# Calculate the kinetic energy of a proton which moves into a field of induction B=167mT, on a circle with radius r=50cm.

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Student Comments

giorgiana1976 | Student

Lorentz force is acting on electron, so *lex secunda* is giving the followings:

module (q*vxB)=e*v*B=m*ac=m*v^2/r, so v=(r/m)*e*B

Ec=m*v^2/2=(e*B*r)^2/(2*m)=53.4fJ=0.334MeV