## A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60-T magnetic

Question

A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60-T magnetic field. What is the radius of the resulting path

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3 years 2021-07-31T23:45:16+00:00 2 Answers 39 views 0

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2×10^5m/s

= 1.67*10^-27×6.92*10^6/1.6*10^-16 x0.6

=0.012m

=1.2cm

0.012m

Explanation:

To find the radius of the path you can use the formula for the radius od the trajectory of a charge that moves in a constant magnetic field:

m: mass of the proton 9.1*10^{-31}kg

q: charge of the proton 1.6*10^{-19}C

B: magnitude of the magnetic field 0.60T

v: velocity of the proton

In order to use the formula you need to calculate the velocity of the proton. This can be made by using the potential difference and charge, that equals the kinetic energy of the proton:

Then, by replacing in the formula for the radius you obtain: