At time t = 0, a proton is moving along +j^Hat at some instant with a speed equa

Question

At time t = 0, a proton is moving along +j^Hat at some instant with a speed equal to 3.0 times 10^6 at a location where the magnetic field from a big magnet is 0.75 T directed along -k^Hat·The proton has a charge of +1.60 times 10^-19 C and a mass of 1.67 times 10^-27kg. (a) what are the direction and magnitude of the force on the proton at t = 0? (b) Assuming no other forces significantly affect the motion of the proton what is the radius of curvature of its trajectory, assuming the field given above to be uniform? (c) What is the direction and magnitude of the uniform electric field that must applied at t = 0 so that the trajectory of the proton is a straight line? Find the field:

Explanation / Answer

1. v = 3 x 10^6 m/s j

B = - 0.75 T k

q = +1.6 x 10^-19 C

(A) F = q (v X B)

F = (1.6 x 10^-19) [ (3 x 10^6 j) x (-0.75k)]

= - 3.6 x 10^-13 N i

magnitude = 3.6 x 10^-13 N

direction = along -i

(B) magntic force = m v^2 / r

3.6 x10^-13 = (1.67 x 10^-27) (3 x 10^6)^2 / r

r = 0.042 m

(C) Fe + Fb = 0

q E + (-3.6 x 10^-13i) = 0

E = 3.6 x 10^-13 i / (1.6 x 10^-19)

E = 2.25 x 10^6 N/C i

magnitude = 2.25 x 10^6 N/C OR V/m

direction = along i

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