Question

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.285 T. If the kinetic energy of the electron is 2.10 10-19 J, find the speed of the electron and the radius of the circular path. (a) the speed of the electron

Answer 1

The speed of the electron is 6.79 x 10⁵ m/s

The radius of the circular path is 1.357 x 10⁻⁵ m

Step-by-step explanation:

Given;

magnetic field, B = 0.285 T

energy of electron, E = 2.10 x 10⁻¹⁹ J

The kinetic energy of the electron is calculated as;

`K.E = (1)/(2) m_eV^2`

Where;

`m_e` is the mass of electron = 9.11 x 10⁻³¹ kg

V is the speed of the electron

`K.E = (1)/(2) m_eV^2\\\\V^2 = (2.K.E)/(m_e) \\\\V = \sqrt{(2K.E)/(m_e) } \\\\V = \sqrt{(2*(2.1*10^(-19)))/(9.11*10^(-31)) }\\\\V = 6.79 *10^(5) \ m/s`

The radius of the circular path is given by;

`R = (M_eV)/(qB)`

where;

q is the charge of the electron = 1.6 x 10⁻¹⁹ C

`R = (M_eV)/(qB) \\\\R = (9.11 *10^(-31)*6.79 *10^(5))/(1.6*10^(-19)*0.285) \\\\R = 1.357 *10^(-5) \ m`