Physic Labs

Problem 2

A proton of charge q=+1.60×10−19q=+1.60\times10^{-19} C and mass m=1.67×10−27m=1.67\times10^{-27} kg moves in a uniform magnetic field B=0.50B=0.50 T. Its initial velocity has magnitude v=2.0×106v=2.0\times10^6 m/s and is perpendicular to B⃗\vec B; neglect gravity. (a) Find the magnetic-force magnitude and orbit radius. (b) Find the angular frequency, ordinary frequency, and period of the motion. (c) The proton is accelerated until its kinetic energy quadruples in the same field; find the new speed, radius, and period.
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qvB=mv2r⇒r=mvqB=(1.67×10−27)(2.0×106)(1.60×10−19)(0.50)=4.18×10−2 mqvB=\frac{mv^2}{r}\Rightarrow r=\frac{mv}{qB}=\frac{(1.67\times10^{-27})(2.0\times10^6)}{(1.60\times10^{-19})(0.50)}=4.18\times10^{-2}\ \mathrm m
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The magnetic force remains perpendicular to velocity, leaving speed constant and bending the path into a circle. Equating the Lorentz force with mv2/rmv^2/r gives r=mv/(qB)r=mv/(qB).

problems.proof.pitfall. The magnetic force does no work and cannot change kinetic energy; the radius must use the same instantaneous vv as the field experienced by the particle.