A proton travels with a speed of 5.05 ✕ 106 m/s at an angle of 64° with the direction of a magnetic field of magnitude 0.160 T in the positive x-direction.(a) What is the magnitude of the magnetic force on the proton? N
(b) What is the proton's acceleration? m/s2

A. 1.19 * 10^(-13) N

B. 7.12 * 10^(15) m/s²

Explanation:

Parameters given:

Speed, v = 5.05 * 10^6 m/s

Angle, A = 64°

Magnetic field strength, B = 0.160T

Mass of proton, m = 1.673 * 10^(-27) kg

Charge of proton, q = 1.6023 * 10^(-19) C

A. Magnetic force is given as:

F = q*v*B*sinA

F = 1.6023 * 10^(-19) * 5.05 * 10^6 * 0.160 * sin64

F = 1.19 * 10^(-13) N

B. Force is generally given as:

F = m*a

Hence, we can find acceleration, a, by making it the subject of formula:

a = F/m

a = (1.19 * 10^(-13))/(1.673 * 10^-27)

a = 7.12 * 10^15 m/s²

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