Step 1: Contrast the currents.
(i) Conduction current needs a material carrier and moving charges; displacement current needs only a changing electric field and can exist in vacuum.
(ii) Conduction current \(I_c=nqAv_d\) (drift of carriers); displacement current \(I_d=\varepsilon_0\,\dfrac{d\Phi_E}{dt}\) (rate of change of electric flux).
(iii) Between capacitor plates \(I_c=0\) but \(I_d\ne0\), and together they keep the total current continuous.
Step 2: Identify wave parameters.
Writing the given field as \(B_y=B_0\sin(kx+\omega t)\) gives \(B_0=2\times10^{-7}\) T, \(k=500\) m\(^{-1}\), \(\omega=1.5\times10^{11}\) s\(^{-1}\).
Step 3: Use \(c=E_0/B_0=\omega/k\).
\(c=\dfrac{\omega}{k}=\dfrac{1.5\times10^{11}}{500}=3\times10^{8}\) m/s, so \(E_0=cB_0=3\times10^{8}\times2\times10^{-7}=60\) V/m.
Step 4: Fix the axes by the rule \(\hat{n}=\hat{E}\times\hat{B}\).
The phase \((kx+\omega t)\) travels toward \(-x\). With \(\hat{B}=\hat{y}\), we need \(\hat{E}\times\hat{y}=-\hat{x}\), which is satisfied by \(\hat{E}=\hat{z}\). So \(\vec E\) oscillates along \(+z\).
Step 5: Write \(\vec E\) and \(\lambda\).
\(E_z=60\sin(500x+1.5\times10^{11}t)\) V/m; wavelength \(\lambda=\dfrac{2\pi}{k}=\dfrac{6.28}{500}=0.0126\) m.
\[\boxed{E_z=60\sin(500x+1.5\times10^{11}t)\ \text{V/m},\ -x\ \text{direction},\ \lambda\approx1.26\times10^{-2}\ \text{m}}\]