An electron is accelerated from rest by a potential differen

An electron is accelerated from rest by a potential difference of 432 V. It then enters a uniform magnetic field of magnitude 228 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.

Solution

Solution:

a) Using the conservation of energy

=> 1/2*m*v^2 = V*q

=> v = sqrt(2*V*q/m)

=> v = sqrt( 2 x 432 x 1.6 x 10^-19 / 9.1 x 10 ^ -31)

=> v = 12.32 x 10^6 m/s

b) r = mv/Bq

=> r = 9.1 x 10^(-31) x 12.32 x 10^6 / 228 x 10^(-3) x 1.6 x 10^(-19)

=> r = 0.307 x 10^(-3) m

=> r = 3.1 x 10^(-4) m (approx)

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