Question

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

Answer 1

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;

$F = m(\frac{v^2}{r})\\\\F = m(a_c) \\\\a_c \ is \ the \ centripetal \ acceleration$

$qvBsin\theta = ma_c\\\\ac = \frac{qvBsin(90)}{m}$

where;

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

m is mass of proton = 1.67 x 10⁻²⁷ kg

$ac = \frac{(1.602*10^{-19})(7.7)(1.9)sin(90)}{1.67*10^{-27}}\\\\a_c = 1.403*10^{9} \ m/s^2$

Therefore, the acceleration of the proton is 1.403 x 10⁹ m/s²