You plan to build a large, cheap cyclotron using the Earth's magnetic fields(1x10^-4 T) and orbiting just above the Earth's atmosphere (radius 5.7 x10 ^6 m) What energy, in volts, should protons be given to just circle the earth? q= 1.6 x10^-19 C, m= 1.67x10^-27
1 have used the formula,
Kinetic energy of orbiting proton
= B^2q^2r^2/2m joules and then converted it into MeV. Am i right?
No.
velcity=qBR/m
KE=1/2 m v^2= 1/2 (qBR)^2 m
I have simplified it as (B^2q^2r^2) / 2m
Well, lets do the algebra, I can make errors.
v=qBR/m\
v^2 = (qBR)^2 /m^2
1/2 mv^2=1/2.... you are correct.
Yes, you are correct in using the formula for kinetic energy of an orbiting proton in a magnetic field. The formula is:
Kinetic energy = (B^2 * q^2 * r^2) / (2 * m)
where B is the magnetic field strength, q is the charge of the proton, r is the radius of the orbit, and m is the mass of the proton.
To express the energy in volts, you need to convert it from joules to electron volts (eV). The conversion factor is 1 joule = 6.242 × 10^18 eV.
First, let's plug in the given values into the formula:
B = 1 × 10^-4 T (tesla)
q = 1.6 × 10^-19 C (coulombs)
r = 5.7 × 10^6 m (meters)
m = 1.67 × 10^-27 kg (kilograms)
Plugging these values into the formula, we get:
Kinetic energy = (1 × 10^-4)^2 * (1.6 × 10^-19)^2 * (5.7 × 10^6)^2 / (2 * 1.67 × 10^-27) joules
Now, simplify this value and convert it to electron volts (eV) by multiplying by the conversion factor:
Kinetic energy in eV = (Kinetic energy in joules) * 6.242 × 10^18 eV
After performing the calculations, you will get the energy in volts that should be given to protons to just circle the Earth.