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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) 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 magnetic field is needed to hold antiprotons, moving at $${\rm{5}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{7}}}{\rm{ m/s}}$$ in a circular path $${\rm{2}}{\rm{.00 m}}$$ in radius? Antiprotons have the same mass as protons but the opposite (negative) charge. (b) Is this field strength obtainable with today’s technology or is it a futuristic possibility?

(a) The magnetic field strength is obtained as: $${\rm{0}}{\rm{.261 T}}$$.

(b) The magnetic field strength is nowadays obtainable.

See the step by step solution

## Step 1: Effect of magnetic field on the charged particle

When a charged particle moves in the presence of a magnetic field, a force acts on the mong charge particle given by Fleming’s right-hand rule and which deviates the path of the moving charged particle.

## Step 2: Production of the magnetic field

Nowadays the magnetic field in the laboratories is produced using electromagnets and a value of 2-3 T can be easily produced without many efforts, i.e., without the use of superconducting coils.

## Step 3: Evaluating the field strength(a)

To obtain the magnetic field strength for an anti-proton a particular circular path is followed.

The equation used is: $${\rm{r = }}\frac{{{\rm{mv}}}}{{{\rm{qB}}}}$$.

Solving it for the value of $${\rm{B}}$$ where the value of $${\rm{q}}$$ is the antiproton’s charge and the value of $${\rm{m}}$$ is the antiproton’s mass.

\begin{align}{}{\rm{B}}& = \frac{{{\rm{mv}}}}{{{\rm{qr}}}}\\ &= \frac{{\left( {{\rm{1}}{\rm{.67 \times 1}}{{\rm{0}}^{{\rm{ - 27}}}}\;{\rm{kg}}} \right){\rm{ \times }}\left( {{\rm{5}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{7}}}{\rm{ m/s}}} \right)}}{{\left( {{\rm{1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}\;{\rm{C}}} \right){\rm{ \times }}\left( {{\rm{2}}{\rm{.00 m}}} \right)}}\\ &= {\rm{0}}{\rm{.261 T}}\end{align}

## Step 4: Explaining that Is this field strength obtainable with today’s technology or is it a futuristic possibility?(b)

The magnetic field strength is small and is easily obtainable these days.

Therefore, we get:

1. The magnetic field strength is: $${\rm{0}}{\rm{.261 T}}$$.
2. The magnetic field strength is nowadays obtainable.