Question

Let \(\hat{i}\) and \(\hat{j}\) be the unit vectors along \(x\) and \(y\) axes, respectively, and let \(A\) be a positive constant. Which one of the following statements is true for the vector fields: \[ \vec{F_1} = A(\hat{i}y + \hat{j}x) \quad {and} \quad \vec{F_2} = A(\hat{i}y - \hat{j}x)? \]

• A. Both \(\vec{F_1}\) and \(\vec{F_2}\) are electrostatic fields.
• B. Only \(\vec{F_1}\) is an electrostatic field.
• C. Only \(\vec{F_2}\) is an electrostatic field.
• D. Neither \(\vec{F_1}\) nor \(\vec{F_2}\) is an electrostatic field.

Hint:

To check if a vector field is electrostatic, compute its curl. If the curl is zero, the field is electrostatic. Use determinants for precise and systematic calculations.

Answer 1

The Correct Option is B

Step 1: Establish the condition for an electrostatic field.
A vector field \(\vec{F}\) qualifies as an electrostatic field if its curl is zero: \[ \nabla \times \vec{F} = 0. \]

Step 2: Verify the curl of \(\vec{F_1}\).
The given field is \(\vec{F_1} = A(\hat{i}y + \hat{j}x)\). Its curl is: \[ \nabla \times \vec{F_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ Ay & Ax & 0 \end{vmatrix}. \] Expanding the determinant gives: \[ \nabla \times \vec{F_1} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}\left(\frac{\partial Ax}{\partial y} - \frac{\partial Ay}{\partial x}\right). \] The \(z\)-derivatives vanish as the terms are independent of \(z\), and the \(x\)- and \(y\)-derivatives cancel: \[ \nabla \times \vec{F_1} = \hat{k}(A - A) = 0. \] Thus, \(\vec{F_1}\) satisfies the electrostatic condition.

Step 3: Verify the curl of \(\vec{F_2}\).
The given field is \(\vec{F_2} = A(\hat{i}y - \hat{j}x)\). Its curl is: \[ \nabla \times \vec{F_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ Ay & -Ax & 0 \end{vmatrix}. \] Expanding the determinant gives: \[ \nabla \times \vec{F_2} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}\left(\frac{\partial (-Ax)}{\partial x} - \frac{\partial Ay}{\partial y}\right). \] Simplifying: \[ \nabla \times \vec{F_2} = \hat{k}(-A - A) = \hat{k}(-2A). \] Since the curl is nonzero, \(\vec{F_2}\) is not an electrostatic field.

Final Answer: \[ \boxed{(2) Only \(\vec{F_1}\) is an electrostatic field.} \]