Question

Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. What strength magnetic field is needed to hold antiprotons, moving at 5.00×10^7 m/s in a circular path 2.00 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.

a) 1.00 T
b) 1.25 T
c) 1.50 T
d) 2.00 T

Answer 1

The strength of the magnetic field needed to hold the antiprotons in a circular path with a radius of 2.00 m and a velocity of 5.00 × 10^7 m/s is approximately 1.04 T.

Step-by-step explanation:

To calculate the strength of the magnetic field needed to hold the antiprotons in a circular path, we can use the equation for the centripetal force:

Centripetal force = Magnetic force

mv^2/r = QvB

Where m is the mass of the antiproton, v is its velocity, r is the radius of the circular path, Q is the charge of the antiproton, and B is the magnetic field strength.

Since the antiprotons have the same mass as protons, we can use the known values:

m = 1.67 × 10^-27 kg

v = 5.00 × 10^7 m/s

r = 2.00 m

Q = -1.60 × 10^-19 C (negative charge)

Substituting these values into the equation, we can solve for B:

B = mv / (Qr)

B = (1.67 × 10^-27 kg)(5.00 × 10^7 m/s) / (-1.60 × 10^-19 C)(2.00 m)

Calculating the value, we get:

B ≈ 1.04 T

So, the strength of the magnetic field needed to hold the antiprotons in a circular path is approximately 1.04 T.