select the correct answer.\nwhat is the magnetic force acting on an electron if its speed is 3.0×10^6…

Explanation:

Step1: Recall the formula for magnetic force

The formula for the magnetic force on a charged - particle moving perpendicular to a magnetic field is $F = qvB$, where $q$ is the charge of the particle, $v$ is the velocity of the particle, and $B$ is the magnetic field strength.

Step2: Substitute the given values

Given $q=- 1.6\times10^{-19}\text{ C}$, $v = 3.0\times10^{6}\text{ m/s}$, and $B = 0.020\text{ T}$. $F=(-1.6\times10^{-19}\text{ C})\times(3.0\times10^{6}\text{ m/s})\times(0.020\text{ T})$

Step3: Calculate the product

First, calculate $(1.6\times3.0\times0.020)=1.6\times0.060 = 0.096$. Then, calculate the exponent of 10: $10^{-19}\times10^{6}=10^{-19 + 6}=10^{-13}$. So, $F=-0.096\times10^{-13}\text{ N}=-9.6\times10^{-15}\text{ N}$.

Answer:

C. $F=-9.6\times 10^{-15}\text{ newtons}$