Final answer:
The radius of the circular orbit of the proton is 0.165 meters.
Step-by-step explanation:
To find the radius of the circular orbit of the proton, we can use the equation for the centripetal force experienced by a charged particle in a magnetic field. The centripetal force, which is provided by the magnetic force, is given by the equation F = qvB, where q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength. In this case, the charge of the proton is q = +e (the elementary charge), the velocity is v = 2.60×10⁶ m/s, and the magnetic field strength is B = 0.170 T. We can rearrange the equation to solve for the radius of the circular orbit:
F = qvB = (mv² / r)
qvB = mv² / r
r = mv / qB
Substituting the given values, we have:
r = (m)(2.60×10⁶ m/s) / (+e)(0.170 T)
where m is the mass of the proton. The mass of the proton is approximately 1.67×10^-27 kg, and the elementary charge is e = 1.60×10^-19 C.
Plugging in the values:
r = (1.67×10^-27 kg)(2.60×10⁶ m/s) / (1.60×10^-19 C)(0.170 T)
r = 0.165 m
Therefore, the radius of the circular orbit of the proton is 0.165 meters.