A proton moves perpendicular to a uniform magnetic field at a speed of 115 times

Question

A proton moves perpendicular to a uniform magnetic field at a speed of 115 times 10 7 m/s and experiences an acceleration of 2.65 times 10 13 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field. Step 1 By Newton's second law, . The magnetic force is the only force accelerating the proton, so we have the following. We have a magnetic force in the + x-direction of Step 2 Solving for the magnitude of the magnetic field, we have the following.

Explanation / Answer

B=(4.43*10^-14)/[(1.6*10^-19)(1.15*10^7)]=0.02407 T=24.07 mT

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