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 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.)

Answer 1

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:

`F_m=|q|vB`

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

`F_m=(mv^2)/(R)`

If we equal both of them:

`|q|vB=(mv^2)/(R)\Rightarrow B=(mv)/(|q|R)`

Replacing our values:

`B=(1.67*10^(-27)*6.1*10^7)/(1.6*10^(-19)*1.7)=0.3745T`

Then, our magnetic field will be B=0.3745T