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

Answer:

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

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;

$K.E = \frac{1}{2} m_eV^2$

Where;

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

V is the speed of the electron

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

The radius of the circular path is given by;

$R = \frac{M_eV}{qB}$

where;

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

$R = \frac{M_eV}{qB} \\\\R = \frac{9.11 *10^{-31}*6.79 *10^{5}}{1.6*10^{-19}*0.285} \\\\R = 1.357 *10^{-5} \ m$