A proton with kinetic energy 100 keV is moving in a plane perpindicular to a uniform magnetic field of magnitude 2.3 T. Find the radius of its orbit.
Not sure what equation to use or how to solve.
To find the radius of the proton's orbit, we can use the equation for the cyclotron frequency:
ω = qB/m
where ω is the angular velocity, q is the charge of the particle, B is the magnetic field strength, and m is the mass of the proton.
The cyclotron frequency is related to the angular velocity as:
ω = 2πf
where f is the frequency. And the relationship between angular velocity, frequency, and velocity is given by:
ω = v/r
where v is the velocity of the proton and r is the radius of the orbit.
To find the velocity of the proton, we can use the kinetic energy:
KE = (1/2)mv^2
To solve for the radius, we need to relate all of these equations. Here's how you can do it:
1. Calculate the velocity of the proton using the given kinetic energy:
KE = (1/2)mv^2
100 keV = (1/2)(1.602 x 10^-19 C)v^2
v^2 = (2 * 100 keV) / (1.602 x 10^-19 C)
Convert keV to joules: 1 keV = 1.602 x 10^-16 J
v^2 = (2 * 100 * 1.602 x 10^-16 J) / (1.602 x 10^-19 C)
v^2 = 2 x 10^-14 J / C
v = √(2 x 10^-14 J / C)
2. Calculate the cyclotron frequency:
ω = qB/m
Here, q = 1.602 x 10^-19 C (charge of a proton) and B = 2.3 T (given magnetic field strength). m is the mass of the proton, which is approximately 1.67 x 10^-27 kg.
ω = (1.602 x 10^-19 C)(2.3 T) / (1.67 x 10^-27 kg)
3. Relate the angular velocity, velocity, and radius:
ω = v/r
Substituting the values calculated above, we get:
(1.602 x 10^-19 C)(2.3 T) / (1.67 x 10^-27 kg) = √(2 x 10^-14 J / C) / r
Simplify and solve for r:
r = √((2 x 10^-14 J / C) / [(1.602 x 10^-19 C)(2.3 T) / (1.67 x 10^-27 kg)])
Calculate the result to find the radius of the proton's orbit.