Constants PartA A beam of protons traveling at 1.40 km/s enters a uniform magnet

Question

Constants PartA A beam of protons traveling at 1.40 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (the figure (Figure 1)). The beam travels a distance of 1.80 cm while in the field What is the magnitude of the magnetic field? B 1.33.10-3 Submit Previous Answers Request Answer X Incorrect Try Again; 5 attempts remaining Provide Feedback Next > Figure 1 of 1>

Explanation / Answer

Given

Charge of proton is q = 1.6 * 10^-19 C

mass   m = 1.67 * 10^-27 kg

velocity of proton is v = 1.40 km/s = 1400 m/s

Distance travelled is d = 1.8 cm = 0.018 m

This distance is quarter cycle so

d = (pi * r)/2 = 0.018

so radius r = 0.0114 m

Here the centripetal force is provided by magnetic force so

m * v^2 / r = q * v * B

we get

B = (m * v) / (q * r)

B = (1.67 * 10^-27 * 1400) / (1.6 * 10^-19 * 0.0114)

B = 1.28 * 10^-3 T

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