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 3.20 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge. (Enter the magnitude.)

Answer 1

Answer:

The magnetic field strength  required to hold anti-protons, moving at 5.70 ✕ 10⁷ m/s in a circular path of 3.20 m in radius is 0.186 T.

Explanation:

To solve the question we note that the magnetic force on a moving charge is given by

F = q·v·B

Where

q = Charge

v = Velocity of the charge =5.70 ✕ 10⁷ m/s

B = Magnetic field

Based on Newton's second law,

Force = Mass, m × Acceleration, a = m × a

Where:

a = Acceleration

m = Mass of anti-proton = Mass of proton = 1.6726219 × 10⁻²⁷ kg

We note that for circular motion, acceleration a is given by

$\alpha = \frac{v^{2} }{r}$ .

Where:

r = Radius = 3.20 m

Therefore, for the circular motion, force, F = $\frac{m\cdot v^{2} }{r}$

Equating the magnetic force equation to the circular force equation, we have

$\frac{m\cdot v^{2} }{r}$ = q·v·B So that, we find B by making the subject of the formula as follows

$B= \frac{m\times v^{2} }{r\times q \times v}$ . Which gives

$B= \frac{m\times v }{r\times q} = \frac{(1.6726219 \times 10^{-27}) \times (5.7\times 10^{7}) }{(3.20)\times (1.602\times 10^{-19}) }$ =  0.186 T

The magnetic field strength is

B = 0.186 T