024 (part 1 of 2) 10.0 points In a nuclear research laboratory, a proton moves i

Question

024 (part 1 of 2) 10.0 points In a nuclear research laboratory, a proton moves in a particle accelerator through a mag-netic field of intensity 0.226 T at a speed of 4.48 × 107 m/s. The charge of a proton is 1.60218 × 1019 C. If the proton is moving perpendicular to the field, what force acts on it? Answer in units of N.

Part 2

If the proton of mass 1.67262 × 1027 kg con- tinues to move in a direction that is consis- tently perpendicular to the field, what is the radius of curvature of its path?.

Answer in units of m.

Explanation / Answer

Part 1:

The force = charge * velocity* Magnetic field = 1.6e-19 * 4.48e7 * .226 = 1.62e-12 N

Part 2:

qvB = mv^2/r

r = mv/qB = 2.069 m

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