Question

An electron is traveling at 1.2×1000m/s when it enters an electric field the shows it the field has a magnitude of 5200N/C how far does the electron penetrate the field before it stops

Answer 1

SOLUTION-;

• Let’s start by focusing on the alpha-particle entering the field near the bottom of the picture. First, point your thumb up the page. In order for your palm to open to the left where the centripetal force (and hence the magnetic force) points, your fingers need to change orientation until they point into the page. This is the direction of the applied magnetic field.
• The period of the charged particle going around a circle is calculated by using the given mass, charge, and magnetic field in the problem. This works out to be
• T=\frac{2\pi m}{qB}=\frac{2\pi \left(6.64\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-27}\text{kg}\right)}{\left(3.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\text{C}\right)\left(0.050\phantom{\rule{0.2em}{0ex}}\text{T}\right)}=2.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\text{s.}
• However, for the given problem, the alpha-particle goes around a quarter of the circle, so the time it takes would be
• t=0.25\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2.61\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\text{s}=6.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\text{s.}