Question

Let the vector \(\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}\) be a differentiable vector function of Cartesian coordinates x, y and z. The curl of the vector \(\vec{v}\) is given by \(\text{curl } \vec{v} =\)

• A. \(\left(\dfrac{\partial v_2}{\partial z} - \dfrac{\partial v_3}{\partial y}\right)\hat{i} + \left(\dfrac{\partial v_3}{\partial x} - \dfrac{\partial v_1}{\partial z}\right)\hat{j} + \left(\dfrac{\partial v_1}{\partial y} - \dfrac{\partial v_2}{\partial x}\right)\hat{k}\)
• B. \(\left(\dfrac{\partial v_3}{\partial z} - \dfrac{\partial v_2}{\partial y}\right)\hat{i} + \left(\dfrac{\partial v_1}{\partial x} - \dfrac{\partial v_3}{\partial z}\right)\hat{j} + \left(\dfrac{\partial v_2}{\partial y} - \dfrac{\partial v_1}{\partial x}\right)\hat{k}\)
• C. \(\left(\dfrac{\partial v_3}{\partial y} - \dfrac{\partial v_2}{\partial z}\right)\hat{i} + \left(\dfrac{\partial v_1}{\partial z} - \dfrac{\partial v_3}{\partial x}\right)\hat{j} + \left(\dfrac{\partial v_2}{\partial x} - \dfrac{\partial v_1}{\partial y}\right)\hat{k}\)
• D. \(\left(\dfrac{\partial v_2}{\partial y} - \dfrac{\partial v_3}{\partial z}\right)\hat{i} + \left(\dfrac{\partial v_3}{\partial z} - \dfrac{\partial v_1}{\partial x}\right)\hat{j} + \left(\dfrac{\partial v_1}{\partial x} - \dfrac{\partial v_2}{\partial y}\right)\hat{k}\)

Answer 1

The Correct Option is C

To find the curl of the vector function \(\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}\), we use the formula for the curl of a vector field in Cartesian coordinates. The curl of a vector is given by:

\(\text{curl } \vec{v} = \nabla \times \vec{v} = \left(\dfrac{\partial v_3}{\partial y} - \dfrac{\partial v_2}{\partial z}\right)\hat{i} + \left(\dfrac{\partial v_1}{\partial z} - \dfrac{\partial v_3}{\partial x}\right)\hat{j} + \left(\dfrac{\partial v_2}{\partial x} - \dfrac{\partial v_1}{\partial y}\right)\hat{k}\)

This is derived from the determinant of the following symbolic matrix:

\(\hat{i}\)	\(\hat{j}\)	\(\hat{k}\)
\(\dfrac{\partial}{\partial x}\)	\(\dfrac{\partial}{\partial y}\)	\(\dfrac{\partial}{\partial z}\)
\(v_1\)	\(v_2\)	\(v_3\)

Expanding the determinant gives the expression for the curl as mentioned above. Let's break it down for clarity:

1. The component along \(\hat{i}\) is calculated as: \(\dfrac{\partial v_3}{\partial y} - \dfrac{\partial v_2}{\partial z}\)
2. The component along \(\hat{j}\) is calculated as: \(\dfrac{\partial v_1}{\partial z} - \dfrac{\partial v_3}{\partial x}\)
3. The component along \(\hat{k}\) is calculated as: \(\dfrac{\partial v_2}{\partial x} - \dfrac{\partial v_1}{\partial y}\)

Thus, the correct expression for the curl of the vector \(\vec{v}\) is:

\(\left(\dfrac{\partial v_3}{\partial y} - \dfrac{\partial v_2}{\partial z}\right)\hat{i} + \left(\dfrac{\partial v_1}{\partial z} - \dfrac{\partial v_3}{\partial x}\right)\hat{j} + \left(\dfrac{\partial v_2}{\partial x} - \dfrac{\partial v_1}{\partial y}\right)\hat{k}\)

This matches the third option given in the question.