Answer to Question #308912 in Electricity and Magnetism for Elle

Question

A high-speed electron having a mass of 9 x 10^ -31 kg is moving at right angle to 0.75-T magnetic field and has a speed of 2.5 x 10^7m/s. What is the size of the force acts on the high-speed electron? What is the magnitude of acceleration of the particle?

Answer 1

Answer

the size of the force acts on the high-speed electron

$F=qvBsin\theta\\=1.6*10^{-19}*2.5*10^7*0.75*1\\=3*10^{-19} N$

Acceleration

$a=\frac{F}{m}\\=\frac{3*10^{-19}}{9.1 *10^{-31} }\\=0.33*10^{12}m/s^2$