Problem 3.367

In an inertial reference frame $K$ there is only a uniform electric field $E=8 \mathrm{kV} / \mathrm{m}$ in strength. Find the modulus and direction (a) of the vector $\mathbf{E}^{\prime}$ , (b) of the vector $\mathbf{B}^{\prime}$ in the inertial reference frame $K^{\prime}$ moving with a constant velocity $\mathbf{v}$ relative to the frame $K$ at an angle $\alpha=45^{\circ}$ to the vector $\mathbf{E}$ . The velocity of the frame $K^{\prime}$ is equal to a $\beta=0.60$ fraction of the velocity of light.

\begin{aligned} &\text { 3.367. (a) } E^{\prime}=E \sqrt{\frac{1-\beta^{2} \cos ^{2} \alpha}{1-\beta^{2}}}=9 \mathrm{kV} / \mathrm{m} ; \quad \tan \alpha^{\prime}=\frac{\tan \alpha}{\sqrt{1-. \beta^{2}}} \text { , }\\ &\text { whence } \alpha \approx 51^{\circ} ; \quad \text { (b) } B^{\prime}=\frac{\beta E \sin \alpha}{c \sqrt{1-\beta^{2}}}=14 \mu \mathrm{T} \text { . } \end{aligned}

Electromagnetic Induction. Maxwells Equations