Правка разделов «Statement», «Answer»
en/9.4.3.md
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| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $9.4.3.$ | ||
| + | $9.4.3.$ Show that the magnetic flux generated by a plane with linear current density $i$ through any closed surface is zero. | ||
| ### Solution | |||
| We know that a plane with linear current density $i$ creates a uniform field around itself, its value $$B=\frac{\mu_0 i}{2}$$ | |||
| , but we don't really need it. | |||
| Consider an arbitrary closed surface. The stream is by definition equal to: | |||
| $$\oint_S\hat{B}\cdot d\hat{S}$$ | |||
| The definition of vector divergence will be useful to us.: | |||
| $$div \hat{a} = {lim}_{\Delta V\to0} \frac{1}{\Delta V}\oint_S \hat{a}\cdot d\hat{S}=\frac{\partial a}{\partial x}+\frac{\partial a}{\partial y} + \frac{\partial a}{\partial z}$$ | |||
| Note for now that the second definition and uniformity of the field implies $$div\hat{B}=0$$ | |||
| in the whole space | |||
| Now let's break down the volume that borders our surface into many infinitely small volumes $dV$. From the first definition of divergence, an alternative expression for the flow can be obtained: | |||
| $$\Phi = \int_V div \hat{B} dV$$ | |||
| Because the divergence is zero at any point in space.: $\to$ $$\Phi=0$$ | |||
| @@ -23,4 +23,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | Q.E.D. | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $9.4.3.$ |
$9.4.3.$ Show that the magnetic flux generated by a plane with linear current density $i$ through any closed surface is zero. | ||
| ### Solution | ### Solution | ||
| We know that a plane with linear current density $i$ creates a uniform field around itself, its value $$B=\frac{\mu_0 i}{2}$$ | We know that a plane with linear current density $i$ creates a uniform field around itself, its value $$B=\frac{\mu_0 i}{2}$$ | ||
| , but we don't really need it. | , but we don't really need it. | ||
| Consider an arbitrary closed surface. The stream is by definition equal to: | Consider an arbitrary closed surface. The stream is by definition equal to: | ||
| $$\oint_S\hat{B}\cdot d\hat{S}$$ | $$\oint_S\hat{B}\cdot d\hat{S}$$ | ||
| The definition of vector divergence will be useful to us.: | The definition of vector divergence will be useful to us.: | ||
| $$div \hat{a} = {lim}_{\Delta V\to0} \frac{1}{\Delta V}\oint_S \hat{a}\cdot d\hat{S}=\frac{\partial a}{\partial x}+\frac{\partial a}{\partial y} + \frac{\partial a}{\partial z}$$ | $$div \hat{a} = {lim}_{\Delta V\to0} \frac{1}{\Delta V}\oint_S \hat{a}\cdot d\hat{S}=\frac{\partial a}{\partial x}+\frac{\partial a}{\partial y} + \frac{\partial a}{\partial z}$$ | ||
| Note for now that the second definition and uniformity of the field implies $$div\hat{B}=0$$ | Note for now that the second definition and uniformity of the field implies $$div\hat{B}=0$$ | ||
| in the whole space | in the whole space | ||
| Now let's break down the volume that borders our surface into many infinitely small volumes $dV$. From the first definition of divergence, an alternative expression for the flow can be obtained: | Now let's break down the volume that borders our surface into many infinitely small volumes $dV$. From the first definition of divergence, an alternative expression for the flow can be obtained: | ||
| $$\Phi = \int_V div \hat{B} dV$$ | $$\Phi = \int_V div \hat{B} dV$$ | ||
| Because the divergence is zero at any point in space.: $\to$ $$\Phi=0$$ | Because the divergence is zero at any point in space.: $\to$ $$\Phi=0$$ | ||
| @@ -23,4 +23,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | Q.E.D. | ||