Motion of charged particles

Question 1

Direction of conventional current and deflection of a beam of electrons as it crosses a magnetic field (Figure 25.3 pg. 510)

Answer 1

In Figure 25.3, a beam of electrons is moving from right to left, into a region where a magnetic field is directed into the plane of the paper. Since electrons are negatively charged, they represent a conventional current from left to right. Fleming’s left-hand rule predicts that, as the electrons enter the field, the force on them will be upwards and so the beam will be deflected up the page.

Question 2

How does a current-carrying wire experiences force?

Answer 2

The force due to the magnetic field is always at 90° to the velocity of the electrons. It is this force that gives rise to the motor effect. The electrons in a wire experience a force when they flow across a magnetic field, and they transfer the force to the wire itself.

Question 3

Magnetic force on a moving charged particle

Answer 3

The factors that determine the size of the
force on a charged particle depend on:
1- the magnetic flux density B (strength of the magnetic field)
2- the charge Q on the particle the speed ν of the particle
3- the speed ν of the particle.
The magnetic force F on a moving particle at right angles to a magnetic field is given by the equation:
F = BQν
The direction of the force can be determined from Fleming’s left-hand rule. The force F is always at 90° to the velocity of the particle. Consequently, the path described by the particle will be an arc of a circle. If the charged particle is moving at an angle θ to the magnetic field, the component of its velocity at right angles to B is v sin θ. Hence, the equation becomes:
F = BQνsinθ
where θ is the angle between the magnetic field and the velocity of the particle.

Question 4

Prove that force on a moving charged particle: “Beν force” is the force acting on all the electrons in a wire that gives rise to the “BIL force”

Answer 4

The two equations F = BIL and F = BQν are consistent with one another.
Substituting Q/t in the equation:
F = BQL / t
Now, L / t is the speed ν of the moving particle, so we can write:
F = BQν
For an electron, with a charge of −e, the magnitude of the force is:
F = Beν (e = 1.60 × 10^−19 C)
Hence proven that the force on a moving charged particle: “Beν force” is the force acting on all the electrons in a wire that gives rise to the “BIL force”.

Question 5

IMPT!! - What path does a charged particle follow in a uniform magnetic field?

Answer 5

A charged particle of mass m and charge Q moving at right angles to a uniform magnetic field will describe a circular path because the magnetic force F is always perpendicular to its velocity (the charged particle’s). The magnetic force F, provides the centripetal force on the particle – the direction of the force is always towards the centre of the circle.
Example ➡ Figure 25.8 (pg. 512) shows a fine-beam tube. In this tube, a beam of fast-moving electrons is produced by an electron gun. The beam is directed vertically downwards as it emerges from the gun. It enters the spherical tube, which has a uniform horizontal magnetic field. The beam is at right angles to the magnetic field and the Beν force pushes it round in a circle.

Question 6

Finding the radius of the circular path followed by a charged particle

Answer 6

The fact that the centripetal force is provided by the magnetic force BQv, gives us a clue as to how we can calculate the radius r of the orbit of a charged particle in a uniform magnetic field. The centripetal force is given by:
centripetal force = mv^2 / r
Therefore,
F magnetic force = F centripetal
BQv = mv^2/r
Cancelling and rearranging gives:
r = mv / BQ
If the charged particles are electrons, then Q is numerically equal to e. The equation then becomes:
r = mv /Be

Question 7

Momentum of the charged particle moving in a circular path due to BQv force acting as a centripetal force

Answer 7

r = mv / BQ
The momentum p of the particle is mv. You can therefore write the equation as:
p = BQr
In case of the charged particle being an electron:
p = Ber

Question 8

What does the equation r = mv / BQ prove?

Answer 8

The equation r = mv / BQ shows that:

♦ faster-moving particles move in bigger circles because r ∝ v.
♦ particles with greater masses also move in bigger circles because r ∝ m.
♦ particles with greater charge move in tighter (smaller) circles because r ∝ 1/Q
♦ a stronger field (greater magnetic flux density) makes the particles move in tighter circles because r ∝ 1/B (greater magnetic flux density applies greater force on the charged particle).

Question 9

What is charge-to-mass ratio of an electron?

Answer 9

Experiments to find the mass of an electron first involve finding the charge-to-mass ratio e/m. This is known as the specific charge on the electron – the word ‘specific’ here means ‘per unit mass’. Using the equation for an electron travelling in a circle in a magnetic field, we have e/m = v/Br. Clearly, measurements of v, B and r are needed to determine e/m. There are difficulties in measuring B and r. For example, it is difficult to directly measure r with a ruler outside the tube in Figure 25.8 because of parallax error. Velocity v must also be measured.

