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A proton travels with a speed of 4.90 106 m/s at an angle of 64° with the direction of a magnetic field of magnitude 0.250 T in the positive x-direction.

(a) What is the magnitude of the magnetic force on the proton?
Answer - __________ N

(b) What is the proton's acceleration?
Answer - __________ m/s2
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givenB = 0.250Tq = 1.6 * 10-19Cv = 4.90106m/s? = 640magnitude of magnetic forceF = qvB sin?= 1.76 * 10-19Nproton acclerationradius of proton pathr = mv/qB= 1.67 * 10-27kg * 4.90106m/s/1.6 * 10^-19 * 0.250T = 20.46 * 10-8maccleration = a = v2/r= 1.17 * 108 m/s2
answered by: tlyer
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