Question

Let \(\mathrm{J}=400 \sin \theta /\left(r^{2}+4\right)\) a \(_{r} \mathrm{~A} / \mathrm{m}^{2} \cdot(a)\) Find the total current flowing through that portion of the spherical surface \(r=0.8\), bounded by \(0.1 \pi<\theta<0.3 \pi, 0<\phi<2 \pi .(b)\) Find the average value of \(\mathbf{J}\) over the defined area.

Short answer

Question: Find the total current flowing through a portion of a spherical surface and the average value of the current density J over the defined area, given the current density function \(J = \frac{400 \sin{\theta}}{(r^2 + 4)}\) A/m², where r is fixed at 0.8 meters, and the bounds for θ and φ are \(0.1\pi\) to \(0.3\pi\) and 0 to \(2\pi\), respectively. Answer: The total current flowing through the portion of the spherical surface is approximately 55.172 A, and the average value of the current density J over the defined area is approximately 17.56 A/m².

Step-by-step solution

Convert J to spherical coordinates

Given the current density function \(J = \frac{400 \sin{\theta}}{(r^2 + 4)}\) A/m², we see that it is already in spherical coordinates, and we have that \(r = 0.8\).

Find the total current flowing (Part A)

In spherical coordinates, we have the current density vector \(\mathbf{J} = J\mathbf{e}_r = \frac{400 \sin{\theta}}{(0.8^2 + 4)}\mathbf{e}_r\) A/m². The total current I flowing through the portion of surface is given by the integral of the current density over the surface. We have \(\displaystyle I= \iiint_{\text{sphere}} \mathbf{J} \cdot d\mathbf{S}\), where dS is the area element in spherical coordinates given by \(r^2 \sin{\theta} dr d\theta d\phi\). We will now integrate the given function \(\displaystyle I = \int_0^{2\pi}\int_{0.1\pi}^{0.3\pi}\int_0^{0.8} \frac{400(0.8)^2\sin{\theta}}{(0.8^2 + 4)}dr d\theta d\phi\). Since r is fixed at 0.8, we only need to perform the integration with respect to \(\theta\) and \(\phi\). So, \(I = (0.8)^2\int_0^{2\pi}\int_{0.1\pi}^{0.3\pi}\frac{400\sin{\theta}}{(0.8^2 + 4)} d\theta d\phi\). Now we can separate the integrals: \(I = \frac{400(0.8)^2}{(0.8^2 + 4)}\left(\int_{0.1\pi}^{0.3\pi}\sin{\theta} d\theta\right) \left(\int_0^{2\pi} d\phi\right)\). Calculating the integrals, we find: \(\displaystyle I = \frac{400 \cdot 0.64}{4.64}\left[\left(-\cos{(0.3\pi)}+\cos{(0.1\pi)}\right)\left(2\pi\right)\right] \approx 55.172\) A. So, the total current flowing through the portion of the spherical surface is approximately \(55.172\) A.

Calculate the average value of J (Part B)

To find the average value of J over the defined area, we will divide the total current obtained in part (a) by the total area of the surface. The area of the surface element in spherical coordinates is given by: \(dA= r^2 \sin{\theta} d\theta d\phi\). For the given bounds, the total area A can be calculated as: \(A = \int_0^{2\pi} \int_{0.1\pi}^{0.3\pi} (0.8)^2\sin{\theta} d\theta d\phi\). Separating the integrals, \(A = (0.8)^2\left(\int_{0.1\pi}^{0.3\pi} \sin{\theta}d\theta\right)\left(\int_0^{2\pi} d\phi\right)\). Calculating the integrals, we find: \(A = (0.8)^2\left[-\cos{(0.3\pi)}+\cos{(0.1\pi)}\right]\left(2\pi\right) \approx 3.1416\) m². Now, to find the average value of J, we take the total current from Part A, and divide it by the total area: \(\displaystyle \bar{J} = \frac{I}{A} =\frac{55.172}{3.1416} \approx 17.56\) A/m². Thus, the average value of the current density J over the defined area is approximately \(17.56\) A/m².