Question 10

Calculating charge-to-mass ratio (direct-way)

Answer 10

The potential difference (p.d.) Vca between the cathode and the anode causes each electron to accelerate as it moves from the cathode to the anode. An individual electron has charge −e, therefore an amount of work is done on each electron is e × Vca. This is equivalent to the kinetic energy of the electron as it leaves the anode - we assume that the electron has zero kinetic energy at the cathode. Therefore:
e × Vca = 1/2 mv^2
[Remember: The work done W in moving charge Q through a constant p.d V is W = VQ]

If the values of B and r are provided, then after finding velocity v from the above eqn, the value for e/m can be calculated by e/m = v/Br.
↓
↓
↓
↓
[r = mv / Be (from F magnetic = F centripetal)
Ber = mv
e/m = v/Br]

Question 11

Calculating charge-to-mass ratio (round-about way)

Answer 11

Eliminating v from the equations to find e/m ratio:
e × Vca = 1/2mv^2 and r = mv/Be—>v = Ber/m
gives,
e/m = 2Vca / r^2 B^2
↓
↓
[Derivation:
1/2mv^2 = e × Vca and v = Ber/m
1/2m(B^2e^2r^2 / m^2) = e × Vca
B^2er^2 / 2m = Vca
e/m = 2Vca / r^2B^2]

A voltmeter can be used to measure Vca, and if r and B are known, the ratio e/m can be calculated.

[Any method can be used to calculate e/m depending on what quantities the Q provides - but recall these equations again and again as the Qs will be confusing]

Question 12

Question asked : An equation relating e and p.d (Vca)?

Answer 12

e × Vca = 1/2 mv^2

Question 13

Worked Example 1 - An electron is travelling at right angles to a uniform magnetic field of flux density 1.2 mT. The speed of the electron is 8.0 × 10^6 m s−1. Calculate the radius of circle described by this electron.
e = 1.60 × 10^−19 C and m = 9.11 × 10−^31 kg

Answer 13

Step 1: Calculate the magnetic force on the electron.
F = Bev = (1.2 × 10^-3) × (1.6 × 10^-19) × (8 ×10^6)
F = 1.536 × 10^-15 N
Step 2: Use your knowledge of motion in a circle to determine the radius r.
F = mv^2/r
Therefore:
r = mv^2 / F = 9.11 × 10^-31 × (8 ×10^6)^2 / = 3.8 × 10^2 m
Note: You can get the same result by using the equation:
r = mv/Be

Question 14

Practical Activity 25.2 - The deflection tube

Answer 14

A deflection tube (Figure 25.9) is designed to show a beam of electrons passing through a combination of electric and magnetic fields. The magnetic field is provided by two vertical coils, called Helmholtz coils which give a very uniform field in the space between them. When the electron beam remains straight, it follows that the electric and magnetic forces on each electron must have the same magnitude and act in opposite directions.
Therefore: electric force (upwards ‘towards the positive plate’) = magnetic force (downwards ‘using Fleming’s’ left hand rule)
eE = Bev

Therefore: electric force (upwards) = magnetic force (downwards) electrons is simply related to E and B because e in the expression cancels out. Therefore:
v = E / B
The electric field strength E is given by:
E = V / d
where V is the p.d. between the plates and d is the distance between the plates. Therefore:
v = V / Bd

Question 14

Velocity selection

Answer 14

In a device called a velocity selector, charged particles of a specific velocity are selected using both electric and magnetic fields. i.e. This is used in devices such as mass spectrometers where it is essential to produce a beam of charged particles all moving with the same velocity.
Fig. 25.11 Pg. 515:
The apparatus is very similar to the deflection tube. Two oppositely charged horizontal plates are situated in an evacuated chamber. These plates provide a uniform electric field of strength E in the space between the plates. The region between the plates is also occupied by a uniform magnetic field of flux density B that is at right angles to the electric field. Negatively charged particles (electrons or ions) enter from the left. They all have the same charge −Q but are travelling at different speeds. The magnitude of the electric force EQ will be the same on all particles as it does not depend on their speed. However, the magnitude of the magnetic force BQv will be greater for those particles that are travelling faster. Hence, for particles travelling at the desired speed v, the electric force and the magnetic force must have the same value, but be in opposite directions. The resultant force on the charged particles in the vertical direction must be zero, and all the charged particles with the speed v will emerge undeflected from the slit S. Therefore:
QE = BQν
ν = E / B
If a charged particle has a speed greater than E / B, the downward magnetic force on it will be greater than the upward electric force. Thus, it will be deflected downwards and it will hit below slit S. i.e. if E/B = 15000 ms^-1 and v > 15000 ms^-1 than Magnetic force will be greater than Electric force as Fmagnetic ∝ ν
[Note that we do not have to concern ourselves with the gravitational force mg acting on the charged particles as this will be negligible compared with the electric and magnetic forces.]

Question 15

Define Hall effect (& Brief explanation)

Answer 15

Definition: “The production of a potential difference across an electrical conductor when an external magnetic field is applied in a direction perpendicular to the direction of the current.”
If the conductor is a metal, then the current is due to the flow of electrons. These electrons will experience a magnetic force, which will make them drift towards one side of the conductor, where they will gather. The opposite side of the slice is deficient of electrons. A potential difference, known as the Hall voltage, will be developed across the conductor. As you will see later, the Hall voltage VH for the slice is constant for a given current and is directly proportional to the magnetic flux density B of the external magnetic field.

