Let's work through the problem step by step: **Given:** - A thick slab, infinite in \(x\) and \(y\), extending from \(z = -a\) to \(z = +a\). - The slab carries a **uniform volume current**: \(\mathbf{J} = J \hat{x}\). We are to find: 1. **(a) The vector potential \(\mathbf{A}(z)\)** 2. **(b) The magnetic field \(\mathbf{B}(z)\)** **Assumptions:** - The system is infinite in \(x\) and \(y\), so all quantities depend only on \(z\). - The current is only within the slab (\(|z| < a\)). Outside (\(|z| > a\)), \(\mathbf{J} = 0\). --- ## (a) The Vector Potential \(\mathbf{A}(z)\) Recall the general equation for the vector potential in magnetostatics: \[ \nabla^2 \mathbf{A} = -\mu_0 \mathbf{J} \] Given symmetry, \(\mathbf{A}\) can only depend on \(z\), and will point in the \(x\)-direction (since \(\mathbf{J}\) is in \(x\)-direction): \[ \mathbf{A} = A_x(z) \hat{x} \] So, \[ \nabla^2 \mathbf{A} = \frac{d^2 A_x(z)}{dz^2} \hat{x} \] Thus, \[ \frac{d^2 A_x}{dz^2} = -\mu_0 J \quad \text{for } |z| < a \] \[ \frac{d^2 A_x}{dz^2} = 0 \quad \text{for } |z| > a \] --- ### **1. Inside the slab (\(|z| < a\)):** Integrate twice: \[ \frac{d^2 A_x}{dz^2} = -\mu_0 J \implies \frac{d A_x}{dz} = -\mu_0 J z + C_1 \implies A_x(z) = -\frac{\mu_0 J}{2}z^2 + C_1 z + C_2 \] ### **2. Outside the slab (\(|z| > a\)):** \[ \frac{d^2 A_x}{dz^2} = 0 \implies \frac{dA_x}{dz} = C_3 \implies A_x(z) = C_3 z + C_4 \] But for \(z > a\), call the constants \(C_3^+\), \(C_4^+\); for \(z < -a\), \(C_3^-\), \(C_4^-\). --- ### **3. Determine the constants (Boundary conditions)** #### **Physical requirements:** - By symmetry, the system is even in \(z\), so \(A_x(z)\) is an even function (since \(J\) is odd in \(x\), but the geometry is even). - The vector potential at infinity should not diverge, so the linear term must vanish far away. Let us check the derivative (related to \(B\)), which must be continuous at \(z = \pm a\). #### **Let us write the solutions:** - For \(|z| < a\): \[ A_x(z) = -\frac{\mu_0 J}{2}z^2 + C_1 z + C_2 \] - For \(z > a\): \[ A_x(z) = C_3^+ z + C_4^+ \] - For \(z < -a\): \[ A_x(z) = C_3^- z + C_4^- \] Because the system is infinite, let’s **set \(A_x \to 0\) as \(z \to \infty\)** (gauge choice). So for \(z \to +\infty\): \[ A_x(z) \to C_3^+ z + C_4^+ \to 0 \implies C_3^+ = 0,~ C_4^+ = 0 \] But let's be careful: the field must vanish at infinity, so \(A_x(z)\) is constant at infinity, but the derivative must be zero, so \(C_3^+ = 0\). We can set the constant to zero by gauge choice. Similarly for \(z \to -\infty\). #### **Match at \(z = a\) and \(z = -a\):** **At \(z = a\):** - \(A_x(a^-) = A_x(a^+)\) - \(\left. \frac{dA_x}{dz} \right|_{a^-} = \left. \frac{dA_x}{dz} \right|_{a^+}\) Let's use the inside expression at \(z=a\): \[ A_x(a^-) = -\frac{\mu_0 J}{2} a^2 + C_1 a + C_2 \] \[ \frac{dA_x}{dz}(a^-) = -\mu_0 J a + C_1 \] For \(z > a\), \(A_x(z) = C_3^+ z + C_4^+\), so: \[ A_x(a^+) = C_3^+ a + C_4^+ \] \[ \frac{dA_x}{dz}(a^+) = C_3^+ \] Set \(C_3^+ = 0\) (no field at infinity), so \(A_x(a^+) = C_4^+\). Thus, \[ A_x(a^-) = A_x(a^+) \implies -\frac{\mu_0 J}{2} a^2 + C_1 a + C_2 = C_4^+ \] \[ \frac{dA_x}{dz}(a^-) = \frac{dA_x}{dz}(a^+) \implies -\mu_0 J a + C_1 = 0 \implies C_1 = \mu_0 J a \] Similarly, for \(z < -a\), \(A_x(z) = C_3^- z + C_4^-\), and \(C_3^-\) must be zero to prevent divergence at \(z \to -\infty\). Thus, \(A_x(z) = C_4^-\) for \(z < -a\). At \(z = -a\): \[ A_x(-a^+) = -\frac{\mu_0 J}{2} a^2 - C_1 a + C_2 \] \[ A_x(-a^-) = C_4^- \] \[ \frac{dA_x}{dz}(-a^+) = +\mu_0 J a + C_1 \] \[ \frac{dA_x}{dz}(-a^-) = 0 \] Set derivative continuous: \[ +\mu_0 J a + C_1 = 0 \implies C_1 = -\mu_0 J a \] But above, for \(z=a\), we found \(C_1 = +\mu_0 J a\). This cannot be, unless \(C_1 = 0\), which is only possible if \(J = 0\). So, the function is **odd** in \(z\), actually, because the current is in \(\hat{x}\), but the geometry is even in \(z\). Let's check the symmetry. But let's try to write \(A_x(z)\) in terms of \(|z|\) for outside. ### **Alternative approach (using symmetry):** We expect, by symmetry, that the vector potential inside is quadratic, and outside is linear (since the field is constant outside): Let’s write: - For \(|z| < a\): \(A_x(z) = -\frac{\mu_0 J}{2} z^2 + C_2\) - For \(|z| > a\): \(A_x(z) = C_3 |z| + C_4\) Let’s match at \(z = a\): Match \(A_x\): \[ A_x(a^-) = -\frac{\mu_0 J}{2} a^2 + C_2 \] \[ A_x(a^+) = C_3 a + C_4 \] Match derivatives: \[ \frac{dA_x}{dz}(a^-) = -\mu_0 J a \] \[ \frac{dA_x}{dz}(a^+) = C_3 \] Set them equal: \[ -\mu_0 J a = C_3 \implies C_3 = -\mu_0 J a \] Set \(A_x(a^-)=A_x(a^+)\): \[ -\frac{\mu_0 J}{2} a^2 + C_2 = -\mu_0 J a^2 + C_4 \implies C_2 = -\frac{\mu_0 J}{2} a^2 + C_4 \] We are free to set the constant \(C_4 = 0\) (gauge choice), so: \[ C_2 = -\frac{\mu_0 J}{2} a^2 \] Thus, **Final answers:** ### **Inside (\(|z| < a\)):** \[ \boxed{ A_x(z) = -\frac{\mu_0 J}{2} (z^2 - a^2) } \] ### **Outside (\(|z| > a\)):** \[ \boxed{ A_x(z) = -\mu_0 J a (|z| - a) } \] So the vector potential is: \[ \boxed{ \mathbf{A}(z) = \begin{cases} -\frac{\mu_0 J}{2} (z^2 - a^2)~\hat{x} & |z| < a \\ -\mu_0 J a (|z| - a)~\hat{x} & |z| > a \end{cases} } \] --- ## (b) The Magnetic Field \(\mathbf{B}(z)\) Recall: \[ \mathbf{B} = \nabla \times \mathbf{A} \] Given \(\mathbf{A} = A_x(z) \hat{x}\): \[ (\nabla \times \mathbf{A})_y = -\frac{dA_x}{dz} \] So: \[ \mathbf{B} = -\frac{dA_x}{dz} \hat{y} \] ### Compute derivative: #### **Inside (\(|z| < a\)):** \[ A_x(z) = -\frac{\mu_0 J}{2} (z^2 - a^2) \] \[ \frac{dA_x}{dz} = -\frac{\mu_0 J}{2} (2z) = -\mu_0 J z \] So, \[ \mathbf{B}(z) = -(-\mu_0 J z) \hat{y} = \mu_0 J z \hat{y} \] for \(|z| < a\). #### **Outside (\(|z| > a\)):** \[ A_x(z) = -\mu_0 J a (|z| - a) \] \[ \frac{dA_x}{dz} = -\mu_0 J a \frac{d}{dz}(|z| - a) = -\mu_0 J a\, \text{sgn}(z) \] So, \[ \mathbf{B}(z) = -(-\mu_0 J a\, \text{sgn}(z)) \hat{y} = \mu_0 J a\, \text{sgn}(z) \hat{y} \] for \(|z| > a\). --- ## **Summary** ### **(a) Vector Potential** \[ \boxed{ \mathbf{A}(z) = \begin{cases} -\frac{\mu_0 J}{2} (z^2 - a^2)~\hat{x} & |z| < a \\ -\mu_0 J a (|z| - a)~\hat{x} & |z| > a \end{cases} } \] ### **(b) Magnetic Field** \[ \boxed{ \mathbf{B}(z) = \begin{cases} \mu_0 J z~\hat{y} & |z| < a \\ \mu_0 J a~\text{sgn}(z)~\hat{y} & |z| > a \end{cases} } \] where \(\text{sgn}(z)\) is the sign function (\(+1\) for \(z > 0\), \(-1\) for \(z < 0\)). --- ### **Physical explanation:** - **Inside the slab (\(|z| < a\)):** The magnetic field grows linearly with \(z\), similar to the field inside a uniformly current-carrying slab. - **Outside the slab (\(|z| > a\)):** The field is constant in magnitude, pointing in \(\pm \hat{y}\) direction depending on the sign of \(z\). --- **Let me know if you would like a diagram or further clarification!**