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!
I wonder if you are limited to two significant digits because of the 5.1cm, and the 6.4KVolts
What answer do you get when you calculate? Am I doing something wrong in my calculations?
You have to change the potential difference to potential energy difference, first. Multiply the potential difference by the charge.
It seems like you are on the right track but made some mistakes in your calculations. Let's go through the problem step by step to find the correct answer.
First, let's calculate the velocity of the electron using the potential difference:
V = 6.4 kV = 6.4 * 10^3 V
Mass of the electron, m = 9.10939 * 10^-31 kg
Charge of the electron, q = 1.60218 * 10^-19 C
Using the formula V = 1/2 mv^2, we can solve for v:
v = √(2V/m)
v = √((2 * 6.4 * 10^3) / (9.10939 * 10^-31))
v ≈ 5.1402 * 10^6 m/s
Now, we can use this velocity to calculate the magnetic field experienced by the electron:
B = (m/q) * (v/r)
Plugging in the values:
B = (9.10939 * 10^-31 kg) / (1.60218 * 10^-19 C) * (5.1402 * 10^6 m/s) / (5.1 * 10^-2 m)
B ≈ 1.7591 * 10^-5 T
So, the magnetic field experienced by the electron in a circular path of radius 5.1 cm is approximately 1.7591 * 10^-5 T.