An electron is accelerated from rest by a potential difference of 350 V. It then enters a uniform magnetic field of magnitude 200 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.

Given Information:

potential difference = V = 350 V

Magnetic field = B = 200 mT = 0.200 T

Mass of electron = m = 9.11x10⁻³¹ kg

charge on electron = e = 1.60x10⁻¹⁹ C

Required Information:

velocity = v = ?

radius = r = ?

Answer:

velocity = 11.1x10⁶ m/s

radius = r = 0.000316 m

Explanation:

As we know when the potential energy of the electron becomes equal to kinetic energy of electron

eV = ½mv²

re-arranging for the speed

v² = 2eV/m

v = √ (2eV) / m

Where e is the charge on electron and m is the mass of electron

v = √ (2*1.60x10⁻¹⁹*350) / 9.11x10⁻³¹

v = 11.1x10⁶ m/s

When the electron enters in the magnetic field, then the radius of the path in which electron moves is given by

r = mv/eB

where B is the magnetic field

r = 9.11x10⁻³¹*1.11x10⁷/1.60x10⁻¹⁹*0.200

r = 0.000316 m

a) 1.11*10^7m/s

b) radius = 3.16*10^-4m

Explanation:

Assume we have an electron accelerated using potential difference of 350v which gives the ion this speed. We set the potential energy to be equal to the kinetic energy.

eV = 1/2mv^2

Rearranging the formular gives

V = Sqrt (2eV/m)

Substituting the given values

V = Sqrt (2 (1.6*10^-19) (350) / (9.11*10^-31)

V = 1.11*10^7m/s

b) The electron enters a region of uniform magnetic feild. It moves in a circular path with raduis of:

r = mv/eB

r = (9.11*10^-31kg) (1.11*10^7m/s) / (1.6*10^'19C) (200*10^-3T)

r = 3.16*10^-4m