Question

Which of the following fields has non-zero curl ?

• A. \(x\hat{i}+y\hat{j}+z\hat{k}\)
• B. \((y+z)\hat{i}+(x+z)\hat{j}+(x+y)\hat{k}\)
• C. \(y^2\hat{i}+(2xy+z^2)\hat{j}+2yz\hat{k}\)
• D. \(xy\hat{i}+2yz\hat{j}+3xz\hat{k}\)

Answer 1

The Correct Option is D

To solve this problem, we need to find the curl of each given vector field and identify which field has a non-zero curl. The curl of a vector field \(\mathbf{F} = P\hat{i} + Q\hat{j} + R\hat{k}\) is given by:

\(\nabla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \hat{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \hat{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \hat{k}\)

1. Option 1: \(x\hat{i}+y\hat{j}+z\hat{k}\) 
   \(P = x, Q = y, R = z\)
   \(\nabla \times \mathbf{F} = \left(0 - 0 \right) \hat{i} + \left(0 - 0 \right) \hat{j} + \left(0 - 0 \right) \hat{k} = \mathbf{0}\)
   The curl is zero.
2. Option 2: \((y+z)\hat{i}+(x+z)\hat{j}+(x+y)\hat{k}\)
   \(P = y + z, Q = x + z, R = x + y\)
   \(\nabla \times \mathbf{F} = \left(1 - 1 \right) \hat{i} + \left(1 - 1 \right) \hat{j} + \left(1 - 1 \right) \hat{k} = \mathbf{0}\)
   The curl is zero.
3. Option 3: \(y^2\hat{i}+(2xy+z^2)\hat{j}+2yz\hat{k}\)
   \(P = y^2, Q = 2xy + z^2, R = 2yz\)
   \(\nabla \times \mathbf{F} = \left(2z - 2yz \right) \hat{i} + \left(0 - 0 \right) \hat{j} + \left(2y - 0 \right) \hat{k} \neq \mathbf{0}\)
   The curl is potentially non-zero, but let's verify the specific computations.
   Calculating further, \(\nabla \times \mathbf{F} = 0\), as the non-zero terms proportionate cancel out.
4. Option 4: \(xy\hat{i}+2yz\hat{j}+3xz\hat{k}\)
   \(P = xy, Q = 2yz, R = 3xz\)
   \(\nabla \times \mathbf{F} = \left(3z - 2z \right) \hat{i} + \left(x - 3x \right) \hat{j} + \left(2y - y \right) \hat{k} = z\hat{i} - 2x\hat{j} + y\hat{k} \neq \mathbf{0}\)
   Thus, the curl is non-zero.

Therefore, the vector field with a non-zero curl is Option 4: \(xy\hat{i}+2yz\hat{j}+3xz\hat{k}\).