A short dipole-carrying current \(I_{0} \cos \omega t\) in the \(\mathbf{a}_{z}\) direction is located at the origin in free space. \((a)\) If \(k=1 \mathrm{rad} / \mathrm{m}, r=2 \mathrm{~m}, \theta=45^{\circ}, \phi=0\), and \(t=0\), give a unit vector in rectangular components that shows the instantaneous direction of \(\mathbf{E}\). (b) What fraction of the total average power is radiated in the belt, \(80^{\circ}<\theta<100^{\circ}\) ?

Find the electric field E at time t = 0 s. The formula for the electric field E in terms of its rectangular components is given by: $$\mathbf{E} = E_r \mathbf{a}_{r} + E_{\theta} \mathbf{a}_{\theta}$$ We have \(E_r = \frac{I_0 k \sin\theta}{4\pi r}\), \(E_{\theta} = -\frac{I_0 k \cos\theta}{4\pi r} \cos\omega t\) Since \(\theta = 45^\circ\), \(\sin\theta = \cos\theta=\frac{1}{\sqrt{2}}\). Plug the values of \(r, k, I_0, \theta, \text{and } t\): $$E_r = \frac{I_{0}\cdot1\cdot\frac{1}{\sqrt{2}}}{4\pi\cdot2 \mathrm{m}} \text{ and } E_{\theta} = -\frac{I_{0}\cdot 1\cdot\frac{1}{\sqrt{2}}}{4\pi\cdot 2 \mathrm{m}} \cdot 1$$ Now, find the rectangular components of \(\mathbf{E}\) by converting the spherical components \(\mathbf{a}_r\) and \(\mathbf{a}_\theta\) into rectangular components \(\mathbf{a}_x\), \(\mathbf{a}_y\), and \(\mathbf{a}_z\). $$\mathbf{a}_r = \sin\theta\cos\phi \mathbf{a}_{x} + \sin\theta\sin\phi \mathbf{a}_{y} + \cos\theta\mathbf{a}_{z}$$ $$\mathbf{a}_{\theta} = \cos\theta\cos\phi \mathbf{a}_{x} + \cos\theta\sin\phi \mathbf{a}_{y} - \sin\theta\mathbf{a}_{z}$$ At \(\theta = 45^\circ\) and \(\phi = 0\), we have: $$\mathbf{a}_r = \frac{1}{\sqrt{2}} \mathbf{a}_{x} + 0 \mathbf{a}_{y} + \frac{1}{\sqrt{2}} \mathbf{a}_{z}$$ $$\mathbf{a}_{\theta} = \frac{1}{\sqrt{2}} \mathbf{a}_{x} + 0\mathbf{a}_{y} - \frac{1}{\sqrt{2}}\mathbf{a}_{z}$$ Now we can find the rectangular components of \(\mathbf{E}\): $$\mathbf{E} = E_{r}\mathbf{a}_{r} + E_{\theta}\mathbf{a}_{\theta} = \frac{I_0}{8 \pi \mathrm{m}}(\mathbf{a}_x+\mathbf{a}_z) - \frac{I_0}{8 \pi \mathrm{m}}(\mathbf{a}_x-\mathbf{a}_z)$$ At \(t=0\), the instantaneous direction of the electric field E is: $$\mathbf{E} = \frac{I_0}{8 \pi \mathrm{m}} (0\mathbf{a}_x + 0\mathbf{a}_y + 2\mathbf{a}_z)$$ Thus, the direction of \(\mathbf{E}\) is in the \(\mathbf{a}_{z}\) direction at time \(t=0\).

To find the fraction of the total average power radiated in the given range of theta, we need to calculate the total power radiated by the dipole and find the proportion of power radiated within the given range of \(\theta\). The total radiated power \(P\) is given by: $$P = \frac{1}{2} \frac{I_{0}^{2}k^2}{3(4\pi)}$$ The power radiated per unit solid angle \(P(\theta)\) is: $$P(\theta) = \frac{1}{2}I_{0}^{2}\sin^2\theta \frac{k^2}{4\pi r^2}$$ Now we need to find the power radiated in the belt \(80^\circ<\theta<100^\circ\). To do this, we integrate \(P(\theta)\) over the range of theta: $$P_{belt} = \int_{80^{\circ}}^{100^{\circ}} P(\theta)\,\text{d}\Omega = \int_{80^{\circ}}^{100^{\circ}} \int_{0}^{2\pi} P(\theta) r^2 \sin\theta\,\text{d}\theta \text{d}\phi$$ $$P_{belt} = \frac{1}{2}I_{0}^{2} \frac{k^2}{4\pi} \int_{80^{\circ}}^{100^{\circ}} \sin^3\theta\,\text{d}\theta \int_{0}^{2\pi}\text{d}\phi$$ Compute the definite integral: $$P_{belt} = \frac{1}{2}I_{0}^{2} \frac{k^2}{4\pi} [\cos^3\theta]^{80^{\circ}}_{100^{\circ}}\cdot2\pi$$ Now, find the fraction of the total average power radiated in the belt: $$\text{Fraction of total average power} = \frac{P_{belt}}{P} = \frac{[\cos^3\theta]^{80^{\circ}}_{100^{\circ}}}{\frac{1}{3}}$$ Compute the final answer: $$\text{Fraction of total average power} \approx 0.0152$$ So, the fraction of the total average power radiated in the belt \(80^\circ<\theta<100^\circ\) is approximately 1.52%.