A proton (q = 1.60×1019 C , m = 1.67×1027 kg )moves in a uniform magnetic field

Question

A proton (q = 1.60×1019 C , m = 1.67×1027 kg )moves in a uniform magnetic field B =( 0.510 T )i^. At t = 0 the proton has a velocity components vx = 1.50×105 m/s , vy=0, and vz = 2.50×105 m/s .

Part A

What is the magnitude of the magnetic force acting on the proton? N=?

Part B

What is the direction of the magnetic force acting on the proton? in the +x-direction in the +y-direction in the +z-direction in the y-direction

Part C

In addition to the magnetic field there is a uniform electric field in the +x-direction, E =( 2.50×104 V/m )i^. Will the proton have a component of acceleration in the direction of the electric field? ax = ? m/s2

Part D

This question will be shown after you complete previous question(s). dont worry about it

Part E

At t=T/2, where T is the period of the circular motion of the proton, what is the x-component of the displacement of the proton from its position at t = 0?

Explanation / Answer

(A) F = q ( v x B)

= (1.6 x 10^-19)[ (1.50 x 10^5 i + 2.50 x 10^5 k) X (0.510 i)]

= 2.04 x 10^-14 N j

magnitude = 2.04 x 10^-14 N

(B) direction -> +y axis

(C) Yes:

ax =q E / m = (1.6 x 10^-19)(2.50 x 10^4) / (1.67 x 10^-27)
= 2.40 x 10^12 m/s^2

(E) F = m v^2 / r

2.04 x 10^-14 = 1`.67 x 10^-27 x (2.50x 10^5)^2 / r

r = 4.90 x 10^-3 m

T = 2 pi r / v = 1.23 x 10^-7 s

d = (1.50 x 10^5)(T/2) + (2.40 x 10^12)(T/2)^2 / 2

d = 0.0138 m

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