Energy density of the magnetic field:
"u_B=\\frac{B^2}{2\\mu_0}."According to the problem, the electric and magnetic field must provide 1150 km/s per kilogram. Apply conservation of energy per unit mass:
"u_B+u_E=\\frac{v^2}{2},\\\\\n\\space\\\\\nu_E=\\frac{v^2}{2}-u_B=\\frac{v^2}{2}-\\frac{B^2}{2\\mu_0}." On the other hand, the electric field energy density is
"u_E=\\frac{\\epsilon_0E}{2},\\\\\n\\space\\\\E=\\frac{2u_E}{\\epsilon_0}=\\frac{1}{\\epsilon_0}\\bigg(v^2-\\frac{B^2}{\\mu_0}\\bigg)=1.49\\cdot10^{23}\\text{ V\/m}."
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