Answer:
[tex]F=1.6\times 10^{-15}\ N[/tex]
Explanation:
Given that,
Speed of the electron, [tex]v=5\times 10^4\ m/s[/tex]
Magnetic field, B = 0.20 T
We need to find the magnitude of magnetic force on the electron. The formula for the magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
[tex]F=qvB[/tex], q is the charge on electron
[tex]F=1.6\times 10^{-19}\times 5\times 10^4\times 0.2[/tex]
[tex]F=1.6\times 10^{-15}\ N[/tex]
So, the magnitude of the magnetic force on the electron is [tex]1.6\times 10^{-15}\ N[/tex]. Hence, this is the required solution.
