21.8 MEASUREMENT OF THE RATIO q/m
21.8.1 Motion of Charged Particle in Uniform Magnetic Field
1 Path of motion of charged particle in uniform magnetic field
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Figure 21.44 (a) A positively charged particle moves perpendicular to a magnetic field and follows a circular path due to magnetic force. |
(a) Consider a positively charged particle moving in a uniform magnetic field which points vertically into the page, as shown in Figure 21.44(a). The velocity of the particle is perpendicular to the field, meaning that the velocity vector lies on the page.
(b) The particle is acted on by a magnetic force which is always perpendicular to the velocity as well as the field. Hence, the magnetic force would force the particle to move in a circle, with the force always pointing towards the centre of the circle.
(c) For a positively charged particle, the motion of the particle is in the anti-clockwise direction. On the other hand, if the particle is negatively charged, the motion of the particle is in the clockwise direction, as shown in Figure 21.44(b).
2 Radius of the circular path
Since the charged particle is moving in a circle, the force acting on it is given by
centripetal force Fc = magnetic force Fm
The centripetal force is given by
Fc = mv2 / r
where m is the mass of the particle
v is the tangential speed of the particle
r is the radius of the circle
The magnetic force is given by
Fm = Bqv sin 90°
where B is the magnetic field strength
q is the charge carried by the particle
Hence, we have
mv2 / r = Bqv
r = mv / Bq
Notice that,
r ∝ mv (momentum of particle)
r ∝ 1 / B
r ∝ 1 / q
3 Period of revolution
The time T taken by the charged particle to complete a circle is given by
T = 2Ï€r / v
T = 2Ï€ ( m / Bq )
Notice that,
T ∝ 1 / B
T ∝ m
T ∝ 1 / q
T is independent of (a) the radius r and (b) the tangential speed v
EXAMPLE 21.18
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| Figure 21.45 Two positively charged particles, X and Y, enter a uniform magnetic field at point A and move in circular paths on the same plane, with the magnetic field perpendicular to the page. |
Two positively charged nuclear particles, X and Y, enter a uniform magnetic field at point A at the same time in a bubble chamber, as shown in Figure 21.45. They both lie on the same plane which is parallel to the page. Each of them follows a circular path. The field is perpendicular to the page.
(a) State the direction of the magnetic field.
(b) Each particle carries the same amount of charge. What can we deduce about the linear momentum of particle X if we compare it with the linear momentum of particle Y?
Answer
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Figure 21.46 Magnetic field directed perpendicularly out of the page, indicating the direction of the field affecting the motion of charged particles. |
(a) The magnetic field points perpendicularly out of the page, as shown in Figure 21.46.
(b) The radius of each particle is given by
r = mv / Bq
Linear momentum of the particle X
px = mxvx = (Bq)rx
Linear momentum of the particle Y
py = myvy = (Bq)ry
Referring to Figure 21.45, we get
rx ry
Hence,
px py
EXAMPLE 21.19
A charged particle enters a uniform magnetic field at point X. It is given that OX = OY = 2.0 cm and OX is perpendicular to OY, as shown in Figure 21.47. The velocity of the particle is parallel to OY.
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| Figure 21.47 A charged particle enters a uniform magnetic field at point X with velocity parallel to OY, moving along a circular path where OX = OY and OX is perpendicular to OY. |
(a) State the type of charge carried by the particle.
(b) The magnitude of the magnetic field strength which can force the particle to pass through Y is B1 while the strength which can force the particle to pass through O is B2. Determine the ratio B1 / B2.
Answer
(a) The particle carries positive charge.
(b)
X to Y:B1qv = mv2 / OX
X to O:B2qv = mv2 / (1/2 OX)
B1 / B2 = 2
EXAMPLE 21.20
A proton and an alpha particle travelling at the same speed enter a uniform magnetic field. The velocity of each particle is perpendicular to the field. Determine the ratio of the radius of the circular path of the proton to the radius of the circular path of the alpha particle.
Answer
Proton:charge = +e, mass = m, radius = r1, speed = v
Alpha particle:charge = +2e, mass = 4m, radius = r2, speed = v
We have
r1 = mv / Be
r2 = (4m)v / B(2e)
r1 / r2 = (m / 4m)(2e / e)
= 1 / 2
EXAMPLE 21.21
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| Figure 21.48 A proton enters a uniform magnetic field at point X and is deflected along a circular path to point Y, where XY is perpendicular to its initial velocity and the magnetic field is perpendicular to the page. |
A proton enters a uniform magnetic field of strength 300 mT through point X as shown in Figure 21.48. The magnetic field deviates the path of the ion to a point Y which is 5.0 cm from X. XY is perpendicular to the direction of the velocity of the ion when it just passes through X. The field points in a direction which is perpendicular to the page.
(a) State the direction of the magnetic field.
(b) Determine
(i) the speed of the proton at the moment it passes through X if the mass of the ion is 1.66 × 10−27 kg
(ii) the time taken by the ion to reach Y from X
(iii) the kinetic energy, in eV, of the proton when it reaches Y.
