Answer to Question #113027 in Electric Circuits for James

Question #113027
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
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Expert's answer
2020-05-04T12:50:12-0400

R=5.49mR=5.49 m

v=1.99×107m/sv=1.99\times10^7m/s

Lorentz force equals centripetal force

mv2R=qvBm{\frac {v^2} R}=qvB

Then B=mvqR=3.8×102TB={\frac {mv} {qR}}=3.8\times10^{-2} T


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