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 1.5 T?

Answer 1

The acceleration of the proton is 9.353 x 10⁸ m/s²

Step-by-step explanation:

Given;

speed of the proton, u = 6.5 m/s

magnetic field strength, B = 1.5 T

The force of the proton is given by;

F = ma = qvB(sin90°)

ma = qvB

where;

m is mass of the proton, = 1.67 x 10⁻²⁷ kg

charge of the proton, q = 1.602 x 10⁻¹⁹ C

The acceleration of the proton is given by;

`a = (qvB)/(m)\\\\a = ((1.602*10^(-19))(6.5)(1.5))/(1.67*10^(-27))\\\\a = 9.353*10^8 \ m/s^2`

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²