a) 0.261 T
b) this field strength is obtainable with today's technology
Explanation:
a)
The force experienced by a charged particle moving perpendicular to a magnetic field is given by
[tex]F=qvB[/tex]
where
q is the charge
v is the velocity
B is the strength of the field
This force is perpendicular to the motion of the particle, which therefore moves in a circular path; and so, this force acts as centripetal force, so we can write:
[tex]qvB=m\frac{v^2}{r}[/tex]
where
m is the mass of the particle
r is the radius of the circle
In this problem, we have:
[tex]q=1.6\cdot 10^{-19}C[/tex] (magnitude of the charge of antiprotons)
[tex]v=5.00 \cdot 10^7 m/s[/tex] (velocity)
[tex]m=1.67\cdot 10^{-27}kg[/tex] (mass of antiprotons)
r = 2.00 m (radius)
Therefore, we can re-arrange the equation and solve to find B, the magnetic field strength:
[tex]B=\frac{mv}{qr}=\frac{(1.67\cdot 10^{-27})(5.00\cdot 10^7)}{(1.6\cdot 10^{-19})(2.00)}=0.261 T[/tex]
B)
The strength of the magnetic field calculated in part A) is
[tex]B=0.261 T[/tex]
This is indeed a very strong magnetic field. In fact, by comparison, the Earth's magnetic field has a strength of about
[tex]B_{earth}=5\cdot 10^{-5} T[/tex]
However, there are current technologies available that are able to produce such strong fields. For instance, the superconducting magnets in the LHC (Large Hadron Collider) are able to produce magnetic fields of strength up to 8 Tesla (8 T).
Therefore, we can say that this field strength is obtainable with today's technology.
