Question

What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.4 T?

Answer 1

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

$a = \frac{qvB}{m} \\\\a = \frac{1.602 \times 10^{-19} \times6.5 \times 1.4}{1.673\times10^{-27}}\\\\a = 8.714 \times 10^8 \ m/s ^2$

Therefore, the acceleration of the proton is 8.714 x 10⁸ m/s².