Motion of a Charged Particle in Magnetic Field

  • Thread starter Ignitia
  • Start date
  • #1
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Homework Statement


Viewers of Star Trek have heard 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 normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.0×107m/s in a circular path 2.00 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.

Homework Equations


F = ma = m * (v2/r)
F = qvB

The Attempt at a Solution


Okay, this is pretty straightforward:

v = 5.0x10-7 m/s
m = 1.672x10-27 kg
q = -1.6x10-19C
r = 2m

F = qvB
F/(qv) = B
[(m*(v2) / r ] * 1/(qv) = B
(mv/rq) = B

Plugging in the values, answer becomes - 0.26T, but the correct answer is 0.26T. What am I missing?
 

Answers and Replies

  • #2
kuruman
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Strength of magnetic field is another name for magnitude which is a positive number.
 
  • #3
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Strength of magnetic field is another name for magnitude which is a positive number.

So I just take the absolute value? That makes sense, thanks.
 

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