To solve for the value of $n$ for which the vector $\mathbf{r}^n \mathbf{r}$ is solenoidal, we need to compute its divergence. A vector field is solenoidal if its divergence is zero. Let's proceed step-by-step:

Given:

$\mathbf{r} = x \hat{i} + y \hat{j} + z \hat{k}$

$\mathbf{r}^n = (x^2 + y^2 + z^2)^{n/2}$

The vector field under consideration is: $\mathbf{F} = \mathbf{r}^n \mathbf{r} = \mathbf{r}^n (x \hat{i} + y \hat{j} + z \hat{k})$

Compute divergence of $\mathbf{F}$:

The divergence of a vector field $\mathbf{F}$ is given by: $\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(F_x) + \frac{\partial}{\partial y}(F_y) + \frac{\partial}{\partial z}(F_z)$

Here, $F_x = \mathbf{r}^n x, \quad F_y = \mathbf{r}^n y, \quad F_z = \mathbf{r}^n z$

Step 1: Differentiate $F_x$ with respect to $x$:

$\frac{\partial}{\partial x}(F_x) = \frac{\partial}{\partial x}(\mathbf{r}^n x)$ Using the product rule: $\frac{\partial}{\partial x}(\mathbf{r}^n x) = x \frac{\partial}{\partial x}(\mathbf{r}^n) + \mathbf{r}^n$

The derivative of $\mathbf{r}^n = (x^2 + y^2 + z^2)^{n/2}$ with respect to $x$ is: $\frac{\partial}{\partial x}(\mathbf{r}^n) = \frac{n}{2} (x^2 + y^2 + z^2)^{(n/2)-1} \cdot 2x = n x (x^2 + y^2 + z^2)^{(n/2)-1}$

Thus: $\frac{\partial}{\partial x}(F_x) = x \cdot n x (x^2 + y^2 + z^2)^{(n/2)-1} + \mathbf{r}^n$ $= n x^2 (x^2 + y^2 + z^2)^{(n/2)-1} + \mathbf{r}^n$

Step 2: Compute $\frac{\partial}{\partial y}(F_y)$ and $\frac{\partial}{\partial z}(F_z)$:

Similarly: $\frac{\partial}{\partial y}(F_y) = n y^2 (x^2 + y^2 + z^2)^{(n/2)-1} + \mathbf{r}^n$ $\frac{\partial}{\partial z}(F_z) = n z^2 (x^2 + y^2 + z^2)^{(n/2)-1} + \mathbf{r}^n$

Step 3: Add the components to get $\nabla \cdot \mathbf{F}$:

$\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(F_x) + \frac{\partial}{\partial y}(F_y) + \frac{\partial}{\partial z}(F_z)$ [ \nabla \cdot \mathbf{F} = [n x^2 (x^2 + y^2 + z^2)^{(n/2)-1} + \mathbf{r}^n] + [n y^2 (x^2 + y^2 + z^2)^{(n/2)-1} + \mathbf{r}^n] + [n z^2 (x^2 + y^2 + z^2)^{(n/2)-1} + \mathbf{r}^n] ] $\nabla \cdot \mathbf{F} = n (x^2 + y^2 + z^2)^{(n/2)-1} (x^2 + y^2 + z^2) + 3 \mathbf{r}^n$ $\nabla \cdot \mathbf{F} = n (x^2 + y^2 + z^2)^{n/2} + 3 \mathbf{r}^n$

Since $\mathbf{r}^n = (x^2 + y^2 + z^2)^{n/2}$, we simplify: $\nabla \cdot \mathbf{F} = (n + 3) \mathbf{r}^n$

Step 4: Solve for $n$ such that $\nabla \cdot \mathbf{F} = 0$:

For $\mathbf{F}$ to be solenoidal, $\nabla \cdot \mathbf{F} = 0$. Thus: $n + 3 = 0$ $n = -3$

Final Answer:

$n = -3$

Would you like further clarification on any step or additional examples?

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Related Questions:

1. How is the divergence of a vector field computed for general cases?
2. What physical significance does a solenoidal vector field have in physics?
3. Can you derive the condition for a curl-free vector field?
4. What does the divergence theorem state and how does it relate to solenoidal fields?
5. If $\mathbf{r}^n$ were scalar, how would the solenoidal condition change?

Tip:

Always verify whether a vector field is solenoidal by checking its divergence systematically; this ensures correctness in physical and mathematical applications