I need help on the data section of the circuit design lab on Edg.

An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field

lawyer [7]

Answer:

a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

Em₀ = U = e DV

final point. Where they hit the target

Em_f = K = ½ m v2

energy is conserved

Em₀ = Em_f

e ΔV = ½ m v²

ΔV = $\frac{1}{2}$ mv²/e (1)

If the speed of light is c and this is 100% then 1% is

v = 1% c = c / 100

v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

ΔV = $\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }$

ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

$K_e = K_p$

K_p = ½ m v_e²

K_p = $\frac{1}{2}$ 9.1 10⁻³¹ (3 10⁶)²

K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

ΔV = Kp / M

ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

e ΔV = ½ M v²

v² = 2 e ΔV / M

v² = $\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }$

v² = 49,035 10⁸

v = 7 10⁴ m / s

Relation

v/c = $\frac{7 \ 10^4 }{ 3 \ 10^8}$

v/c= 2.33 10⁻⁴

8 0

3 years ago

The electron drift speed in a 1.2-mm-diameter gold wire is 5.00 10-5 m/s. how long does it take 1 mole of electrons to flow thro

Vlada [557]

Twenty four minutes per second

7 0

3 years ago

A long, straight wire lies in the plane of a circular coil with a radius of 0.018 m. the wire carries a current of 5.6 a and is

iris [78.8K]

(a) The net flux through the coil is zero.
In fact, the magnetic field generated by the wire forms concentric circles around the wire. The wire is placed along the diameter of the coil, so we can imagine as it divides the coil into two emisphere. Therefore, the magnetic field of the wire is perpendicular to the plane of the coil, but the direction of the field is opposite in the two emispheres. Since the two emispheres have same area, then the magnetic fluxes in the two emispheres are equal but opposite in sign, and so they cancel out when summing them together to find the net flux.

(b) If the wire passes through the center of the coil but it is perpendicular to the plane of the wire, the net flux through the coil is still zero.
In fact, the magnetic field generated by the wire forms concentric lines around the wire, so it is parallel to the plane of the coil. But the flux is equal to
$\Phi = BA \cos \theta$
where $\theta$ is the angle between the direction of the magnetic field and the perpendicular to the plane of the coil, so in this case $\theta=90^{\circ}$ and so the cosine is zero, therefore the net flux is zero.