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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F=qvBsin⁡90∘=(1.60×10−19)(2.0×106)(0.50)=1.60×10−13 NF=qvB\sin90^\circ=(1.60\times10^{-19})(2.0\times10^6)(0.50)=1.60\times10^{-13}\ \mathrm N
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The magnetic part of the Lorentz force is qv⃗×B⃗q\vec v\times\vec B. Since v⃗⊥B⃗\vec v\perp\vec B, sin⁡θ=1\sin\theta=1 and the magnitude is maximal, qvBqvB.