2aprova-gaba-manha-1

Problema 4\nB = 4,00 x 10^-3 T\nF = 4,0 x 10^-15 N\na)\nF_m = q v x B\ncom q > 0, t \u2265 0 \nF_m\napontando para o centro\nlogo a carga da part\u00edcula e\u00e9 positiva\nm = 10 x 10^-27 kg\nF_m = q v B \u2192 v = F_m/qB\nr = m v^2/qB\nr = (m/qB)F_m\nr = 10 x 10^-27 kg x 4,0 x 10^-15 N/(1,6 x 10^-19 x 4 x 10^-3)^2\nr = 97,65 m\nb)\n\u03b3 = 2 \u03c0 r/v^2 = 2 \u03c0 qB/F_m\n\u03b3 = 9,8 x 10^-5 s ~ 1 x 10^-4 s ~ 0,910 m Problema 2\nR_0 - R_2 = 2\u03c1n + k - 2\u03c1l + 2\u03c1n - R_0\nR_0 - R_1 = 4\u03c1n - 2\u03c1l\nR_0 - R_1 + 2\u03c1l = 4\u03c1n\nx = (R_0 - R_1 + 2\u03c1l)/4C\nx ~ 8,11 km\na)\nR = 2x0 + 110 \u03a9 - 2,13 km - 12 km\nR = 9,0 \u03a9\nProblema 3\nR = 1,40 x 10^6 \u03a9\nC = 1,80 x 10^-6 F\n\u03ce = 12V\nQ_max = CE = 1,80 x 10^-6 F. 12,0 V = 2,16 x 10^-5 C\na)\nQ = Q_max (1 - e^-t/tau)\nQ/Q_max = 1 - e^-t/tau\nb)\nQ = 16 x 10^-6 C\nt = 1,40 x 10^6 \u03a9 . 1,80 x 10^-6 F ln[-1/(4x10^-6)] - t/tau = ln[-Q/Q_max] \nQ = 2,10 x 10^-10\nk\nt = 0,22 s