Question

If $\overrightarrow{H}=2y{\hat{a}}_x+(z^2-x^2){\hat{a}}_y+3y{\hat{a}}_z$ , then the curl of vector field $\overrightarrow{H}$ at origin is

• A. $\sqrt{12}$
• B. $\sqrt{13}$
• C. $\sqrt{14}$
• D. $\sqrt{15}$

Answer 1

The correct option is B $\sqrt{13}$

$▽\times \overrightarrow{H}=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 2y& (z^2-x^2)& 3y\end{array}\right|$

= $\hat{i}(3-2z)+\hat{k}(-2x-2)$

= $(3-2z)\hat{i}-(2x+2)\hat{k}$

At the origin , x =0, z= 0

$▽\times \overrightarrow{H}=3\hat{i}-2\hat{k}$

|$▽\times \overrightarrow{H}$| = $\sqrt{3^2+2^2}$

= $\sqrt{13}$