Asked by cr
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 usind B=(1/r)Sqrt(2mV/q) and I'm not sure what I am doing wrong.
Answer in units of T.
*** I tried usind B=(1/r)Sqrt(2mV/q) and I'm not sure what I am doing wrong.
Answers
Answered by
bobpursley
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)
So it appears you need to square q in the sqrt sign. Check my work.
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)
So it appears you need to square q in the sqrt sign. Check my work.
Answered by
cr
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!
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!
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