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

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Consider the Hamiltonian.

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_What is D?_ _Why is dH set to zero?_


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<math>H = \int_{-D}^{D} \left [A_i(x) \left ( \frac{d \theta}{dx} \right )^2 + K_i(x) \sin^2 \theta - HM \cos \theta \right]  dx</math>

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

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<math>\delta H = -2A_i \frac{d^2 \theta}{d x^2} + K_i \sin 2 \theta + HM \sin \theta</math>

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

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_How does multiplying by this term produce equations below?_ 


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<math>\int \frac{d \theta}{d x} dx</math>

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<math>-A_i \left ( \frac{d \theta_i}{dx} \right )^2 + K_i \sin^2 \theta - HM \cos \theta = D_i</math>

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<math>-A_1 \left ( \frac{d \theta_1}{dx} \right )^2 + K_1 \sin^2 \theta - HM \cos \theta = D_1</math>

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<math>-A_2 \left ( \frac{d \theta_2}{dx} \right )^2 + K_2 \sin^2 \theta - HM \cos \theta = D_2</math>

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<math>-A_3 \left ( \frac{d \theta_3}{dx} \right )^2 + K_3 \sin^2 \theta - HM \cos \theta = D_3</math>

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

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<math>D_1 = -HM</math>

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<math>D_3 = HM</math>

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Determine the boundary conditions  at the grain boundary.

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<math>A_1 \frac{d \theta}{d x}|_{x=x_1} = A_2 \frac{d \theta}{d x}|_{x=x_2}</math>

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<math> A_1 \left ( \frac{d \theta}{d x} \right )^2 + A_1 k_1 \sin^2 \theta_1 - kMA_1 \cos \theta_1 = AD_1</math>

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<math> A_2 \left ( \frac{d \theta}{d x} \right )^2 + A_2 k_2 \sin^2 \theta_1 - kMA_1 \cos \theta_1 = AD_1</math>

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

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<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>

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<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>

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

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<math>-A_2 \left ( \frac{d \theta}{dx} \right )^2 +K_2 \sin^2 \theta - HM \cos \theta = D_2</math>

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<math>-A \left ( \frac{d \theta}{dx} \right )^2 = D_2 + K_2 \sin^2 \theta - HM \cos \theta</math>

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Use the computer to solve integrals.

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_What is the meaning of_ <math>\theta_2</math> _and_ <math>\theta_2</math>_?_

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_Why is_ <math>\partial H / \partial \theta</math> _set equal to zero?_

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_What is the connection between the integral and_ <math>H</math>_?_

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_How is_ <math>H</math> _used to find an expression of the coercive force?_

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<math>W = \sqrt { \frac{A}{K} } \int_{\theta_1}^{\theta_2} \frac{d \theta}{\sin^2 \theta - h \cos \theta + b \sin^2 \theta - a h \cos \theta_1 + h(a+1)}</math>

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<math>H = \frac{1}{2} \frac{K_1}{S} \left (\frac{A_1}{A_2} - \frac{K_2}{K_1} \right ) \left ( \sin \theta^2 \cos \theta \right )</math>

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<math>\frac{ \partial H }{ \partial \theta } = 0</math>

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