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An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 1.95 mT. If the speed of the electron is 1.44 107 m/s, determine the following. (a) the radius of the circular path _____ cm (b) the time interval required to complete one revolution ____ s

Respuesta :

Answer: a) 42 *10^-3 m= 42 mm; b) 10.40 *10^-6 s=10.40 μs

Explanation: In order to response this problem we have to consider the Newton law for the circular movement,

Fm=m*ac where Fm is the magnetci force and ac the centripetal acceleration which is equal to v^2*R ( the raduis of teh curcular trajectory)

Fm=e*v*B considering that v and B are perpendicular

then we have:

e*v*B=m*v^2/R so

R=m*v/(e*B)= 9.1*10^-31*1.44*10^7/(1.6*10^-19*1.95*10^-3)= 42 mm

Then to calculate the time to complete one revolution ( period)

we know that ω=v*R and T= 2π/ω

then we have:

ω=1.44*10^7*42*10^-3=604.8 *10^3 rad/s

T=2*π/604.8 *10^3 = 10.4 μs

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