Recent content by Dom Tesilbirth

If the question had mentioned ##\overrightarrow{E}## instead of ##E_\bot##, then we could have used ##\overrightarrow{B}=\dfrac{1}{v}\widehat{k}\times \overrightarrow{E}## to get the direction of the magnetic field. But the question had only mentioned ##E_\bot##. To my understanding, knowing...

Attempt at a solution: \begin{aligned}Z=\sum ^{N}_{r=0}C\left( N,r\right) e^{-\beta \left[ -NJ+2rJ\right] }\\ \Rightarrow Z=e^{\beta NJ}\sum ^{N}_{r=0}C\left( N,r\right) e^{-2\beta rJ}\end{aligned} Let ##e^{-2\beta J}=x##. Then ##e^{-2\beta rJ}=x^{r}##. \begin{aligned}\therefore Z=e^{\beta...

I wish I could edit my question again. The actual question reads, “Consider the interference of two coherent electromagnetic waves...” Thank you for your reply.

My Try: The resultant field is given by $$\begin{aligned}\overrightarrow{E}=\overrightarrow{E}_{1}+\overrightarrow{E}_{2}=\widehat{i} E_{0}\cos \omega t+\widehat{j}E_{0}\cos \left( \omega t+\phi \right) \\ \Rightarrow E^{2}=E_{0}^{2}\cos ^{2}\omega t+E_{0}^{2}\cos ^{2}\left( \omega t+\phi...