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An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 1.98 mT. If the speed of the electron is 1.53 107 m/s, determine the following.(a) the radius of the circular path

Respuesta :

Answer:

4.4 cm

Explanation:

B = 1.98 mT = 1.98 x 10^-3 T, v = 1.53 x 10^7 m/s, m = 9.1 x 10^-31 kg

q = 1.6 x 10^-19 C

(a) The force due to the magnetic field is balanced by the centrpetal force

mv^2 / r = q v B

r = m v / q B

r = (9,1 x 10^-31 x 1.53 x 10^7) / (1.98 x 10^-3 x 1.6 x 10^-19)

r = 0.044 m = 4.4 cm

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