Radiative Electric and Manetic field from an accelerating electron Note: I've started a meta post to discuss this type of question. Yes, it looks like it's straight out of a worksheet from a physics class. That's because it is, and it's also why I posted the answer as well. Ok, now on to the question.



 A: a) The time it will take for the detectors to detect it will always be
$$\frac{\text{Distance from detectors}}{\text{Speed of light}} = \frac{10m}{3*10^8 \frac{m}{s}}  \approx\boxed{ 3.3333× 10^-8}$$

b) We can start out using the equation:
$$\vec E_{rad} = \frac{1}{4 \pi \epsilon_0}\frac{-q \vec a_{\perp}}{c^2r}$$
$$\text{since $\vec a$ is pointing $126^{\circ}$ away from $A$, $\hat a_{\perp}$ is going to be $\left <1,0,0 \right>$}$$
$$\implies \vec E_{rad} = \frac{1}{4 \pi \epsilon_0}\frac{-q \left <5*10^9*cos(126-90),0,0 \right>}{c^2r} = \frac{1}{4 \pi \varepsilon_0}\frac{1.6 * 10^{-19} * \left <cos(36)5*10^9,0,0 \right>}{(3*10^8)^2(10)}$$
$$= \boxed{\left< 6.47214^{-18}, 0, 0\right>\frac{N}{C}}$$
c) We will use our previous answer to get the radiative magnetic field using the equation:
$$\vec E_{rad} = c \vec B_{rad}$$
$$\text{Since $\vec E=\vec v \times \vec B$, the direction of propagation is in the $-y$ direction, $\vec E_{rad}$ is in the $+x$ direction, from the right hand rule we can find that $B$ must be in the +z direction.}$$
$$\text{Therefore, $\vec B_{rad}$ is going to be: }$$
$$\implies \vec B_{rad} = \frac{\vec E_{rad}}{c}\hat r = \frac{6.47214^{-18}\frac{N}{C}}{3*10^8 \frac{m}{s}} = \boxed{\left<0, 0,2.15738 *10^{-26}\right>}$$
