In a cloud chamber experiment, a proton enters a uniform 0.260 magnetic field directed perpendicular to its motion. You measure the proton's path on a photograph and find that it follows a circular arc of radius 6.38 . How fast was the proton moving?
To find how fast the proton was moving in the cloud chamber experiment, we can use the principles of centripetal force and magnetic force acting on a charged particle.
The centripetal force acting on a charged particle moving in a magnetic field can be written as:
F_c = (mv^2) / r
Where:
F_c is the centripetal force
m is the mass of the particle (in this case, the proton)
v is the velocity of the proton
r is the radius of the circular arc
In this case, we know the radius, r = 6.38, and we are asked to find the velocity, v.
The magnetic force acting on a charged particle moving in a magnetic field can be written as:
F_m = qvB
Where:
F_m is the magnetic force
q is the charge of the particle (in this case, the charge of the proton)
v is the velocity of the proton
B is the magnetic field
Since the proton is moving perpendicular to the magnetic field, the magnetic force is equal to the centripetal force:
qvB = (mv^2) / r
Simplifying the equation, we can solve for the velocity:
v = (qrB) / m
Given:
q = the charge of the proton = 1.602 × 10^-19 C (Coulombs)
B = magnetic field = 0.260 T (Tesla)
m = mass of the proton = 1.67 × 10^-27 kg (kilograms)
Now we can substitute the known values into the equation to find the velocity:
v = ((1.602 × 10^-19 C) * (0.260 T)) / (1.67 × 10^-27 kg)
Calculating this expression will give us the velocity of the proton.