Question

Find the radius of the path described by a proton moving at 15 km/s in a plane perpendicular to a 400-g magnetic field.

Answer 1

Answer:

The radius described the proton moving is

r=3.91mm

Explanation:

The magnetic field that moving the proton with the velocity

v=15 km/s

To find the radius combinate the magnetic field with the mass of the proton and the charge of the proton

r = mv=eB

r = mv/eB

$r=\frac{(1.67*10^{-27})*(15\frac{km}{s})}{(1.6*10^{-19})(4*10^-2kg)}$

r = 3.9mm