Question 16

Equation for the Hall Voltage Vᵸ

Answer 16

Using what we know about electric current and the forces on electric charges produced by electric and magnetic fields, we can derive an expression for the Hall voltage Vᵸ.
Figure 25.13 shows a current-carrying slice of a metal. The Hall voltage is the voltage that appears between the two opposite sides of the slice.
As we have seen, this voltage arises because electrons accumulate on one side of the slice. There is a corresponding lack of electrons on the opposite side – this opposite side may be considered to have a positive charge. As a result, there is an electric field set up within the slice between the two sides. The two charged sides may be treated as oppositely charged parallel plates. Therefore, the electric field strength E is related to the Hall voltage VH by:
E = Vᵸ / d
where d is the width of the slice.
Now, imagine a single electron as it travels with drift velocity v through the slice. The magnetic field is into the plane of the paper, so this electron will experience a magnetic force Bev to the right. It will also experience an electric force Ee to the left. When the current first starts to flow, there is no Hall voltage and so electrons are pushed to the right by the magnetic field. However, as the charge on the right-hand side builds up, so does the internal electric field and this pushes the electrons in the opposite direction to the magnetic force. Soon, an equilibrium situation is reached, the resultant force on each electron is zero and the electrons are undeflected. Now we can equate the two forces:
eE = Bev
Substituting for E we have:
eVᵸ / d = Bev
Now recalling that the current I is related to the mean drift velocity v of the electrons by 𝙸 = nAve, where A is the cross-sectional area of the conductor and n is the number density of charge carriers (in this case, electrons). So, we can substitute for v to get:
eVᵸ / d = Be𝙸 / nAe
Making VH the subject of the equation (and cancelling e) gives:
Vᵸ = B𝙸d / nAe
The cross-sectional area A of each side-face of the slice is:
A = d × t
​where t is the thickness of the slice.
Substituting and cancelling gives:
Vᵸ = B𝙸 /nte
This equation for the Hall voltage shows that Vᵸ is directly proportional to the magnetic flux density B for a given slice and current. That is what makes the Hall effect so useful for measuring B.
General version of the equation:
Vᵸ = B𝙸 /ntq

Answer 17

To get a large voltage, it would be desirable to have a material with a smaller value for n compared with metals. Hall probes use a very thin slice of semiconductor. Semiconductors have a number density many thousands of times smaller than metals, hence the Hall voltage will be thousands of times larger. In some semiconductors, the charge carriers are not electrons, but positively charged particles referred to as ‘holes’. We can write a more general equation for the Hall voltage replacing e with q, where q is the charge of an individual charge carrier. This gives
Vᵸ = B𝙸 /nte
Positive charges will be deflected in the opposite direction to negative charges, and so we can determine whether the charge carriers are positive or negative by the sign of the Hall voltage.

Question 18

Q9 (P.g 517) - Suggest how the Hall effect could be used to determine the number density of charge carriers n in a semiconducting material.

Answer 18

The Hall voltage is given by the equation:
Vᵸ = B𝙸 /nte
Using a single sample of a semiconducting material of measured thickness t, and using a constant magnetic field of known flux density B, the Hall voltage could be measured for different values of current. By plotting a graph of Hall voltage against current, the graph should form a straight line passing through the origin, and have a gradient = B / nte, from which the number density n of the semiconductor can be determined.

Question 19

Summary of observations and conclusions made by J.J. Thompson on electrons

Answer 19

• The beam in his tube was deflected towards a positive plate and away from a negative plate, so the particles involved must have negative charge. This was confirmed by the deflection of the beam by a magnetic field.
• When the beam was deflected, it remained as a tight, single beam rather than spreading out into a broad beam. This showed that, if the beam consisted of particles, they must all have the same mass, charge and speed. (Lighter particles would have been deflected more than heavier ones; particles with greater charge would be deflected more, and faster particles would be deflected less)
• By applying both electric and magnetic fields, Thomson was able to balance the electric and magnetic forces so that the beam in the tube remained straight. He could then calculate the charge-to-mass ratio e/mₑ for the particles he had discovered.

Question 20

0E.O.C.Q. 8b -The Hall voltage is measured using the same slice of semiconductor, the same current and the same magnetic field, but with the laboratory at two temperatures, one significantly higher than the other. Describe and explain the changes in the magnitude of the number density, the drift velocity of the charge carriers and the Hall voltage in the two experiments.

Answer 20

Ans:
When the hall voltage is measured using the same slice of semiconductor, the same current and the same magnetic field, but with the laboratory at two different temperatures, one significantly higher than the other. Then, the number density n becomes large and more electrons have enough energy to set free from valence shells and conduct. However the drift velocity v becomes smaller as same charge flows per second and as there are large number of charge carriers so hall voltage Vᵸ becomes smaller.
Smaller force on each charge carrier due to lower drift velocity needs a smaller electric field E to cancel force due to magnetic field.
[ref : Vᵸ = B𝙸 /ntq & Fₘ.𝒻 = Bqv]