A proton with a speed of 2.20 times 10^6 m/s is shot a region between two plates

Question

A proton with a speed of 2.20 times 10^6 m/s is shot a region between two plates that are separated by a distance of 0.165 m. As the drawing shows, a magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field (in T), so the proton just misses colliding with the opposite plate? (in T) A 9.60 times 10^-2 B 1.39 times 10^-1 C 2.02 times 10^-1 D 2.93 times 10^-1 E 4.24 times 10^-1 F 6.15 times 10^-1 G 8.92 times 10^-1 H 1.29 I 1.88 J 2.72

Explanation / Answer

magnetic foec fb = qvB = mv^2/r

B = mv/qr

B = 1.67*10^-27*2.2*10^6/1.6*10^-19*0.165

B = 0.139 T

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