The length of an antenna is $l = \frac{1}{2}m\lambda$, where $\lambda$ is the wavelength and $m$ is an even number. The antenna is fed at the center, with the center as the origin, and the $z$-axis is along the antenna. The current distribution on the antenna is
$$I = I_0 \sin\left(k\left(\frac{l}{2} - z\right)\right) \mathrm{e}^{-\mathrm{i}\omega}, \quad z \leqslant l/2$$
Try to find its radiation field is _____.

Options:
A. $$B(r,t)=-\frac{\mu_0I_0}{4\pi}\frac{\mathrm{e}^{\mathrm{i}(kr-at)}}r\frac{\sin\left(\frac m2\pi\mathrm{cos}\theta\right)}{\mathrm{sin}\theta}e_\theta $$
$$E(r,t)=\frac{I_0}{4\pi\varepsilon_0c}\frac{\mathrm{e}^{-\mathrm{i}(kr+at)}}{r}\frac{\cos\left(\frac{m}{2}\pi\mathrm{sin}\theta\right)}{\mathrm{cos}\theta}e_r $$
B. $$B(r,t)=-\frac{\mu_0I_0}{2\pi}\frac{\mathrm{e}^{\mathrm{i}(2kr-at)}}r\frac{\cos\left(\frac m2\pi\mathrm{cos}\theta\right)}{\mathrm{cos}\theta}e_\phi $$
$$E(r,t)=\frac{I_0}{4\pi\varepsilon_0c}\frac{\mathrm{e}^{-\mathrm{i}(kr+at)}}{r}\frac{\cos\left(\frac{m}{2}\pi\mathrm{cos}\theta\right)}{\mathrm{sin}\theta}e_\theta $$
C. $$B(r,t)=-\frac{\mu_0I_0}{2\pi}\frac{\mathrm{e}^{-\mathrm{i}(kr-at)}}r\frac{\sin\left(\frac m2\pi\mathrm{cos}\theta\right)}{\mathrm{cos}\theta}e_r $$
$$E(r,t)=-\frac{I_0}{4\pi}\frac{\mathrm{e}^{\mathrm{i}(kr+2at)}}{r}\frac{\cos\left(\frac{m}{2}\pi\mathrm{sin}\theta\right)}{\mathrm{sin}\theta}e_\phi $$
D. $$B(r,t)=-\frac{\mu_0I_0}{2\pi}\frac{\mathrm{e}^{\mathrm{i}(kr+2at)}}r\frac{\cos\left(\frac m2\pi\mathrm{sin}\theta\right)}{\mathrm{sin}\theta}e_\phi $$
$$E(r,t)=-\frac{I_0}{2\pi\varepsilon_0c}\frac{\mathrm{e}^{\mathrm{i}(2kr-at)}}{r}\frac{\sin\left(\frac{m}{2}\pi\mathrm{cos}\theta\right)}{\mathrm{cos}\theta}e_\theta $$
E. $$B(r,t)=\frac{\mu_0I_0}{2\pi}\frac{\mathrm{e}^{\mathrm{i}(kr-at)}}r\frac{\cos\left(\frac m2\pi\mathrm{sin}\theta\right)}{\mathrm{sin}\theta}e_\theta $$
$$E(r,t)=-\frac{I_0}{2\pi}\frac{\mathrm{e}^{\mathrm{i}(2kr-at)}}{r}\frac{\cos\left(\frac{m}{2}\pi\mathrm{cos}\theta\right)}{\mathrm{sin}\theta}e_\phi $$
F. $$B(r,t)=-\frac{\mu_0I_0}{4\pi}\frac{\mathrm{e}^{\mathrm{i}(kr+at)}}r\frac{\sin\left(\frac m2\pi\mathrm{sin}\theta\right)}{\mathrm{cos}\theta}e_r $$
$$E(r,t)=-\frac{I_0}{4\pi\varepsilon_0c}\frac{\mathrm{e}^{\mathrm{i}(kr+at)}}{r}\frac{\sin\left(\frac{m}{2}\pi\mathrm{sin}\theta\right)}{\mathrm{cos}\theta}e_\theta $$
G. $$B(r,t)=-\frac{\mu_0I_0}{2\pi}\frac{\mathrm{e}^{\mathrm{i}(kr-at)}}r\frac{\cos\left(\frac m2\pi\mathrm{cos}\theta\right)}{\mathrm{sin}\theta}e_\theta $$
$$E(r,t)=\frac{I_0}{2\pi\varepsilon_0c}\frac{\mathrm{e}^{\mathrm{i}(kr+at)}}{r}\frac{\cos\left(\frac{m}{2}\pi\mathrm{cos}\theta\right)}{\mathrm{sin}\theta}e_r $$
H. $$B(r,t)=-\frac{\mu_0I_0}{2\pi}\frac{\mathrm{e}^{\mathrm{i}(kr-at)}}r\frac{\sin\left(\frac m2\pi\mathrm{cos}\theta\right)}{\mathrm{sin}\theta}e_\phi $$
$$E(r,t)=-\frac{I_0}{2\pi\varepsilon_0c}\frac{\mathrm{e}^{\mathrm{i}(kr-at)}}{r}\frac{\sin\left(\frac{m}{2}\pi\mathrm{cos}\theta\right)}{\mathrm{sin}\theta}e_\theta $$
I. $$B(r,t)=-\frac{\mu_0I_0}{2\pi}\frac{\mathrm{e}^{\mathrm{i}(kr-at)}}r\frac{\sin\left(\frac m2\pi\mathrm{cos}\theta\right)}{\mathrm{sin}\theta}e_r $$
$$E(r,t)=-\frac{I_0}{2\pi}\frac{\mathrm{e}^{\mathrm{i}(2kr-at)}}{r}\frac{\sin\left(\frac{m}{2}\pi\mathrm{sin}\theta\right)}{\mathrm{cos}\theta}e_\phi $$
J. $$B(r,t)=-\frac{\mu_0I_0}{4\pi}\frac{\mathrm{e}^{-\mathrm{i}(kr+at)}}r\frac{\sin\left(\frac m2\pi\mathrm{cos}\theta\right)}{\mathrm{sin}\theta}e_\theta $$
$$E(r,t)=-\frac{I_0}{4\pi\varepsilon_0c}\frac{\mathrm{e}^{-\mathrm{i}(kr+at)}}{r}\frac{\sin\left(\frac{m}{2}\pi\mathrm{sin}\theta\right)}{\mathrm{cos}\theta}e_r $$
Answer: H