A proton (m=1.67x10 -27 kg) has a charge of 1.6 * 10 -19 C is emmited from rest

Question

A proton (m=1.67x10-27 kg) has a charge of 1.6 * 10-19 C is emmited from rest out of a tube which is located at the origin of the x-y plane with a velocity of v=6 m/s x-direction in a uniform magnetic field strenght of 50 T in the z-direction. If there is no gravity and electrical field:
(a) What is the magnitud of the force on the proton at t=0
(b) What is the direction of the force on the proton
(c) What is the acceleration on the proton
(d) What is the direction of the acceleration from the last question
(e) What is the radius of the helix about which the proton moves (Please draw a picture of what you are solving)
(f) How long does it take the proton to complete one revolution of its motion?

Explanation / Answer

a) |Fb| = qvB = (1.6*10^-19)(6)(50) = 4.8*10^-17 N

b) Fb= q(v x B) = q ( i x k) = - j

Magnetic force at t=0 is F directed negative y axis.

c) After t=0s, a= Fc /m = mv^2/mr = v^2/r

d) After t=0s proton moves continuously in a circular orbit hence the net force is centripetal force Fc acting along the radius of the circle and directed inward towards the center of it.

e) Applying Newton’s second law to the orbiting proton,

Fb= Fc

qvB=mv^2/r

qB= mv/r

r=mv/qB = (1.67*10-27*6)/( 1.6*10^-19*50) = 1.25*10^-9 m = 1.25nm

e) Period , T = 2r/v = (2*3.14*1.25*10^-9)/(6) = 1.3*10^-9 s = 1.3ns

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