Question

Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength (in T) magnetic field is needed to hold antiprotons, moving at 5.70 ✕ 107 m/s in a circular path 2.60 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge. (Enter the magnitude.)

Answer 1

Answer:

0.23 T

Explanation:

The magnetic force exerted on the antiproton must be equal to the centripetal force, since it is a circular motion, therefore we can write:

$qvB = m\frac{v^2}{r}$

where

$q=1.6\cdot 10^{-19}C$ is the charge of the antiprotons

$v=5.70\cdot 10^7 m/s$ is the speed of the antiprotons

B is the magnitude of the magnetic field

$m=1.67\cdot 10^{-27}kg$ is the antiproton mass

r = 2.60 m is the radius of the orbit

Solving the equation for B, we find the strength of the magnetic field:

$B=\frac{mv}{qr}=\frac{(1.67\cdot 10^{-27} kg)(5.70\cdot 10^7 m/s)}{(1.6\cdot 10^{-19}C)(2.60 m)}=0.23 T$