Question

The curl of vector field f (x, y,z) = x²i+2zi - yk is
-31
-33
-3k
0

Answer 1

Given,

Vector field,$\[f\left( {x,y,z} \right) = {x^2}\hat i + 2z\hat j - y\hat k\]$

Solution,

Calculate the curl of vector field,

$\[ = \begin{array}{*{20}{c}}\vline& {\hat i}&{\hat j}&{\hat k}\vline& \\\vline& {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\vline& \\\vline& {{x^2}}&{2z}&{ - y}\vline& \end{array}\]$

$\[ = \hat i\left( {\frac{{\partial \left( { - y} \right)}}{{\partial y}} - \frac{{\partial \left( {2z} \right)}}{{\partial z}}} \right) - \hat j\left( {\frac{{\partial \left( { - y} \right)}}{{\partial x}} - \frac{{\partial \left( {{x^2}} \right)}}{{\partial z}}} \right) + \hat k\left( {\frac{{\partial \left( {2z} \right)}}{{\partial x}} - \frac{{\partial \left( {{x^2}} \right)}}{{\partial y}}} \right)\]$

$\[ = \hat i\left( { - 1 - 2} \right) - \hat j\left( {0 - 0} \right) + \hat k\left( {0 - 0} \right)\]$

$\[ = - 3\hat i\]$

Hence the curl of vector field is $\[ - 3\hat i\]$

The correct option is (a), i.e. $\[ - 3\hat i\]$.