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**Hint:-**The potential energy is the energy which an object attains at a particular position in its motion. The force due to potential energy is the force required to move the object from the reference point to a position which is at a distance r from the reference point.

Formula used: The formula of the force exerted by a particle in conservative field having a potential energy is given by,

$F = - \nabla U$

Where $\nabla $ is equal to $\nabla = \dfrac{\partial }{{\partial x}}\hat i + \dfrac{\partial }{{\partial y}}\hat j + \dfrac{\partial }{{\partial z}}\hat k$ and $U$ is the potential energy of the particle. Also$\hat i$,$\hat j$ and $\hat k$ are directions representing x-direction ,y-direction and z-direction.

**Complete step-by-step solution**

It is given that the potential energy of a particle is equal to $U = \dfrac{{20xy}}{z}$ and we have to find the force that is exerted on the particle.

As the force exerted on the particle is given by,

$F = - \nabla U$

Where $\nabla $ is equal to $\nabla = \dfrac{\partial }{{\partial x}}\hat i + \dfrac{\partial }{{\partial y}}\hat j + \dfrac{\partial }{{\partial z}}\hat k$ and $U$ is the potential energy of the particle.

Therefore, the force is given by,

$ \Rightarrow F = - \nabla U$

Replace the value of potential energy in the above equation and the differentiating it partially.

$ \Rightarrow F = - \nabla \left( {\dfrac{{20xy}}{z}} \right)$

$ \Rightarrow F = - \left( {\dfrac{\partial }{{\partial x}}\hat i + \dfrac{\partial }{{\partial y}}\hat j + \dfrac{\partial }{{\partial z}}\hat k} \right) \cdot \left( {\dfrac{{20xy}}{z}} \right)$

After differentiating the potential energy we get,

$ \Rightarrow F = - \left( {\dfrac{{20y}}{z}\hat i + \dfrac{{20x}}{z}\hat j - \dfrac{{20xy}}{{{z^2}}}\hat k} \right)$

Solving furthermore we get,

$ \Rightarrow F = - \dfrac{{20y}}{z}\hat i - \dfrac{{20x}}{z}\hat j + \dfrac{{20xy}}{{{z^2}}}\hat k$.

The force applied on the particle is given by$F = - \dfrac{{20y}}{z}\hat i - \dfrac{{20x}}{z}\hat j + \dfrac{{20xy}}{{{z^2}}}\hat k$. The correct answer for this problem is option B.

**Note:-**It is important for students to differentiate the potential energy with respect to x, y and z with care as it is not a normal process of differentiation but this is the partial differentiation of the potential energy. The partial differential is done such that if a given term is differentiated with respect to x then every term except x is taken as constant.

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