At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.80 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.05 ✕ 10−5 T?

The velocity of the proton is 4.25 x 10³ m/s

Explanation:

Given;

speed of electron, Ve = 7.80 x 10⁶ m/s

magnetic field strength, B = 1.05 x 10⁻⁵ T

In cyclotron motion, the radius of a charged particle path in a magnetic field is given as:

where;

m is the mass of the particle

v is the speed of the particle

q is the charge of the particle

B is the strength of the magnetic field

If both electron and proton travel in a path with same radius, our equation becomes;

Me is the mass of electron = 9.1 x 10⁻³¹ kg

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

Therefore, the velocity of the proton is 4.25 x 10³ m/s

we need to know the time interval during which the force acts. impulse of force = momentum change