In a cyclotron (one type of particle accelerator), a deuteron (of mass 2.00 u) reaches a final speed of 10.1% of the speed of light while moving in a circular path of radius 0.540 m. What magnitude of magnetic force is required to maintain the deuteron in a circular path?
To calculate the magnitude of the magnetic force required to maintain the deuteron in a circular path in a cyclotron, we can use the following formula:
F = (m * v^2) / r
Where:
F is the magnitude of the magnetic force
m is the mass of the deuteron
v is the final speed of the deuteron
r is the radius of the circular path
First, let's convert the deuteron's mass from atomic mass units (u) to kilograms (kg). We can use the conversion factor that 1 u = 1.66053906660 x 10^(-27) kg.
m = 2.00 u * (1.66053906660 x 10^(-27) kg/u)
m = 3.32107813320 x 10^(-27) kg
Next, we need to convert the final speed of the deuteron to meters per second (m/s). We're given that the deuteron reaches a final speed of 10.1% of the speed of light. Since the speed of light is approximately 3.00 x 10^8 m/s, we can calculate the final speed as follows:
v = 10.1% * (3.00 x 10^8 m/s)
v = 0.101 * (3.00 x 10^8 m/s)
v = 3.03 x 10^7 m/s
Now, we can substitute the values of m, v, and r into the formula to calculate the magnitude of the magnetic force:
F = (3.32107813320 x 10^(-27) kg * (3.03 x 10^7 m/s)^2) / 0.540 m
Calculating this expression will give us the magnitude of the magnetic force required to maintain the deuteron in a circular path in the cyclotron.