Question

14. is the vector is irrotational.
E=yzi+xzj+xyk
(1 Point)
yes
no​

Answer 1

SOLUTION

TO CHECK

True / False below statement

The vector $\vec{E}$ is irrational Where

$\vec{E} = yz \hat{i} + xz \hat{j} + xy \hat{k}$

CONCEPT TO BE IMPLEMENTED

A vector $\vec{v}$ is called irrational if

$curl \: \vec{v} = \nabla \times \vec{v} = 0$

EVALUATION

Here the given vector is

$\vec{E} = yz \hat{i} + xz \hat{j} + xy \hat{k}$

Now

$\nabla \times \vec{E}$

$= \nabla \times \: (yz \hat{i} + xz \hat{j} + xy \hat{k})$

$= \displaystyle\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \\ \frac{ \partial}{ \partial x} & \frac{ \partial}{ \partial y} & \frac{ \partial}{ \partial z} \\ \\ yz & xz & xy \end{vmatrix}$

$= \hat{i}(x - x) - \hat{j} (y - y) + \hat{k}(z - z)$

$= 0 \hat{i}- 0\hat{j} + 0 \hat{k}$

$= \hat{0}$

Hence the given statement is TRUE

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. write the expression for gradient and divergence

https://brainly.in/question/32412615

2. prove that the curl of the gradient of

(scalar function) is zero

https://brainly.in/question/19412715