Let's go step by step for each part: ### **Given:** \[ \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|} d\tau' \] --- ### **(a) Check that applying the divergence to \(\mathbf{A}(\mathbf{r})\) gives \(\nabla \cdot \mathbf{A} = 0\)** #### **Solution:** Take the divergence: \[ \nabla \cdot \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \nabla \cdot \int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|} d\tau' \] Since \(\mathbf{J}(\mathbf{r}')\) depends only on \(\mathbf{r}'\), and the divergence acts on \(\mathbf{r}\), we have: \[ \nabla \cdot \left( \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|} \right) = \mathbf{J}(\mathbf{r}') \cdot \nabla \left( \frac{1}{|\mathbf{r} - \mathbf{r}'|} \right) \] But, since \(\nabla \cdot \mathbf{J}(\mathbf{r}') = 0\) (steady currents), and using the vector calculus identity: \[ \nabla \cdot \left( f(\mathbf{r}) \mathbf{a} \right) = \mathbf{a} \cdot \nabla f(\mathbf{r}) \quad \text{if } \mathbf{a} \text{ doesn't depend on } \mathbf{r} \] So, \[ \nabla \cdot \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \mathbf{J}(\mathbf{r}') \cdot \nabla \left( \frac{1}{|\mathbf{r} - \mathbf{r}'|} \right) d\tau' \] But since \(\nabla \cdot \mathbf{J}(\mathbf{r}') = 0\) and with the continuity equation for steady currents, this is zero. **Final Answer for (a):** \[ \boxed{\nabla \cdot \mathbf{A} = 0} \] --- ### **(b) Check that \(\mathbf{A}(\mathbf{r})\) gives \(\mathbf{B}(\mathbf{r})\) via the curl** Given: \[ \mathbf{B}(\mathbf{r}) = \nabla \times \mathbf{A}(\mathbf{r}) \] Let's calculate: \[ \nabla \times \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \nabla \times \int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|} d\tau' \] Bring the curl inside the integral (since it's with respect to \(\mathbf{r}\)): \[ = \frac{\mu_0}{4\pi} \int \nabla \times \left( \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|} \right) d\tau' \] Since \(\mathbf{J}(\mathbf{r}')\) is constant with respect to \(\mathbf{r}\): \[ \nabla \times \left( f(\mathbf{r}) \mathbf{a} \right) = \nabla f(\mathbf{r}) \times \mathbf{a} \] So, \[ = \frac{\mu_0}{4\pi} \int \nabla \left( \frac{1}{|\mathbf{r} - \mathbf{r}'|} \right) \times \mathbf{J}(\mathbf{r}') \, d\tau' \] But, \[ \nabla \left( \frac{1}{|\mathbf{r} - \mathbf{r}'|} \right) = -\frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3} \] So, \[ \nabla \times \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \left( -\frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3} \times \mathbf{J}(\mathbf{r}') \right) d\tau' \] But \( \mathbf{a} \times \mathbf{b} = -\mathbf{b} \times \mathbf{a} \), so: \[ = \frac{\mu_0}{4\pi} \int \mathbf{J}(\mathbf{r}') \times \frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3} d\tau' \] Let \( \hat{\mathbf{r}} = \frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|} \), and \( n = |\mathbf{r} - \mathbf{r}'| \): \[ \mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}') \times \hat{\mathbf{r}}}{n^2} d\tau' \] Which matches the given equation. **Final Answer for (b):** \[ \boxed{\nabla \times \mathbf{A} = \mathbf{B}(\mathbf{r})} \] --- ### **(c) Check that applying the Laplacian gives: \(\nabla^2 \mathbf{A} = -\mu_0 \mathbf{J}\)** Apply the Laplacian: \[ \nabla^2 \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \mathbf{J}(\mathbf{r}') \nabla^2 \left( \frac{1}{|\mathbf{r} - \mathbf{r}'|} \right) d\tau' \] But, \[ \nabla^2 \left( \frac{1}{|\mathbf{r} - \mathbf{r}'|} \right) = -4\pi \delta^3(\mathbf{r} - \mathbf{r}') \] Therefore, \[ \nabla^2 \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \mathbf{J}(\mathbf{r}') \left[ -4\pi \delta^3(\mathbf{r} - \mathbf{r}') \right] d\tau' \] \[ = -\mu_0 \int \mathbf{J}(\mathbf{r}') \delta^3(\mathbf{r} - \mathbf{r}') d\tau' \] \[ = -\mu_0 \mathbf{J}(\mathbf{r}) \] **Final Answer for (c):** \[ \boxed{\nabla^2 \mathbf{A} = -\mu_0 \mathbf{J}} \] --- ## **Summary of Answers:** **(a)** \(\nabla \cdot \mathbf{A} = 0\) (divergence is zero) **(b)** \(\nabla \times \mathbf{A} = \mathbf{B}(\mathbf{r})\) (curl gives B-field) **(c)** \(\nabla^2 \mathbf{A} = -\mu_0 \mathbf{J}\) (Laplacian gives -μ₀J) If you need more detailed explanations for any step, let me know!