A proton of charge enters a uniform magnetic field of magnitude with speed

PHYSICS · Multi-part · CBSE · Physics · Moving Charges and Magnetism · Moving Charges and Magnetism

Problem

A proton of charge +e+e enters a uniform magnetic field of magnitude 0.50 T0.50\,\text{T} with speed 2.0×106 m s−12.0\times10^{6}\,\text{m\,s}^{-1}. The angle between v⃗\vec v and B⃗\vec B is slowly reduced from 90∘90^{\circ} to 0∘0^{\circ} while keeping the speed and the field magnitude constant. (a) Derive an expression for the magnitude of the magnetic force on the proton as a function of the angle θ\theta between v⃗\vec v and B⃗\vec B. (b) Calculate the magnetic force when θ=30∘\theta=30^{\circ}, 60∘60^{\circ} and 90∘90^{\circ}. (c) At what angle θ\theta will the magnetic force on the proton be 3.0×10−13 N3.0\times10^{-13}\,\text{N}? Show all working steps. (d) Explain physically why the magnetic force vanishes when θ\theta approaches 0∘0^{\circ} even though both vv and BB are non-zero.

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