Applications of Motion of Charged Particles Problems and Solutions 2

 Problem#1

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 V/m and the magnetic field is 0.0315 T. The ions next enter a uniform magnetic field of magnitude 0.0175 T that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass?

Answer:
Known:
electric field, E = 155 V/m
magnetic field of magnitude, B = 0.0175 T
The ions next enter a uniform magnetic field of magnitude, B’ = 0.0175 T

(a) moving ion speed is,

v = E/B = (155 V/m)/(0.0175 T) = 4920 m/s

(b) If the radius of the path of the ions in the second magnetic field is R = 17.5 cm = 0.175 m, the mass of these ions is given by

R = mv/qB

m = qBR/v = (1.6 x 10^-19 C)(0.0175 T)(0.175 m)/(4920 m/s)

m = 9.96 x 10^-26 kg

Problem#2
In the Bainbridge mass spectrometer (see Fig. 3), the magnetic-field magnitude in the velocity selector is 0.650 T, and ions having a speed of 1.82 x 10⁶ m/s pass through undeflected. (a) What is the electric-field magnitude in the velocity selector? (b) If the separation of the plates is 5.20 mm, what is the potential difference between plates P and P’.

Fig.3

Answer:

Known:
Speed, v = 1.82 x 10⁶ m/s
magnetic-field magnitude, B = 0.650 T

(a) the electric-field magnitude in the velocity selector is

E = vB = (1.82 x 10⁶ m/s)(0.650 T) = 1.18 x 10⁶  V/m

(b) If the separation of the plates is d = 5.20 mm =  5.20 x 10^-3 m, what is the potential difference between plates P and P’.

V = Ed = (1.18 x 10⁶  V/m)(5.20 x 10^-3 m) = 6.14 kV

Problem#3
The amount of meat in prehistoric diets can be determined by measuring the ratio of the isotopes nitrogen-15 to nitrogen-14 in bone from human remains. Carnivores concentrate ¹⁵N, so this ratio tells archaeologists how much meat was consumed by ancient people. For a mass spectrometer that has a path radius of 12.5 cm for ¹²C ions ( mass 1.99 × 10⁻²⁶ kg),find the separation of the 14N and 15N isotopes at the detector. The measured masses of these isotopes are 2.32 x 10^-26 kg (¹⁴N) and  2.49 x 10^-26 kg (¹⁵N).

Answer:
Known:
m ¹⁴N = 2.32 x 10^-26 kg,
m ¹⁵N = 2.49 x 10^-26 kg

For mass spectrometers that have a path radius,

R = mv/qB
R/m = v/qB = constant, then

R₁₄/m₁₄ = R₁₂/m₁₂

R₁₄ = [R₁₂/m₁₂] m₁₄ = 12.5 cm[2.32 x 10^-26 kg/1.99 × 10⁻²⁶ kg] = 14.6 cm

R₁₅/m₁₅ = R₁₂/m₁₂

R₁₅ = [R₁₂/m₁₂] m₁₅ = 12.5 cm[2.49 x 10^-26 kg/1.99 × 10⁻²⁶ kg] = 15.6 cm

So, 2(R₁₅ – R₁₄) = 2(15.6 cm – 14.6 cm) = 2.0 cm

Problem#4
A certain particle is sent into a uniform magnetic field, with the particle’s velocity vector perpendicular to the direction of the field. Figure 4  gives the period T of the particle’s motion versus the inverse of the field magnitude B. The vertical axis scale is set by T_s = 40.0 ns, and the horizontal axis scale is set by B_s^-1 = 5.0 T^-1. What is the ratio m/q of the particle’s mass to the magnitude of its charge?

Fig.4

Answer:

Certain particles are sent to a uniform magnetic field, with the speed of a straight particle vector with the direction of the magnetic field that will produce a magnetic force perpendicular to the velocity, so that the particle will move irregularly with a period given by

T = 2πm/qB

With: T = period particle, m = mass particle, q = particle charge and B = magnetic field, then

T/B^-1 = 2πm_p/q, or

m/q = (T/B^-1)(2π)^-1

from the graph we get

T/B^-1 = 11 x 40 ns/[15 x 5.0 T^-1] = 88/15 ns/T^-1

m/q = (88/15 ns/T^-1)(2π)^-1

m/q = 9.215 x 10^-9 kg/C

Fig.5

Problem#5

In Fig. 6, a charged particle moves into a region of uniform magnetic field , goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 130 ns in the region. (a) What is the magnitude of B ? (b) If the particle is sent back through the magnetic field (along the same initial path) but with 2.00 times its previous kinetic energy, how much time does it spend in the field during this trip?

Fig.6

Answer:

We consider the point at which it enters the field-filled region, velocity vector pointing downward. The field points out of the page so that v x B points leftward, which indeed seems to be the direction it is pushed, therefore, q > 0 (it is a protorn).

(a) the magnitude of B is

T = 2πm_p/eB

2t = 2πm_p/eB

2(130 x 10^-9 s) = 2π(1.67 x 10^-27 kg)/[(1.60 x 10^-19 C)B]

B = 0.25 T

(b) doubling the kinetic energy implies multiplying the speed by √2. Since the period T does not depend on speed, then it remains the same (even thought the radius increases by a factor of √2). Thus, t = T/2 = 130 ns.