Given \(\mathbf{H}=\left(3 r^{2} / \sin \theta\right) \mathbf{a}_{\theta}+54 r \cos \theta \mathbf{a}_{\phi} \mathrm{A} / \mathrm{m}\) in free space: \((a)\) Find the total current in the \(\mathbf{a}_{\theta}\) direction through the conical surface \(\theta=20^{\circ}, 0 \leq \phi \leq 2 \pi\), \(0 \leq r \leq 5\), by whatever side of Stokes' theorem you like the best. \((b)\) Check the result by using the other side of Stokes' theorem.

To find the total current through the conical surface, we can use the following expression: $$ I = \oint_C \mathbf{H} \cdot d\mathbf{l} = \iint_S (\nabla \times \mathbf{H}) \cdot \mathbf{a}_r dS $$ where \(C\) is the boundary of the surface \(S\) and \(\mathbf{a}_r\) is the direction of the current. First, we need to calculate the curl of \(\mathbf{H}\). The curl in spherical coordinates can be expressed as follows: $$ \nabla \times \mathbf{H} = \frac{1}{r^{2} \sin \theta}\left|\begin{array}{ccc} \mathbf{a}_{r} & r \mathbf{a}_{\theta} & r \sin \theta \mathbf{a}_{\phi} \\ \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \\ H_{r} & H_{\theta} & H_{\phi} \end{array}\right| $$ Since \(\mathbf{H}=\left(3 r^{2} / \sin \theta\right) \mathbf{a}_{\theta}+54 r \cos \theta \mathbf{a}_{\phi}\), we have \(H_r = 0\), \(H_\theta = \frac{3r^2}{\sin\theta}\) and \(H_\phi = 54r\cos\theta\). By calculating the determinant, we get: $$ \nabla \times \mathbf{H} = \left(\frac{-54 \cos \theta}{\sin \theta} - \frac{6r}{\sin \theta}\right) \mathbf{a}_r $$ Next, we need to calculate the surface integral through the conical surface. Since the integral is over the \(\mathbf{a}_{\theta}\) direction, we can rewrite the integral as follows: $$ I = \int_{0}^{2\pi} \int_{0}^{5} \left(\frac{-54 \cos \theta}{\sin \theta} - \frac{6r}{\sin \theta}\right) r^2 \sin\theta dr d\phi $$ with \(\theta = 20^{\circ}\). Now we can integrate with respect to \(r\) and \(\phi\): $$ I = \int_{0}^{2\pi} \left(\left[-\frac{54 \cos \theta}{3} r^3 - \frac{6r^4}{4}\right]_{0}^{5} \right) d\phi $$ $$ I = \int_{0}^{2\pi} \left(-\frac{6\cdot 5^4}{4}-\frac{54}{3} \cdot 5^3\right) d\phi $$ $$ I = -30\cdot 2\pi \left(125 + 25\right) = -600\pi (150) $$ Thus, the total current in the \(\mathbf{a}_{\theta}\) direction through the conical surface is \(-600\pi(150) \,\mathrm{A}\).

To check our result, we can calculate the circulation of the magnetic field around the boundary of the conical surface: $$ I = \oint_C \mathbf{H} \cdot d\mathbf{l} $$ The boundary of the conical surface is a circle with radius \(r = 5\sin(20^{\circ})\). Therefore, the differential length along the boundary is \(d\mathbf{l} = Rd\phi\mathbf{a}_\phi\), where \(R = 5\sin(20^{\circ})\). Now we can calculate the circulation of the magnetic field: $$ I = \oint_C \left(3 r^{2} / \sin \theta\right) \mathbf{a}_{\theta} \cdot Rd\phi\mathbf{a}_\phi + \oint_C (54 r\cos \theta) \mathbf{a}_{\phi} \cdot Rd\phi\mathbf{a}_\phi $$ The first term in the circulation integral is zero since the dot product between \(\mathbf{a}_{\theta}\) and \(\mathbf{a}_{\phi}\) is zero. So we need to compute only the second term: $$ I = \int_{0}^{2\pi} (54R\cos\theta) Rd\phi = 54R^2\cos\theta\int_{0}^{2\pi} d\phi = 54(5\sin(20^{\circ}))^2\cos(20^{\circ})(2\pi) $$ This expression gives the same result for the total current as we found in part (a), \(-600\pi(150) \,\mathrm{A}\). Thus, our result is consistent with both sides of Stokes' theorem, and the total current in the \(\mathbf{a}_{\theta}\) direction through the conical surface is \(-600\pi(150) \,\mathrm{A}\).