A proton moves horizontally due east with constant speed

PHYSICS · Multi-part · A Level · Physics · Fields and their consequences · Fields and their consequences

Problem

A proton moves horizontally due east with constant speed v=2.0×105 m s−1v=2.0\times10^{5}\,\text{m\,s}^{-1}. Beginning at t=0t=0, the proton enters a region of uniform magnetic field of magnitude 0.30 T0.30\,\text{T} that is directed vertically upward. Take the mass of the proton to be mp=1.67×10−27 kgm_{p}=1.67\times10^{-27}\,\text{kg} and the magnitude of its charge to be e=1.60×10−19 Ce=1.60\times10^{-19}\,\text{C}. (a) Calculate the magnitude of the magnetic force acting on the proton immediately after it enters the field. (b) Determine the direction of the magnetic force on the proton at that instant. Express your answer using the compass directions north, south, east, west, up, and down. (c) Is the speed of the proton expected to change while it is in this magnetic field? Explain your answer. (d) Assuming the magnetic field extends over a large region, calculate the radius of the circular path followed by the proton.

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