Answer:
The electron's speed is 34007.35 m/s
Explanation:
It is given that,
Magnetic field, B = 0.34 T
Magnetic force on the electron, [tex]F=1.85\times 10^{-15}\ N[/tex]
The electron follows a helical path. We have to find the speed of an electron. The formula for magnetic force is given by :
[tex]F=B\times q\times v[/tex]
q = charge on an electron, [tex]q=1.6\times 10^{-19}\ C[/tex]
v = velocity of an electron
[tex]v=\dfrac{F}{Bq}[/tex]
[tex]v=\dfrac{1.85\times 10^{-15}\ N}{0.34\ T\times 1.6\times 10^{-19}\ C}[/tex]
v = 34007.35 m/s
Hence, this is the required solution.
