Consider a cube having the dimension

x, y, z ∈ [1, 3]. If a B̅ = \(2 x^{2} y \hat{a}_{x}+3 x^{2} y^{2} \hat{a}_{y}\).

Divergence of B̅ at the centre of cube is :

Concept:

The divergence of a vector field \( \vec{B} = B_x \hat{a}_x + B_y \hat{a}_y + B_z \hat{a}_z \) is given by:

\( \nabla \cdot \vec{B} = \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} \)

Given:

\( \vec{B} = 2x^2 y \hat{a}_x + 3x^2 y^2 \hat{a}_y \), and the cube lies in \( x, y, z \in [1,3] \)

Calculation:

Since \( \vec{B} \) has no \( \hat{a}_z \) component, \( \frac{\partial B_z}{\partial z} = 0 \)

Find partial derivatives at the center of the cube: \( x = 2, y = 2 \)

\( B_x = 2x^2 y \Rightarrow \frac{\partial B_x}{\partial x} = 4xy \)

At \( x = 2, y = 2 \Rightarrow \frac{\partial B_x}{\partial x} = 4 \times 2 \times 2 = 16 \)

\( B_y = 3x^2 y^2 \Rightarrow \frac{\partial B_y}{\partial y} = 6x^2 y \)

At \( x = 2, y = 2 \Rightarrow \frac{\partial B_y}{\partial y} = 6 \times 4 \times 2 = 48 \)

Final Result:

\( \nabla \cdot \vec{B} = 16 + 48 = 64 \)

Correct Answer: 4) 64