In a cyclotron (one type of particle accelerator), a deuteron (of mass 2.00 u) reaches a final speed of 8.4% of the speed of light while moving in a circular path of radius 0.551 m. What magnitude of magnetic force is required to maintain the deuteron in a circular path?
centripetal force= magnetic force
m v^2/r=Bqv
B= m v/qr
at .084 c, I would first work it ignoring relativistic changes in mass, then rework it considering them.
mass m is not in u units, but in kg.
OK SO F=MA
M= 2.00U --> kg --> 3.32107773 × 10^-27 kg
A= v^2/r
v=8.4% speed light, => (299792458)*8.4/100...
Plug into F=MA
T=Mg[(v^2/rg)-1]
or
T=[(3.32107773×10^-27)*g][(((299792458)*8.4)^2)/(g*0.551 m))-1]
To find the magnitude of the magnetic force required to maintain the deuteron in a circular path in a cyclotron, we can use the equation for centripetal force.
The centripetal force, F, is given by the equation:
F = (mv²) / r
Where:
F is the centripetal force,
m is the mass of the deuteron,
v is the velocity of the deuteron,
and r is the radius of the circular path.
First, we need to calculate the velocity of the deuteron. The final speed given is 8.4% of the speed of light, which we can express as a fraction of c:
v = (8.4/100) * c
where c is the speed of light in a vacuum (approximately 3 x 10^8 m/s).
Plugging in the values:
v = (8.4/100) * 3 x 10^8 m/s
Now, we can calculate the value of v:
v ≈ 2.52 x 10^7 m/s
Next, we substitute this value into the equation for the centripetal force, along with the mass and radius given:
F = (m * v²) / r
In this case, the mass of the deuteron is given as 2.00 atomic mass units (u). However, we need to convert this to kilograms (kg) for the equation. The atomic mass unit is defined as 1/12th the mass of a carbon-12 atom, approximately equal to 1.66 x 10^-27 kg.
Converting the mass of the deuteron:
m = 2.00 u * (1.66 x 10^-27 kg/u)
Now, we can calculate the value of m:
m ≈ 3.32 x 10^-27 kg
Plugging in the values:
F = (3.32 x 10^-27 kg) * (2.52 x 10^7 m/s)² / (0.551 m)
Finally, we can calculate the magnitude of the magnetic force:
F ≈ 5.96 x 10^-15 N
Therefore, the magnitude of the magnetic force required to maintain the deuteron in a circular path in the cyclotron is approximately 5.96 x 10^-15 N.