A proton moves perpendicular to a uniform magnetic field at a speed of 1.15 time

Question

A proton moves perpendicular to a uniform magnetic field at a speed of 1.15 times 107 m/s and experiences an acceleration of 2.65 times 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field. By Newton's second law, Sigma F = ma. The magnetic force is the only force accelerating the proton, so we have the following. Sigma F = ma = qvB sin 90degree We have a magnetic force in the +x-direction of F = (1.67 times 10-27 kg)( times 1013 m/s2) = times 10-14 N.

Explanation / Answer

As we know

F = Bqv

and F = ma

Therefore

Bqv = ma

Therefore

B = (ma)/(qv)

=(1.67*10^-27*2.65*10^13)/(1.6*10^-19*1.15*10^7)

= 0.02405 N

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