Question

In a circular particle accelerator with a radius of 5.49m,protons are accelerated to 1.99 × 107m/s. Find the magnetic flux density B that’s required to keep the protons following the circular path of the accelerator mp =1.67 ×10-27kg and Qp =1.60 ×10-19J

Answer 1

$R=5.49 m$

$v=1.99\times10^7m/s$

Lorentz force equals centripetal force

$m{\frac {v^2} R}=qvB$

Then $B={\frac {mv} {qR}}=3.8\times10^{-2} T$