Explanation:
It is given that,
Charge on electron, [tex]q=16\times 10^{-20}\ C[/tex]
Mass of the electron, [tex]m=9.1\times 10^{-31}\ kg[/tex]
Speed of the electron, [tex]v=0.5\ Mm/s=0.5\times 10^6\ m/s[/tex]
Magnetic field, B = 1 T (directed out of the page)
Let F is the magnetic force acting on the electron. It is given by :
[tex]F=qvB\ sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
[tex]F=qvB[/tex]
[tex]F=16\times 10^{-20}\ C\times 0.5\times 10^6\ m/s\times 1\ T[/tex]
[tex]F=8\times 10^{-14}\ N[/tex]
Using the right hand rule, the direction of magnetic force is upward to the plane of the paper. Also, the electron will follow the circular path. It is given by :
[tex]r=\dfrac{mv}{qB}[/tex]
[tex]r=\dfrac{9.1\times 10^{-31}\ kg\times 0.5\times 10^6\ m/s}{16\times 10^{-20}\ C\times 1\ T}[/tex]
[tex]r=2.84\times 10^{-6}\ m[/tex]
Hence, this is the required solution.