Physics_A2_Electromagnetism

[ashutoshamatya]

Electrons are accelerated through a potential difference of 220 V. They then pass into a region of uniform magnetic flux of flux desity 0.54 mT. The path os the electron is normal to the magnetic field. Given that the charge of electron is 1.6 x 10^-31 kg calculate:
(a) the speed of the accelerated electron.
(b) the radius of the circular path in the magnetic field.

 

[beacon_of_light]

Hey, charge of an electron is 1.6 x10^-31 KG ??? correct that...
and then use the following formulas...
a) 1/2 mv^2 = qV ... v is the speed of electron while V is the potential difference
b) centripetal force = magnetic force
mv^2/r = BqV

I hope u'll get the correct answers!