Question

An electron follows a helical path in a uniform magnetic field of magnitude 0.340 T. The pitch of the path is 6.00 µm, and the magnitude of the magnetic force on the electron is 1.85 10-15N. What is the electron's speed?

Answer 1

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, $F=1.85\times 10^{-15}\ N$

The electron follows a helical path. We have to find the speed of an electron. The formula for magnetic force is given by :

$F=B\times q\times v$

q = charge on an electron, $q=1.6\times 10^{-19}\ C$

v = velocity of an electron

$v=\dfrac{F}{Bq}$

$v=\dfrac{1.85\times 10^{-15}\ N}{0.34\ T\times 1.6\times 10^{-19}\ C}$

v = 34007.35 m/s

Hence, this is the required solution.