Teaser for Relativistic Electrodynamics

First, determine the velocity of Ineng relative to S, using Einstein velocity addition:

$Vis = \frac{Vit + Vts}{1 + Vit Vts / c^2} = 0.8 c$

Suppose that the line charge has a linear charge density $\lambda$ in S.

In an inertial frame X moving at a speed v relative to S, in a direction parallel to the line charge, the line charge will have linear charge density $\lambda \gamma (v)$ , due to length contraction.

Applying Gauss' Law and Ampere's Law in X, we obtain:

$E (v) = \frac{\lambda \gamma (v)}{2 \pi R \epsilon}$

$B (v) = \frac{\mu v \lambda \gamma (v)}{2 \pi R}$

$B (v) = \frac{v}{c^2} E (v)$

Using the expression for the electric field, we can relate the observed electric field magnitudes in different inertial frames:

$E (0) = 0.5 \sqrt{3} E (0.5 c) = 0.6 E (0.8 c)$

Given that $E (0.5c) = 1 V / m$ , we get, $E (0.8 c) = \frac{2.5}{\sqrt{3}} V / m$ and $c B (0.8 c) = \frac{2}{\sqrt{3}} V / m$

These will yield a final answer of 53.