Final answer:
The speed of the proton moving in a circular path perpendicular to the magnetic field can be found by equating the magnetic force to the centripetal force and solving for velocity.
Step-by-step explanation:
The question involves a proton moving in a circular path perpendicular to a uniform magnetic field. To find the speed of the proton, we can use the formula for the magnetic force that acts as the centripetal force in circular motion.
magnetic force acting on a moving charge in a magnetic field B is given by F = qvB sin(θ), where q is the charge of the proton, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field, which is 90 degrees since the movement is perpendicular to the field.
For circular motion, the centripetal force is also given by F = (mv^2)/r, where m is the mass of the proton, v is its speed, and r is the radius of curvature. By setting the magnetic force equal to the centripetal force (qvB = mv^2/r) and solving for v, we can find the proton's speed. The mass of the proton (m = 1.67 × 10^-27 kg), the magnetic field strength (B = 1.8 T), and the radius of curvature (r = 0.03 m) are given, allowing us to calculate the speed of the proton.
Substituting the known values, we get that the speed of the proton (v) equals to r multiplied by B divided by the charge of the proton (v = (rB)/q). Since the charge of a proton is approximately 1.6 × 10^-19 C, the calculation results in a speed of the proton which can be computed accordingly.