The path of moving proton in a normal magnetic field is circular. If r is the radius of the circular path, then from the figure, From the symmetry of figure, the angle θ = 45°. AC=2rcos45∘=2r×21=2rAsBqv=rmv2 or r=BqmvAC=Bq2mv=1×1.6×10−192×1.67×10−27×107=0.14m