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
R =mv/eB,
m= mass of proton = 1.67•10^-27 kg,
e = charge of proton = 1.6•10^-19 C
1keV =1000•1.6•10^-19 =1.6•10^-16 J
To find the radius of the proton's orbit, we can apply the concept of circular motion in a magnetic field. The equation we will use is known as the magnetic force equation:
F = qvB
Where:
F : Magnetic force acting on the particle (proton)
q : Charge of the particle (proton) = +e (elementary charge)
v : Velocity of the particle (proton)
B : Magnetic field strength
In this case, we are given the kinetic energy of the proton, not the velocity. However, we can use the kinetic energy to find the velocity using the following equation:
KE = 0.5mv^2
Where:
KE : Kinetic energy
m : Mass of the particle (proton)
v : Velocity of the particle (proton)
Given that the kinetic energy of the proton is 100 keV, we need to convert it to joules first.
1 keV = 1.6 x 10^-16 J
So 100 keV = 100 x 1.6 x 10^-16 J = 1.6 x 10^-14 J
Now, we need to find the velocity of the proton using the kinetic energy equation:
1.6 x 10^-14 J = 0.5 x (mass of proton) x v^2
The mass of a proton is approximately 1.67 x 10^-27 kg.
1.6 x 10^-14 J = 0.5 x (1.67 x 10^-27 kg) x v^2
Solving for v:
v^2 = (1.6 x 10^-14 J) / (0.5 x 1.67 x 10^-27 kg)
v^2 ≈ 3.83 x 10^12 m^2/s^2
v = √(3.83 x 10^12 m^2/s^2)
v ≈ 1.96 x 10^6 m/s
Now that we have the velocity of the proton, we can use the magnetic force equation to find the radius of its orbit:
F = qvB
(mv^2 / r) = qvB
mv / r = qB
The mass of a proton is approximately 1.67 x 10^-27 kg, the charge of a proton is approximately 1.6 x 10^-19 C, and the magnetic field strength is given as 2.3 T.
(1.67 x 10^-27 kg)(1.96 x 10^6 m/s) / r = (1.6 x 10^-19 C)(2.3 T)
(1.67 x 10^-27 kg)(1.96 x 10^6 m/s) / (1.6 x 10^-19 C)(2.3 T) = r
Calculating this expression yields the radius of the proton's orbit.