Question

which one of the following is zero?
a)Grad div A
b)Div gradient v
c)Div curl A
d)curl curl A

Answer 1

Answer:

d no is the zero .....byy

Answer 2

C) Div curl A is 'zero'.

Step-by-step explanation:

Given: Vector A

To Prove: Div curl A = 0

Solution:

• Proof for Div curl A = 0

Let the vector A is $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ such that the curl A is given by,

$\vec{\nabla} \times \vec{A} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\A_x&A_y&A_z \end{vmatrix}$

$\Rightarrow \vec{\nabla} \times \vec{A} = \hat{i} \Big( \dfrac{\partial A_z}{\partial y} - \dfrac{\partial A_y}{\partial z} \Big) - \hat{j} \Big( \dfrac{\partial A_z}{\partial x} - \dfrac{\partial A_x}{\partial z} \Big) + \hat{k} \Big( \dfrac{\partial A_y}{\partial x} - \dfrac{\partial A_x}{\partial y} \Big)$

Now, div curl A is given by,

$\vec{\nabla} \cdot ( \vec{\nabla} \times \vec{A}) = \Big( \hat{i}\frac{\partial }{\partial x} + \hat{j}\frac{\partial }{\partial y} + \hat{k}\frac{\partial }{\partial z} \Big) \cdot \Big[ \hat{i} \Big( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \Big) - \hat{j} \Big( \frac{\partial A_z}{\partial x} - \frac{\partial A_x}{\partial z} \Big) + \hat{k} \Big( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \Big) \Big]$

$\Rightarrow \vec{\nabla} \cdot ( \vec{\nabla} \times \vec{A}) = \dfrac{\partial }{\partial x} \Big( \dfrac{\partial A_z}{\partial y} - \dfrac{\partial A_y}{\partial z} \Big) - \dfrac{\partial }{\partial y} \Big( \dfrac{\partial A_z}{\partial x} - \dfrac{\partial A_x}{\partial z} \Big) + \dfrac{\partial }{\partial z} \Big( \dfrac{\partial A_y}{\partial x} - \dfrac{\partial A_x}{\partial y} \Big)$

$\Rightarrow \vec{\nabla} \cdot ( \vec{\nabla} \times \vec{A}) = 0 - 0 + 0 = 0$

Hence, div curl A is 'zero'.