A proton in a particle accelerate has a speed of S O x 10^6 m/s in the positive

Question

A proton in a particle accelerate has a speed of S O x 10^6 m/s in the positive x-direction The protun encounters a magnetic field w hove magnitude is 0.40 T md directum makes an along the plane with respect to the proton's velocity. Find the magnitude of the magnetic force acting on the pro con I Find the direction of the magnetic force acting on the proton i Kind the and direction of the magnetic force if the were an electron instead of a proton from question ft 1 above, the magnitudes of the acceleration for

Explanation / Answer

v = 5*10^6 m/s, B =0.4 T , theta =30 degrees

charge of proton q =1.6*10^-19 C

F =qvBsin(theta)

F = 1.6*10^-19*5*10^6*0.4*sin(30)

F= 1.6*10^-19 N

(b) F = +q ( ix (i+j))

F = k

F is along z direction.

(c) electron charge q = -1.6*10^-19 C

F= 1.6*10^-19 N

direction is -z direction

2) F = ma = Fb

a = qvB/m

(a) mass of electron m =1.67*10^-27 kg

a = (1.6*10^-19)/(1.67*10^-27)

a = 9.58*10^7 m/s^2

direction is along z direction

(b) mass of electron m =9.1*10^-31 kg

a = (1.6*10^-19)/(9.1*10^-31)

a = 1.76*10^11 m/s^2

direction is along negative z direction

(c) earth gravity is along y direction. magnetic force is along z direction.

according to newtons first law we consider force along z direction only

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