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Electromagnetic Induction. Maxwells Equations

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.

Reveal Answer
 3.367. (a) E′=E1−β2cos⁡2α1−β2=9kV/m;tan⁡α′=tan⁡α1−.β2 ,  whence α≈51∘; (b) B′=βEsin⁡αc1−β2=14μT . \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}
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