GATE 2024 PH – Question 45
$\mathbf F=(2xz+3y^2)\hat y+4yz^2\hat z$. The path A → B → C → D → A is the unit square in z = 0. Which option(s) is/are true?
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Correct answer: (A) $\oint\mathbf F\cdot d\mathbf l=0$; (B) F is non-conservative
Explanation
$\mathbf F=(2xz+3y^2)\hat y+4yz^2\hat z$ and the path is the unit square in the plane $z=0$.
- **A. $\oint\mathbf F\cdot d\mathbf l=0$.** At $z=0$, $\mathbf F=3y^2\hat y$. Along the loop, $d\mathbf l$ has components $dx$ and $dy$, so $\oint\mathbf F\cdot d\mathbf l=\oint3y^2\,dy$, the integral of an exact differential around a closed loop. It equals 0. ✓
- **B. $\mathbf F$ is non-conservative.** The curl is
$$\nabla\times\mathbf F=\big(4z^2-2x,\ 0,\ 2z\big)\neq0,$$
so $\mathbf F$ is not conservative (the zero loop integral is only for this particular path). ✓
- **C. $\nabla\cdot\mathbf F=0$.** The divergence is $\dfrac{\partial}{\partial y}(2xz+3y^2)+\dfrac{\partial}{\partial z}(4yz^2)=6y+8yz\neq0$. ✗
- **D. $\mathbf F$ can be written as a gradient.** Not possible, since the curl is non-zero. ✗
Answer **A and B**.