You have a proton with velocity v= 4.0 x 10 7 m/smoving perpendicular to a magne

Question

You have a proton with velocity v= 4.0 x 107 m/smoving perpendicular to a magnetic field B=0.26 T. What force is itexperiencing and what is its acceleration? (mp= 1.673 x10-27 kg and qp = 1.6 x 10-19 C)This may seem to be a very large acceleration, but using that asthe centripetal acceleration (acp) since the particlewill travel in a circle if the field is uniform, calculate theradius of this circle. (acp= v2/r) You have a proton with velocity v= 4.0 x 107 m/smoving perpendicular to a magnetic field B=0.26 T. What force is itexperiencing and what is its acceleration? (mp= 1.673 x10-27 kg and qp = 1.6 x 10-19 C)This may seem to be a very large acceleration, but using that asthe centripetal acceleration (acp) since the particlewill travel in a circle if the field is uniform, calculate theradius of this circle. (acp= v2/r)

Explanation / Answer

velocity v= 4.0 x 107 m/s
magnetic field B=0.26 T In magnetic field   Force on proton due to magneticfield = centripetal force                                                                            B v q = m v 2 / R from this radius R = m v / B q where m = mass of proton = 1.67 * 10 -27kg            q= charge of proton = 1.6 * 10 -19 C subsitute values wget answer

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