Sebuah dipol pendek dialiri arus $I_0 \cos \omega t$ dengan arah $\mathbf{a}_z$ dan terletak pada pusat koordinat (origin) di ruang hampa. (a) Jika $k = 1\text{ rad/m}$, $r = 2\text{ m}$, $\theta = 45^\circ$, $\phi = 0$, dan $t = 0$, tentukan vektor satuan dalam komponen persegi (rektangular) yang menunjukkan arah sesaat dari medan listrik $\mathbf{E}$. (b) Berapa fraksi dari total daya rata-rata yang dipancarkan pada sabuk (belt) wilayah $80^\circ < \theta < 100^\circ$?
Bagian (a): Vektor Satuan Arah Sesaat $\mathbf{E}$
Untuk dipol pendek dengan arus $I_0$ dan panjang $L$, persamaan medan listrik di ruang hampa ($\eta_0 \approx 120\pi$) adalah
$$E_r = \frac{I_0 L \eta_0}{2\pi} \cos\theta \left( \frac{1}{r^2} – \frac{j}{kr^3} \right) e^{-jkr}$$
$$E_\theta = \frac{I_0 L \eta_0}{4\pi} \sin\theta \left( \frac{j k}{r} + \frac{1}{r^2} – \frac{j}{kr^3} \right) e^{-jkr}$$
- Substitusikan nilai pada medan listrik pertama
$$\cos45^\circ \left( \frac{1}{(2)^2} – \frac{j}{(1)(2)^3} \right) e^{-j(1)(2)}$$
$$\frac{1}{\sqrt{2}} \left( \frac{1}{4} – \frac{j}{8} \right) e^{-j2}$$
$$\frac{1}{\sqrt{2}} \left( \frac{1}{4} – \frac{j}{8} \right) (\cos(-2) + j\sin(-2))$$
- Substitusikan nilai pada medan listrik kedua
$$\sin45^\circ \cdot \frac{1}{2} \cdot \left( \frac{j (1)}{(2)} + \frac{1}{(2)^2} – \frac{j}{(1)(2)^3} \right) e^{-j(1)(2)}$$
$$\frac{1}{\sqrt{2}} \cdot \frac{1}{2} \cdot \left( \frac{j}{2} + \frac{1}{4} – \frac{j}{8} \right) e^{-j2}$$
$$\frac{1}{\sqrt{2}} \cdot \left (\frac{j}{4} + \frac{1}{8} – \frac{j}{16} \right) (\cos(-2) + j\sin(-2))$$
Karena $t=0$, kita cukup mengambil bagian riil dari phasor dikalikan dengan konstanta pengali $\frac{I_0 L \eta_0}{2\pi \sqrt{2}}$:
- Kita cari nilai yang pertama
$$\mathcal{R}e\left[{ \left(\frac{1}{4} – \frac{j}{8}\right) (\cos 2 – j\sin 2)}\right]$$
$$\mathcal{R}e\left[\frac{1}{4}\cos2 – \frac{j}{8}\cos2 – \frac{j}{4}\sin2 – \frac{1}{8}\sin2\right]$$
$$0,25 \cdot 0,999 – 0,125 \cdot 0,034$$
$$0,24975 – 0,00425 = 0,2455$$
- Kita cari nilai yang kedua
$$\mathcal{R}e\left[\left (\frac{j}{4} + \frac{1}{8} – \frac{j}{16} \right) (\cos(2) – j\sin(2))\right]$$
$$\mathcal{R}e\left[\frac{j}{4}\cos2 + \frac{1}{4}\sin2 + \frac{1}{8}\cos2 – \frac{j}{16}\sin2 – \frac{j}{16}\cos2 – \frac{1}{16}\sin(2)\right]$$
$$0,25 \cdot 0,034 + 0,125 \cdot 0,999 – 0,062 \cdot 0,034$$
$$0,0085 + 0,1248 – 0,0021 = 0,1312$$
Kita cari nilai dari koordinat polar
$$\mathbf{a}_r = \sin\theta\mathbf{a}_x + \cos\theta\mathbf{a}_z$$
$$\mathbf{a}_r = \frac{1}{\sqrt{2}}\mathbf{a}_x + \frac{1}{\sqrt{2}}\mathbf{a}_z$$
$$\mathbf{a}_\theta = \cos\theta\mathbf{a}_x – \sin\theta\mathbf{a}_z$$
$$\mathbf{a}_\theta = \frac{1}{\sqrt{2}}\mathbf{a}_x – \frac{1}{\sqrt{2}}\mathbf{a}_z$$
Kita gabungkan hasilnya
