Question

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.245 T. If the kinetic energy of the electron is 2.90 ✕ 10−19 J, find the speed of the electron and the radius of the circular path.

Answer 1

Answer

Given,

Magnetic field, B = 0.245 T

KE of the electron = 2.90 x 10⁻¹⁹ J

Speed of electron = ?

$KE = \dfrac{mv^2}{2}$

$v=\sqrt{\dfrac{2(KE)}{m}}$

$v=\sqrt{\dfrac{2\times 2.90\times 10^{-19}}{9.11\times 10^{-31}}}$

v = 7.97 x 10⁵ m/s

radius of the circular path

so,

$\dfrac{mv^2}{r}=evB$

$r=\dfrac{mv}{eB}$

$r=\dfrac{9.11\times 10^{-31}\times 7.97 \times 10^5}{1.6\times 10^{-19}\times 0.245}$

r = 1.85 x 10⁻⁵ m