**1 Answer**

written 6.5 years ago by | • modified 5.4 years ago |

The rate of change of a vector field is complex. The divergence of a vector field indicates how much the vector field spreads out from a certain point. The divergence of a vector is scalar.

If $\vec{B}=B_x\hat{i}_x+B_y\hat{i}_y+B_z\hat{i}_z \tag{1}$

Its divergence is written as $$\vec{\nabla}.\vec{B}=\frac{\partial{B_x}}{\partial{x}}+\frac{\partial{B_y}}{\partial{y}}+\frac{\partial{B_z}}{\partial{z}}\tag{2}$$

In cylindrical coordinates, the field is written as$\vec{B}=B_r\hat{r}_x+B_\phi\hat{i}_\phi+B_z\hat{i}_z \tag{3}$

The divergence of the field is $$\vec{\nabla}.\vec{B}=\frac{1}{r}\frac{\partial}{\partial{r}}(rB_r)+\frac{1}{r}\frac{\partial{B_\phi}}{\partial{\phi}}+\frac{\partial{B_z}}{\partial{z}}\tag{4}$$

In spherical coordinates, $\vec{B}=B_r\hat{r}_x+B_\theta\hat{i}_\theta+B_\phi\hat{i}_\phi \tag{1}$

the divergence of a vector field is given by $$\vec{\nabla}.\vec{B}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2B_r)+\frac{1}{r\sin\theta}\frac{\partial}{\partial{\theta}}(\sin\theta B_{\theta})+\frac{1}{r\sin\theta}\frac{\partial{B_\phi}}{\partial{\phi}}\tag{5}$$