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

Given,

Magnetic field, B = 0.245 T

KE of the electron = 2.90 x 10⁻¹⁹ J

Speed of electron = ?

v = 7.97 x 10⁵ m/s

radius of the circular path

so,

r = 1.85 x 10⁻⁵ m

the components of a vector are the parts of the vectors that are to each other.

b perpendicular