1. A uniform magnetic field B with magnitude 1.2 mT is directed verticallyupward
ID: 1727598 • Letter: 1
Question
1. A uniform magnetic fieldB with magnitude 1.2 mT is directed verticallyupward throughout the volume of a laboratory chamber. A proton withkinetic energy 5.3MeV enters the chamber, moving horizontally fromsouth to north.
(Mass of proton is 1.67x10-27 kg, neglectEarth’s magnetic field)
(a) What magnetic deflecting force acts on theproton as it enters the chamber?
(b) What is the acceleration of the proton in thechamber?
(c) Assume the proton enters the chambervertically downward, opposite the direction of the magnetic field,what magnetic force acts on the proton?
Explanation / Answer
(a) The magnetic deflecting force F = BqVsin =(1.2*10-3 T)(1.6*10-19 C)V sin90 Here v = (2KE/m) =( 5.3*106 (1.6*10-19)/1.67*10-27 kg) =2.25*107 m/s = (1.2*10-3 T)(1.6*10-19 C) (2.25*107 m/s )
= 4.32 *10-15 N (b) Acceleration a = F/m = 4.32*10-15 N / (1.67*10-27 kg) =2.58*1012 m/s2 (c) The magnetic deflecting force F = BqVsin =(1.2*10-3 T)(1.6*10-19 C)V sin 180 = 0 N
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