Answer:
a. [tex]F = 1.32x10^{-12} N[/tex]
b. [tex]a = 1.45x10^{18} m/s^2[/tex]
Explanation:
Knowing and using the speed light as a [tex]3.0 x10^8 m/s[/tex] and the mass and charge of electron are [tex]q = 1.6 x10^{-19}C[/tex], [tex]m= 9.1 x10^{-31} kg[/tex]
So using the equation to determine the force
a.
[tex]F = q*v*\beta[/tex]
[tex]F = 1.6x10^{-19}C * 5.5 *3.0x10^8m/s* 0.0050T[/tex]
[tex]F = 1.32x10^{-12} N[/tex]
b.
Using the equation of force to find the acceleration
[tex]F=m*a[/tex]
[tex]a= \frac{F}{m}[/tex]
[tex]a = \frac{1.32x10^{-12}N}{9.1x10^{-31}kg}[/tex]
[tex]a = 1.45x10^{18} m/s^2[/tex]