Question

An electron is accelerated by a 6.4 kV potential difference. The charge on an electron is 1.60218 × 10^−19 C and its mass is 9.10939 × 10^−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.1 cm?
Answer in units of T.

*** I tried using B=(1/r)Sqrt(2mV/q^2) and I'm not sure what I am doing wrong.



centripetal force= magnetic force
m v^2/r=Bqv

B= m/q * v/r

you have to change the potential difference to velocity first
6.4E3*q=1/2 m v^2

or v=sqrt 2*6.4E3*2/m

1/r * sqrt (m*2*6.4E3/q^2)

What am I doing wrong? Sqrt (9.10939x10^-31)x(2)x(6400)/(1.60218x10^-19)sqrd =673966.5256

ans * (1/.051) = 13215029.91 Wrong answer. I have entered four wrong answers. I can not enter another wrong answer and pass this homework. pls help!

Answer 1

I wonder if you are limited to two significant digits because of the 5.1cm, and the 6.4KVolts

Answer 2

What answer do you get when you calculate? Am I doing something wrong in my calculations?

Answer 3

You have to change the potential difference to potential energy difference, first. Multiply the potential difference by the charge.