(Electronic charge = −1.6 × 10−19 C)
Answer
(a) The magnetic field points perpendicularly out of the page.
(b) (i)
mv2 / r = Bqv
v = Br(q / m)
= (0.300)(0.025)(1.6 × 10−19 / 1.66 × 10−27)
= 7.2 × 105 m s−1
(ii) Time t needed to travel half a circle at constant speed v is given by
t = distance / speed
= πr / v
= Ï€(0.025) / (7.2 × 105) = 1.1 × 10−7 s
(iii) The kinetic energy K of the proton is given by
K = 1/2 mv2
= 1/2 (1.66 × 10−27)(7.2 × 105)2 = 4.3 × 10−16 J
= 4.3 × 10−16 / 1.6 × 10−19 = 2.7 keV
180
EXAMPLE 21.22
An electron in vacuum accelerates through a potential difference of 5.0 kV. After that it enters a uniform magnetic field of strength 040 mT. The velocity is perpendicular to the direction of the field. Neglect the effects of relativity. Determine
(a) the speed of the electron just before it enters the magnetic field
(b) the radius of the circular path followed by the electron
(c) the period of revolution of the electron.
(Specific charge of electron, e/m = −1.76 × 1011 kg C−1)
Answer
(a) KE gained by electron = electric PE lost by electric field
1/2 mv2 = eΔV
v2 = 2(e/m) ΔV
= 2(1.76 × 1011)(5.0 × 103) = 17.6 × 1014
v = 4.2 × 107 m s−1
(b)
r = v / B(e/m)
= 4.2 × 107 / ((0.40 × 10−3)(1.76 × 1011)) = 0.60 m
(c)
T = 2Ï€ / B(e/m)
= 2Ï€ / ((0.40 × 10−3)(1.76 × 1011)) = 9.0 × 10−8 s
EXAMPLE 21.23
An electron, a proton and an alpha particle move in circles inside a uniform magnetic field with the planes of the circles perpendicular to the field. Each particle has the same amount of kinetic energy. The radius of the circular path of the electron is r. Determine the radii of the circular paths of the proton and the alpha particle in terms of r.
(Masses: electron, m; proton, 1 836m; alpha particle, 7 344m)
Answer
Electron:
Ke = 1/2 mve2
We have
mve2 / r = Beve
mve = Bre
m2ve2 = (Bre)2
mve2 = (Bre)2 / m
Hence,
Ke = (Bre)2 / 2m
Proton:
Kp = (Brpe)2 / 2(1836m) = (Brpe)2 / 3672m
But,
Kp = Ke
EXERCISE 21.5
For electron, e / m = −1.76 × 1011 C kg−1.
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An alpha particle of mass 6.68 × 10−27 kg moves inside a uniform magnetic field of strength 1.0 T in a direction which is perpendicular to the field. If the speed of the particle is 2.0 × 107 m s−1, determine
(a) the magnitude of the magnetic force that acts on the particle
(b) the radius of the circular path of the particle. - An electron accelerates through an electric potential difference of 500 V. After that it enters a uniform magnetic field of strength 500 mT in a direction which is perpendicular to the field. Determine the radius of the circular path followed by an electron.
21.8.2 Velocity Selector
1 Principle
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| Figure 21.49 A uniform electric field and a magnetic field are applied perpendicular to each other, forming a velocity selector that allows only particles with a specific velocity to pass straight through. |
A uniform electric field is produced by a pair of parallel charged plates. A uniform magnetic field is then superimposed on the electric field in such a way that the two fields are perpendicular to each other, as shown in Figure 21.49.
Suppose that we allow charged particles travelling at different speeds to enter the crossed fields in a direction which is perpendicular to both fields. In other words, the velocity vector, electric field vector and magnetic field vector are mutually perpendicular to one another.
We find that as the particles travel through the fields some of them are forced to move in curved paths towards the positively charged plate while some are deviated towards the negatively charged plate. The rest of the particles can travel completely through both fields in straight lines and finally emerge from the fields at one constant speed v.
The particles enter the fields in the same direction but at different speeds. However, only those particles moving at some particular speed v can emerge straight from the fields while the rest are deviated away from their initial paths. Hence, we can use this setup consisting of crossed electric and magnetic fields as a ‘velocity selector’.
2 Determining the selected speed v
At the moment a charged particle enters the crossed fields, it is acted on by two forces: electric force Fe and magnetic force Fm. They are given by
Fe = qE
and
Fm = Bqv
The forces must have the following features:
(a) Since the velocity and the fields are mutually perpendicular to one another, the two forces must be directly opposite each other.
(b) The particle travels straight through the fields without being deviated. This means that the net force acting on it must be zero.
If gravitational pull on the particles is neglected, we get
Fm = Fe
Bqv = qE
v = E / B
EXAMPLE 21.24
An electron beam passes through a velocity selector whose structure is as shown in Figure 21.46. The electric field has strength 30 kV m−1 and the magnetic field has strength 15 mT. Neglect gravitational pull on the electrons. Determine the kinetic energy, in eV, of those electrons which pass straight through the fields without being deviated.
