Find the curl of the fields shown on the image below.

Tuesday, June 11, 2013

Find the curl of the fields shown on the image below.

a) A curl is the vector derivative of a vector field. It can be denoted as

$\displaystyle{\vec{\nabla}}\times{\vec{{F}}}$ , where $\displaystyle{\vec{{F}}}$ is the vector field.

The curl is calculated as three-dimensional determinant:

i j k

d/dx d/dy d/dz

$\displaystyle{F}_{{x}}$ $\displaystyle{F}_{{y}}$ $\displaystyle{F}_{{z}}$

This determinant equals the vector quantity

$\displaystyle{\left(\frac{{{d}{F}_{{z}}}}{{\left.{d}{y}}}-\frac{{{d}{F}_{{y}}}}{{{\left.{d}{z}\right.}}}\right)}{\vec{{i}}}-{\left(\frac{{{d}{F}_{{z}}}}{{{\left.{d}{x}\right.}}}-\frac{{{d}{F}_{{x}}}}{{{\left.{d}{z}\right.}}}\right)}{\vec{{j}}}+{\left(\frac{{{d}{F}_{{y}}}}{{{\left.{d}{x}\right.}}}-\frac{{{d}{F}_{{x}}}}{{{\left.{d}{y}\right.}}}\right)}{\vec{{k}}}$ .

To find curl of the given field, let's first find all the required partial derivatives:

$\displaystyle\frac{{{d}{F}_{{y}}}}{{{\left.{d}{x}\right.}}}={{y}}^{{2}}{z}$

$\displaystyle\frac{{{d}{F}_{{x}}}}{{{\left.{d}{y}\right.}}}={{x}}^{{2}}{z}$

$\displaystyle\frac{{{d}{F}_{{z}}}}{{{\left.{d}{y}\right.}}}={x}{{z}}^{{2}}$

$\displaystyle\frac{{{d}{F}_{{y}}}}{{{\left.{d}{z}\right.}}}={x}{{y}}^{{2}}$

$\displaystyle\frac{{{d}{F}_{{z}}}}{{{\left.{d}{x}\right.}}}={y}{{z}}^{{2}}$

$\displaystyle\frac{{{d}{F}_{{x}}}}{{{\left.{d}{z}\right.}}}={{x}}^{{2}}{y}$

Substituting these into the expression above, we get

$\displaystyle{\vec{\nabla}}\times{\vec{{F}}}={x}{\left({{z}}^{{2}}-{{y}}^{{2}}\right)}{\vec{{i}}}+{y}{\left({{x}}^{{2}}-{{z}}^{{2}}\right)}{\vec{{j}}}+{z}{\left({{y}}^{{2}}-{{x}}^{{2}}\right)}{\vec{{k}}}$ .