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

Answer:

$971605.66\ \text{m/s}$

$25.1\ \mu\text{m}$

Explanation:

m = Mass of electron = $9.11\times 10^{-31}\ \text{kg}$

B = Magnetic field = 0.22 T

K = Kinetic energy of electron = $4.3\times 10^{-19}\ \text{J}$

q = Charge = $1.6\times 10^{-19}\ \text{C}$

v = Velocity of electron

r = Radius of curved path

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 4.3\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=971605.66\ \text{m/s}$

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

The force balance of the system is given by

$qvB=\dfrac{mv^2}{r}\\\Rightarrow r=\dfrac{mv}{qB}\\\Rightarrow r=\dfrac{9.11\times 10^{-31}\times 971605.66}{1.6\times 10^{-19}\times 0.22}\\\Rightarrow r=0.0000251=25.1\ \mu\text{m}$

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