Show that the divergence of the curl of a vector is zero.

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written 7.1 years ago by

vermavarsha432 • 1.0k

For any field $\vec{B}$ the divergence of a curl of $\vec{B}$ is written as

$\begin{align} \vec{\nabla}.\left(\vec{\nabla}.\vec{B} \right)&=\left( \hat{i_x}\frac{\partial}{\partial{x}}+\hat{i_y}\frac{\partial}{\partial{y}}+\hat{i_z}\frac{\partial}{\partial{z}}\right)\times \left[ \hat{i_x}\left( \frac{\partial{B_z}}{\partial{y}} -\frac{\partial{B_y}}{\partial{z}} \right)+\hat{i_y}\left( \frac{\partial{B_x}}{\partial{z}} -\frac{\partial{B_z}}{\partial{x}} \right)+\hat{i_z}\left( \frac{\partial{B_x}}{\partial{y}} -\frac{\partial{B_y}}{\partial{x}} \right)\right] \&=\frac{\partial}{\partial{x}}\left( \frac{\partial{B_z}}{\partial{y}} -\frac{\partial{B_y}}{\partial{z}} \right)+\frac{\partial}{\partial{y}}\left( \frac{\partial{B_x}}{\partial{z}} -\frac{\partial{B_z}}{\partial{x}} \right)+\frac{\partial}{\partial{z}}\left( \frac{\partial{B_x}}{\partial{y}} -\frac{\partial{B_y}}{\partial{x}} \right) \\vec{\nabla}.\left(\vec{\nabla}.\vec{B} \right)&=0 \end{align}$

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