A proton with a speed of 3.2 Times l06m/s is shot into a region between two plat

Question

A proton with a speed of 3.2 Times l06m/s is shot into a region between two plates that are separated by a distance of 0.167 m. as the drawing shows, a magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field, so that the proton just misses colliding with the opposite plate? What is the change in kinetic energy of the proton, if any, during this process? (The mass of a proton is 1.67Times10 - 27 kg, and e=1.60Timesl0 - 19 C) (Detailed calculations leading to the answers have to be presented)

Explanation / Answer

m = mass of proton = 1.67 x 10-27 kg

e = charge on proton = 1.6 x 10-19 C

V = speed of proton = 3.2 x 106 m/s

for avoiding the collision , radius of the circle traced by proton has to be equal to the distance between plates

r = 0.167 m

a)  

radius of circular patrh by proton in magnetic field is given as ::

r = mV/qB

so B = = mV/qr

B = (1.67 x 10-27) (3.2 x 106) / (1.6 x 10-19x 0.167)

B = 0.2 T

b) the speed of proton inside the magnetic field does not change , hence there is no change in kinetic energy

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