<br>

<math>-A_3 \left ( \frac

Unknown macro: {d theta_3}

\right )^2 + K_3 \sin^2 \theta - HM \cos \theta = D_3</math>

<br>

</center>

Apply the boundary conditions to find values of <math>D_1</math> and <math>D_2</math>.

<center>

<br>

<math>D_1 = -HM</math>

<br>

<math>D_3 = HM</math>

<br>

</center>

Determine the boundary conditions at the grain boundary.

<center>

<br>

<math>A_1 \frac

Unknown macro: {d theta}

|_

Unknown macro: {x=x_1}

= A_2 \frac

Unknown macro: {d theta}
Unknown macro: {d x}

|_

Unknown macro: {x=x_2}

</math>

<br>

<math> A_1 \left ( \frac

Unknown macro: {d x}

\right )^2 + A_1 k_1 \sin^2 \theta_1 - kMA_1 \cos \theta_1 = AD_1</math>

<br>

<math> A_2 \left ( \frac

Unknown macro: {d theta}

\right )^2 + A_2 k_2 \sin^2 \theta_1 - kMA_1 \cos \theta_1 = AD_1</math>

<br>

</center>

There is an expression of <math>D_2</math> in terms of <math>\theta</math> at the interface.

<center>

<br>

<math>\left ( A_1 k_1 - A_2 k_2 \right ) \sin^2 \theta_1 - HM \left ( A_1 - A_2 \right ) \cos \theta_1 = A_1 D_1 - A_2 D_2</math>

<br>

<math>\left ( A_3 k_3 - A_2 k_2 \right ) \sin^2 \theta_1 - HM \left ( A_3 - A_2 \right ) \cos \theta_1 = A_3 D_3 - A_2 D_2</math>

<br>

</center>

There are pairs of <math>\theta_1</math> and <math>\theta_2</math> corresponding to each value of <math>D_2</math>.

<center>

<br>

<math>-A_2 \left ( \frac

Unknown macro: {d theta}
Unknown macro: {dx}

\right )^2 +K_2 \sin^2 \theta - HM \cos \theta = D_2</math>

<br>

<math>-A \left ( \frac

Unknown macro: {dx}

\right )^2 = D_2 + K_2 \sin^2 \theta - HM \cos \theta</math>

<br>

</center>

Use the computer to solve integrals.

<p>
</p>

What is the meaning of <math>\theta_2</math> and <math>\theta_2</math>?

<p>
</p>

Why is <math>\partial H / \partial \theta</math> set equal to zero?

<p>
</p>

What is the connection between the integral and <math>H</math>?

<p>
</p>

How is <math>H</math> used to find an expression of the coercive force?

<center>

<br>

<math>W = \sqrt { \frac

Unknown macro: {A}
Unknown macro: {K}

} \int_

Unknown macro: {theta_1}

^

Unknown macro: {theta_2}

\frac

Unknown macro: {d theta}
Unknown macro: {sin^2 theta - h cos theta + b sin^2 theta - a h cos theta_1 + h(a+1)}

</math>

<br>

<math>H = \frac

Unknown macro: {1}
Unknown macro: {2}

\frac

Unknown macro: {K_1}
Unknown macro: {S}

\left (\frac

Unknown macro: {A_1}
Unknown macro: {A_2}

- \frac

Unknown macro: {K_2}

\right ) \left ( \sin \theta^2 \cos \theta \right )</math>

<br>

<math>\frac

Unknown macro: { partial H }
Unknown macro: { partial theta }

= 0</math>

<br>

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