The magnetic field intensity is given in a certain region of space as \(\mathbf{H}=\) \(\left[(x+2 y) / z^{2}\right] \mathbf{a}_{y}+(2 / z) \mathbf{a}_{z} \mathrm{~A} / \mathrm{m} .(a)\) Find \(\nabla \times \mathbf{H} .(b)\) Find \(\mathbf{J} .(c)\) Use \(\mathbf{J}\) to find the total current passing through the surface \(z=4,1 \leq x \leq 2,3 \leq z \leq 5\), in the \(\mathbf{a}_{z}\) direction. ( \(d\) ) Show that the same result is obtained using the other side of Stokes' theorem.

To find the curl of \(\mathbf{H}\), use the following formula: $\nabla \times \mathbf{H} = \left| \begin{array}{ccc} \mathbf{a}_{x} & \mathbf{a}_{y} & \mathbf{a}_{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & \frac{x + 2y}{z^2} & \frac{2}{z} \end{array} \right|$ Now calculate the curl: \(\nabla \times \mathbf{H} = \left(\frac{\partial}{\partial y} \left(\frac{2}{z} \right) - \frac{\partial}{\partial z} \left( \frac{x+2y}{z^2} \right) \right) \mathbf{a}_{x} - \left(\frac{\partial}{\partial x} \left(\frac{2}{z} \right) - \frac{\partial}{\partial z} \left( 0 \right) \right) \mathbf{a}_{y} + \left( \frac{\partial}{\partial x} \left( \frac{x+2y}{z^2} \right) - \frac{\partial}{\partial y} \left( 0 \right) \right) \mathbf{a}_{z}\) After evaluating the partial derivatives, we get: \(\nabla \times \mathbf{H} = - \frac{2}{z^3}\mathbf{a}_{x} + 0\mathbf{a}_{y} + \frac{1}{z^2}\mathbf{a}_{z}\)

Using the relation between the magnetization current density \(\mathbf{J}\) and the curl of \(\mathbf{H}\), we have: \(\mathbf{J} = \nabla \times \mathbf{H}\) Thus, \(\mathbf{J} = - \frac{2}{z^3}\mathbf{a}_{x} + 0\mathbf{a}_{y} + \frac{1}{z^2}\mathbf{a}_{z}\)

Now, integrate \(\mathbf{J}\) over the given surface \(z=4,1 \leq x \leq 2,3 \leq z \leq 5\) to find the total current. Since the current is in the \(\mathbf{a}_{z}\) direction, we only need to consider the \(z\) component of \(\mathbf{J}\): \(I = \int_{1}^{2} \int_{3}^{5} \left(\frac{1}{z^2} \right) dz dx\) Evaluate the integral: \(I = \int_{1}^{2} \left[ -\frac{1}{z} \right]_{3}^{5} dx\) \(I = \int_{1}^{2} \left( \frac{1}{3} - \frac{1}{5} \right) dx\) \(I = \left( \frac{1}{3} - \frac{1}{5} \right) \int_{1}^{2} dx\) \(I = \left( \frac{2}{15} \right) \left[ x \right]_{1}^{2} = \frac{2}{15} (2 - 1) = \frac{2}{15} \thinspace \mathrm{A}\)

Stokes' theorem states that: \(\oint_{\partial S} \mathbf{H} \cdot d\boldsymbol{l} = \iint_{S} \nabla \times \mathbf{H} \cdot d\boldsymbol{S}\) We already found the integral on the right-hand side. Now we need to find the integral on the left-hand side. The line integral will follow the boundary of the surface, which consists of four segments: 1. From \((1,3,4)\) to \((2,3,4)\) 2. From \((2,3,4)\) to \((2,5,4)\) 3. From \((2,5,4)\) to \((1,5,4)\) 4. From \((1,5,4)\) to \((1,3,4)\) So, we calculate the line integral for each segment and add the results: 1. \(\int_{1}^{2} \mathbf{H} \cdot d\boldsymbol{l}_{1} = \int_{1}^{2} \left[ (x + 2(3)) \mathbf{a}_{y} + 2\mathbf{a}_{z} \right] \cdot (dx \mathbf{a}_{x}) = 0\) 2. \(\int_{3}^{5} \mathbf{H} \cdot d\boldsymbol{l}_{2} = \int_{3}^{5} \left[ (2 + 2y) \mathbf{a}_{y} + 2\mathbf{a}_{z} \right] \cdot (dy \mathbf{a}_{y}) = \int_{3}^{5} (2 + 2y) dy\) 3. \(\int_{2}^{1} \mathbf{H} \cdot d\boldsymbol{l}_{3} = \int_{2}^{1} \left[ (x + 2(5)) \mathbf{a}_{y} + 2\mathbf{a}_{z} \right] \cdot (-dx \mathbf{a}_{x}) = 0\) 4. \(\int_{5}^{3} \mathbf{H} \cdot d\boldsymbol{l}_{4} = \int_{5}^{3} \left[ (1 + 2y) \mathbf{a}_{y} + 2\mathbf{a}_{z} \right] \cdot (-dy \mathbf{a}_{y}) = -\int_{5}^{3} (1 + 2y) dy\) Now, sum the line integrals: \(I = 0 + \int_{3}^{5} (2 + 2y) dy - \int_{5}^{3} (1 + 2y) dy\) \(I = \left( \frac{16}{5} \right) + \left( - \frac{14}{5} \right) = \frac{2}{15} \thinspace \mathrm{A}\) We can see that the result obtained by using the other side of Stokes' theorem is the same as the result we obtained earlier. Therefore, our solution is correct.