Q4: Radius of curved path of proton in magnetic field
A proton is moving a t 4.4 × 10^5 m//s the plane of the page when it enters a magnetic field of magnitude 0.04 T perpendicular to the page, as shown in the figure above. The radius of curvature of the path of the proton as it moves through the magnetic field is approximately which of the following? F = qvB=ma_c=mv^2/r => r = (mv)/(qB) r = ((1.67 × 10^-27 kg)(4.4 × 10^5 m//s))/((1.66 × 10^-19 C)(0.04 T)) r = 0.11 m