Respuesta :
Answer:
The magnetic field strength required to hold anti-protons, moving at 5.70 ✕ 10⁷ m/s in a circular path of 3.20 m in radius is 0.186 T.
Explanation:
To solve the question we note that the magnetic force on a moving charge is given by
F = q·v·B
Where
q = Charge
v = Velocity of the charge =5.70 ✕ 10⁷ m/s
B = Magnetic field
Based on Newton's second law,
Force = Mass, m × Acceleration, a = m × a
Where:
a = Acceleration
m = Mass of anti-proton = Mass of proton = 1.6726219 × 10⁻²⁷ kg
We note that for circular motion, acceleration a is given by
[tex]\alpha = \frac{v^{2} }{r}[/tex] .
Where:
r = Radius = 3.20 m
Therefore, for the circular motion, force, F = [tex]\frac{m\cdot v^{2} }{r}[/tex]
Equating the magnetic force equation to the circular force equation, we have
[tex]\frac{m\cdot v^{2} }{r}[/tex] = q·v·B So that, we find B by making the subject of the formula as follows
[tex]B= \frac{m\times v^{2} }{r\times q \times v}[/tex] . Which gives
[tex]B= \frac{m\times v }{r\times q} = \frac{(1.6726219 \times 10^{-27}) \times (5.7\times 10^{7}) }{(3.20)\times (1.602\times 10^{-19}) }[/tex] = 0.186 T
The magnetic field strength is
B = 0.186 T
