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**Hint:**In vector algebra, when a force is represented in the form of $\vec F = P\hat i + Q\hat j$ it means the components of force vector in X and Y directions are $P$ and $Q$ respectively. A conservative force is one whose work done on a body is independent of the path taken.

**Complete step by step answer:**

We know that, if a force is conservative then its work done must be independent of path and hence, if a force field $\vec F$ is conservative its curl is always zero which means $\vec \nabla \times \vec F = 0$.Let us check the first given force field \[\vec F = ay\hat i\] as its $Y$ and $Z$ components are zero.

Checking the $\vec \nabla \times \vec F = 0$ for force \[\vec F = ay\hat i\] we get,

$\vec \nabla \times \vec F = \left( {\begin{array}{*{20}{c}}

{\hat i}&{\hat j}&{\hat k} \\

{\dfrac{\partial }{{\partial x}}}&{\dfrac{\partial }{{\partial y}}}&{\dfrac{\partial }{{\partial z}}} \\

{ay}&0&0

\end{array}} \right)$

Finding determinant of above matrix we get,

$\vec \nabla \times \vec F = 0 + ( - a\hat k)$

$\therefore \vec \nabla \times \vec F = - a\hat k$

Hence, the value of $\vec \nabla \times \vec F \ne 0$ for the given force \[\vec F = ay\hat i\].Hence, force \[\vec F = ay\hat i\] is not a conservative force.

Similarly, Let us check for the force $\vec F = ax\hat i + by\hat j$

We have,

$\vec \nabla \times \vec F = \left( {\begin{array}{*{20}{c}}

{\hat i}&{\hat j}&{\hat k} \\

{\dfrac{\partial }{{\partial x}}}&{\dfrac{\partial }{{\partial y}}}&{\dfrac{\partial }{{\partial z}}} \\

{ax}&{by}&0

\end{array}} \right)$

Again, finding determinant of above matrix, we have

$\vec \nabla \times \vec F = 0 + 0 + 0$

$\therefore \vec \nabla \times \vec F = 0$

Hence, the value of $\vec \nabla \times \vec F = 0$ for the given force $\vec F = ax\hat i + by\hat j$. Hence, the force $\vec F = ax\hat i + by\hat j$ is a conservative force.

**So, force $\vec F = ax\hat i + by\hat j$ is a conservative force while force \[\vec F = ay\hat i\] is not a conservative force.**

**Note:**Remember the operator $\vec \nabla $ is known as del operator and its defined as $\dfrac{\partial }{{\partial x}}\hat i + \dfrac{\partial }{{\partial y}}\hat j + \dfrac{\partial }{{\partial z}}\hat k$ and this del operator is widely used in differential calculus in order to find the divergence and curl of a given field the term $\vec \nabla \times \vec F$ for any vector field $\vec F$ is known as curls of the field and if this value is zero it’s said to be vector field is irrotational.

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