Question

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.220 T. If the kinetic energy of the electron is 4.30 ✕ 10−19 J, find the speed of the electron and the radius of the circular path. (a) the speed of the electron m/s (b) the radius of the circular path µm

Answer 1

`971605.66\ \text{m/s}`

`25.1\ \mu\text{m}`

Step-by-step explanation:

m = Mass of electron =
`9.11* 10^(-31)\ \text{kg}`

B = Magnetic field = 0.22 T

K = Kinetic energy of electron =
`4.3* 10^(-19)\ \text{J}`

q = Charge =
`1.6* 10^(-19)\ \text{C}`

v = Velocity of electron

r = Radius of curved path

Kinetic energy is given by

`K=(1)/(2)mv^2\\\Rightarrow v=\sqrt{(2K)/(m)}\\\Rightarrow v=\sqrt{(2* 4.3* 10^(-19))/(9.11* 10^(-31))}\\\Rightarrow v=971605.66\ \text{m/s}`

The speed of the electron is
`971605.66\ \text{m/s}`

The force balance of the system is given by

`qvB=(mv^2)/(r)\\\Rightarrow r=(mv)/(qB)\\\Rightarrow r=(9.11* 10^(-31)* 971605.66)/(1.6* 10^(-19)* 0.22)\\\Rightarrow r=0.0000251=25.1\ \mu\text{m}`

The radius of the curved path is
`25.1\ \mu\text{m}`