Answer:
The acceleration of the proton is 8.714 x 10⁸ m/s².
Explanation:
Given;
speed of the proton, v = 6.5 m/s
magnetic field strength, B = 1.4 T
The magnetic force on the proton is given as;
F = qvB
The force of the moving proton is given as;
F = ma
Thus, ma = qvB
where;
m is mass of proton = 1.673 x 10⁻²⁷ kg
q is charge of proton = 1.602 x 10⁻¹⁹ J
a is the acceleration of the proton
[tex]a = \frac{qvB}{m} \\\\a = \frac{1.602 \times 10^{-19} \times6.5 \times 1.4}{1.673\times10^{-27}}\\\\a = 8.714 \times 10^8 \ m/s ^2[/tex]
Therefore, the acceleration of the proton is 8.714 x 10⁸ m/s².
