a. Find the radius of the circular path r = mv/Bq (answer in cm) b. What is the magnitude of the acceleration of the proton? q/m = 2Vacc/B^2r^2 A proton moves at a speed of 8.4x10^4 m/s as it passes through a magnetic field of 12.0 mT.
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4x10^4 m/s - Magnetic field strength, B = 12.0 mT = 12.0x10^-3 T Show more…
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