$$\mathbf{E} \propto (0,2455)\left(\frac{1}{\sqrt{2}}\mathbf{a}_x + \frac{1}{\sqrt{2}}\mathbf{a}_z\right) + (0,1312)\left(\frac{1}{\sqrt{2}}\mathbf{a}_x – \frac{1}{\sqrt{2}}\mathbf{a}_z\right)$$
$$\mathbf{E} \propto (0,2455 + 0,1313) \cdot \frac{1}{\sqrt{2}}\mathbf{a}_x – (0,2455 – 0,1313) \cdot \frac{1}{\sqrt{2}}\mathbf{a}_z$$
$$\mathbf{E} \propto (0,3767 \cdot 0,7071)\mathbf{a}_x – (0,1312)(0,7071)\mathbf{a}_z$$
$$\mathbf{E \propto 0,2663\mathbf{a}_x – 0,0927\mathbf{a}_z}$$
Kita cari magnitudonya
$$|\mathbf{E}| = \sqrt{(0,2663)^2 + (0,0927)^2}$$
$$|\mathbf{E}| = \sqrt{0,07091 + 0,00859}$$
$$|\mathbf{E}| = \sqrt{0,0795}$$
$$\mathbf{|E| = 0,2819}$$
Kita cari vektor satuan
$$\frac{0,2663\mathbf{a}_x – 0,0927\mathbf{a}_z}{0,2819}$$
$$\mathbf{0,9329\mathbf{a}_x – 0,3288\mathbf{a}_z}$$
Bagian (b): Fraksi Daya pada Sabuk $80^\circ < \theta < 100^\circ$
Kita integralkan $\sin^3\theta \ d\theta$ dahulu
$$\int \sin^3\theta \ d\theta$$
$$\int \sin^2\theta \cdot \sin\theta \ d\theta$$
$$\int (1 – \cos^2(\theta)) \cdot \sin(\theta) \ d\theta$$
Kita gunakan substitusi
- $u = \cos(\theta)$
- $du = -\sin(\theta) \ d\theta$
- $-du = \sin(\theta) \ d\theta$
Kita selesaikan integrasi
$$\int (1 – u^2) \cdot (-du)$$
$$\int (u^2 – 1) \cdot (du)$$
$$\int u^2 \cdot (du) – \int 1 \ du$$
$$\frac{1}{3}u^3 – u + C$$
$$\frac{1}{3}\cos^3(\theta) – \cos(\theta) + C$$
Kerapatan daya rata-rata dipol memiliki ketergantungan sudut terhadap $\sin^3\theta$ saat diintegrasikan terhadap luas bola ($dS = r^2 \sin\theta \, d\theta \, d\phi$). Maka, daya rata-rata total sebanding dengan:
$$W_{\text{total}} \propto \int_{0}^{\pi} \sin^3\theta \, d\theta$$
$$W_{\text{total}} \propto \left[\frac{1}{3}\cos^3(\theta) – \cos(\theta)\right]_{0}^{\pi}$$
$$W_{\text{total}} \propto \left[\frac{1}{3}\cos^3(\pi) – \cos(\pi) – \frac{1}{3}\cos^3(0) + \cos(0)\right]$$
$$W_{\text{total}} \propto \left[-\frac{1}{3} – (-1) – \frac{1}{3} + 1 \right]$$
$$\mathbf{W_{\text{total}} \propto \frac{4}{3}}$$
Daya pada sabuk tertentu ($W_{\text{sabuk}}$) dengan batas $\theta_1 = 80^\circ$ dan $\theta_2 = 100^\circ$:
$$W_{\text{sabuk}} \propto \int_{80^\circ}^{100^\circ} \sin^3\theta \, d\theta$$
$$W_{\text{sabuk}} \propto \left[\frac{1}{3}\cos^3(\theta) – \cos(\theta)\right]_{80^\circ}^{100^\circ}$$
$$W_{\text{sabuk}} \propto \left[\frac{1}{3}\cos^3(100^\circ) – \cos(100^\circ) – \frac{1}{3}\cos^3(80^\circ) + \cos(80^\circ)\right]$$
$$W_{\text{sabuk}} \propto [0,33333 \cdot (-0,173)^3 – (-0,173) – 0,33333 \cdot (0,173)^3 + 0,173]$$
$$W_{\text{sabuk}} \propto [0,33333 \cdot (-0,00517) + 0,173 – 0,33333 \cdot 0,00517 + 0,173]$$
$$W_{\text{sabuk}} \propto -0,00172 + 0,173 – 0,00172 + 0,173$$
$$W_{\text{sabuk}} \propto 0,3425$$
Menghitung Fraksi Daya:
$$\text{Fraksi} = \frac{W_{\text{sabuk}}}{W_{\text{total}}}$$
$$\text{Fraksi} = \frac{0.3425}{4/3}$$
$$\text{Fraksi} = \frac{0.3438}{1.3333}$$
$$\mathbf{\text{Fraksi} = 0.2578}$$
