Answer:
The velocity of the proton is approximately 9.28 meters per second.
Step-by-step explanation:
To find the velocity of the proton moving in a circle in a magnetic field, we can use the equation for the magnetic force on a charged particle moving perpendicular to a magnetic field:
F = q * v * B
where:
F is the magnetic force on the particle,
q is the charge of the particle (in Coulombs),
v is the velocity of the particle (in m/s), and
B is the magnetic field strength (in Tesla).
The magnetic force provides the necessary centripetal force to keep the proton moving in a circle. The centripetal force is given by:
F_c = (m * v^2) / r
where:
m is the mass of the proton (in kg), and
r is the radius of the circle (in meters).
Since the magnetic force (F) is equal to the centripetal force (F_c), we can set these two equations equal to each other and solve for the velocity (v):
q * v * B = (m * v^2) / r
Now, let's plug in the given values:
q (charge of the proton) = +1.60x10^-19 C
B (magnetic field strength) = 0.800 T
m (mass of the proton) = 1.67x10^-27 kg
r (radius of the circle) = 0.0150 m
Now, we can solve for v:
1.60x10^-19 C * v * 0.800 T = (1.67x10^-27 kg * v^2) / 0.0150 m
Let's solve for v:
1.60x10^-19 C * 0.800 T * 0.0150 m = 1.67x10^-27 kg * v^2
v^2 = (1.60x10^-19 C * 0.800 T * 0.0150 m) / 1.67x10^-27 kg
v^2 = 1.44x10^-25 C T m / 1.67x10^-27 kg
v^2 ≈ 8.632x10^1 m^2/s^2
v ≈ √(8.632x10^1 m^2/s^2)
v ≈ 9.28x10^0 m/s
v ≈ 9.28 m/s
So, the velocity of the proton is approximately 9.28 meters per second.