(Electron charge, e = −1.6 × 10−19 C; electron mass, m = 9.1 × 10−31 kg)
Answer
The speed of those electrons not deviated by the fields is given by
v = E / B
= 30 × 103 / 15 × 10−3 = 2.0 × 106 m s−1
Kinetic energy K of the electron is given by
K = 1/2 mv2
= 1/2 (9.1 × 10−31)(2.0 × 106)2
= 1.8 × 10−18 J
= (1.8 × 10−18) / (1.6 × 10−19) = 11.3 eV
EXERCISE 21.6
An ion moving at speed 2.0 × 106 m s−1 enters a velocity selector whose structure is as shown in Figure 21.49. A p.d. of 500 V is applied to the pair of horizontal plates, which are separated by a distance of 1.0 cm. The ion emerges from the fields at the same speed without changing direction. Determine the strength of the magnetic field.
21.8.3 Determination of the Specific Charge e/m of Electron
Apparatus
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| Figure 21.50 A uniform electric field and a magnetic field are applied perpendicular to each other, forming a velocity selector that allows only particles with a specific velocity to pass straight through. |
The apparatus (as shown in Figure 21.50 consists of
(a) a glass tube in which we can find the following components:
(i) An electron gun, consisting of a filament and anodes, to produce an electron beam.
(ii) A vertical screen coated with a fluorescent substance. The substance interacting with the moving electron will produce a glow. The bright line or curve traced out on the screen indicates the trajectory of the electrons in the electron beam.
(iii) A pair of small horizontal metal plates, P1 and P2, attached to the horizontal sides of the screen. A uniform electric field can be produced between the plates.
(b) a pair of circular Helmholtz coils to produce a magnetic field.
Method
(a) A low voltage is applied to the filament in order to heat it. The hot filament will eject electrons into the glass tube.
(b) A high voltage V is applied between the filament and the anodes. The potential difference will cause the electrons to accelerate towards the anodes, hence producing an electron beam. The beam reaches the screen and, moving on the surface of the screen, will produce a glowing coloured horizontal line on the screen.
(c) The same voltage V is then applied to the pair of horizontal metal plates, P1 and P2, in order to produce a uniform vertical electric field in the region of the screen. The electrons moving in the field will be attracted towards the positively charged plate, hence producing a glowing curve on the screen.
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Figure 21.51 The glass tube is placed between two Helmholtz coils (C1 and C2). A current I produces a horizontal magnetic field, which is adjusted until the electron beam becomes straight, indicating balanced electric and magnetic forces. |
(d) The glass tube is then sandwiched between a pair of Helmholtz coils, C1 and C2, as shown in Figure 21.51. A current I is allowed to flow through both coils in order to produce a horizontal magnetic field at the region of the screen. The magnitude and direction of current I is adjusted until the glowing curve seen on the screen becomes straight and horizontal again.
Principle
Any electron moving through both the electric and magnetic fields is acted on by an electric force Fe and magnetic force Fm. The two forces are given by
Fe = eE
= e (V / d)
where e and d represent the electronic charge and the separation of plates P1 and P2 respectively.
Fm = Bev
where B and v represent the magnetic field strength and the speed of the electron in the fields respectively. The two fields are set up in such a way that the electron velocity and the fields are mutually perpendicular to one another. When an electron travels through both fields in a straight line, it means that the forces Fe and Fm oppose each other and are of equal magnitude. The net force acting on the electron is zero. Hence, we have
Fm = Fe
Bev = e (V / d)
v = (1 / B) (V / d) ........ (1)
An electron accelerates from the filament to the anode through a p.d. of V volts. Its final kinetic energy K is given by
K = lost in electric potential energy
1/2 mu2 = eV
u2 = 2 (e/m) V …… (2)
Combining (1) and (2),
2 (e/m) V = 1/B2 (V/d)2
e/m = 1/2 V (1/Bd)2
The quantities V, B and d can be measured. Hence, the charge-to-mass ratio e/m can be calculated.
Applications of Charge-to-Mass Ratio (q/m) and Magnetic Field
| No | Application | Description | Usage |
|---|---|---|---|
| 1 | Mass Spectrometer | Identifies particles by measuring mass-to-charge ratio. | Chemistry, forensics, pharmaceuticals |
| 2 | e/m Measurement | Determines charge-to-mass ratio of electrons. | Fundamental physics experiments |
| 3 | Velocity Selector | Selects particles with a specific velocity using crossed fields. | Particle accelerators, labs |
| 4 | Cyclotron | Accelerates charged particles using magnetic fields. | Nuclear research, cancer therapy |
| 5 | CRT Technology | Controls electron beams using magnetic deflection. | Old TVs, oscilloscopes |
| 6 | MRI Scanner | Uses strong magnetic fields to create body images. | Medical imaging |
| 7 | Aurora | Charged particles interact with Earth's magnetic field. | Natural phenomenon |
| Conclusion |
| The concepts of charge-to-mass ratio (q/m) and magnetic fields are essential in physics, enabling the control of particle motion and supporting applications in science, medicine, and advanced technology. |
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