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Answer:
The acceleration of the proton is 1.403 x 10⁹ m/s²
Explanation:
Given;
speed of proton, v = 7.7 m/s
magnitude of magnetic field, B = 1.9 T
Magnetic force of moving proton is given by;
F = qvBsinθ
Centripetal force on the moving proton is given by;
where;
q is charge of the proton = 1.602 x 10⁻¹⁹ C
m is mass of proton = 1.67 x 10⁻²⁷ kg
Therefore, the acceleration of the proton is 1.403 x 10⁹ m/s²