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 6.10 ✕ 107 m/s in a circular path 1.70 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge. (Enter the magnitude.)

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AdricG29295 AdricG29295 · Answer 1 · 2022-11-03T10:15:40+01:00

Data:

B=?, v=6.10*10^7 m/s, R=1.70m, m=1.67*10^(-27)kg, q=-1.6*10^(-19)C

Answer:

Firstly, we need to remember the formula for the Magnetic force, which is:

[tex]F_m=|q|vB[/tex]

However, in this scenario the magnetic force will act as a centripetal force, thus:

[tex]F_m=\frac{mv^2}{R}[/tex]

If we equal both of them:

[tex]|q|vB=\frac{mv^2}{R}\Rightarrow B=\frac{mv}{|q|R}[/tex]

Replacing our values:

[tex]B=\frac{1.67*10^{-27}*6.1*10^7}{1.6*10^{-19}*1.7}=0.3745T[/tex]

Then, our magnetic field will be B=0.3745T