Question

You must give your answer starting basic physics formula and step by step in detail.

1) A magnetic field B=0.6T, and an electron with a velocity of 50,000,000 m/s enters the magnetic field with an angle of 90 degrees perpendicular to the magnetic field. Calculate the radius of the curvature of the path of the electron.

Answer 1

The magnetic field strength = B = 0.6 T

The velocity of the electron = v = 50,000,000 m/s = 5 x 10⁷ m/s

The mass of electron = m = 9.1 x 10^-31 kg

The magnitude of the charge of the electron = q = 1.6 x 10^-19 C

The radius of curvature of the electron in the magnetic field is given by the formula:

where m is the mass of the electron, v is the velocity, q is the charge of electron and B is the magnetic field strength.