Asked by TJ
a proton moves in a circle in a uniform magnetic field of 1T, pointing into the page. Its speed is 1.5 x 10 ^5 m/s. What is the radius of its path?
B q v = m v^2/r -------->
r = m v/(B q)
M=(1.67 * 10-27 )
V= (1.5 * 10^ 5)
Q= (1T)
B= (1.6 * 10-19 )
is this correct because im not getting the right answer which is 1.5 mm
B q v = m v^2/r -------->
r = m v/(B q)
M=(1.67 * 10-27 )
V= (1.5 * 10^ 5)
Q= (1T)
B= (1.6 * 10-19 )
is this correct because im not getting the right answer which is 1.5 mm
Answers
Answered by
bobpursley
I get 1.6mm, to two significant digits.
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