Final answer:
The radius of curvature for a 25.0-MeV proton moving perpendicular to a 1.20-T magnetic field is found using the formula r = mv/(qB), where the mass (m) and charge (q) of the proton are constants and the velocity (v) is derived from the proton's kinetic energy.
Step-by-step explanation:
To find the radius of curvature of the path of a 25.0-MeV proton moving perpendicular to a 1.20-T magnetic field of a cyclotron, we apply the principles of circular motion and magnetic force.
The formula that relates the magnetic force exerted on a moving charge to the radius of curvature (r) of its circular path is given by:
F = qvB = mv²/r
where:
q is the charge of the proton,
v is the velocity of the proton,
B is the magnetic field strength,
m is the mass of the proton,
r is the radius of curvature.
To solve for r, rearrange the formula to get:
r = mv/(qB).
The energy (E) of the proton is given as 25.0 MeV, which can be converted to joules (J). Using the relationship E = (1/2)mv², we can solve for the velocity (v) of the proton:
v = √(2E/m)
The mass (m) and charge (q) of a proton are known constants (m = 1.6726219 × 10⁻²⁷ kg, q = 1.602176634 × 10⁻¹⁹ C).
After calculating velocity (v), substitute the values of v, m, q, and B into the radius formula to determine the radius of curvature (r) for the proton's path in the cyclotron.