What is the acceleration of a proton moving with a speed of 7.7 m/s at right angles to a magnetic field of 1.9 T

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