Topic 6 Grand Challenge: The Particle Accelerator Power & Analysis System

Topic 6 Grand Challenge: The Particle Accelerator Power & Analysis System

To master Topic 6: Electric and Magnetic Fields, this comprehensive challenge integrates the electrostatic dynamics of Electric Fields (6A), the energy transformations and timing mechanics of Capacitors (6B), and the induction-based features of Electromagnetic Effects (6C).

Context: A medical research facility uses a high-voltage system to accelerate electrons for X-ray production. The system uses a capacitor bank to provide high-energy pulses, and a magnetic deflection system to monitor the beam parameters.

Part 1: Electric Fields (6A)

Question 1: Uniform Accelerating Fields

The electrons are accelerated between two parallel horizontal plates separated by a distance of $1.5\text{ cm}$. A potential difference (p.d.) of $12\text{ kV}$ is applied across the plates.

• Calculate the electric field strength ($E$) between the plates and state its direction.
• Determine the magnitude of the electrostatic force acting on a single electron while it is between the plates.

Worked Solution:

1. Identify Given Parameters & Convert to SI Base Units:

• Plate separation distance ($d$) = $1.5\text{ cm}=0.015\text{ m}$
• Potential Difference ($V$) = $12\text{ kV}=12000\text{ V}$
• Elementary charge ($e$) = $1.60\times {10}^{-19}\text{ C}$

2. Uniform Electric Field Calculation:

Using the uniform field equation connecting voltage gradient to field intensity:

$$E=\frac{V}{d}=\frac{12000}{0.015}=800000\text{ V m}{-}^1$$

Direction: By convention, electric field lines always point from the positive plate to the negative plate (vertically downwards if the positive plate is positioned on top).

3. Electrostatic Force Magnitude Evaluation:

Applying the base definition relating field intensity to force density ($F=Eq$):

$$F=Ee=8.0\times {10}^5\times 1.60\times {10}^{-19}=1.28\times {10}^{-13}\text{ N}$$

Electric Field Strength (E) = 8.0 × 10⁵ V m⁻¹ (pointing positive to negative)
Electrostatic Force (F) = 1.3 × 10⁻¹³ N

Question 2: Radial Fields & Potential

In another part of the machine, a spherical electrode is charged to $+5.0\mu \text{C}$.

• Define electric potential.
• Calculate the electric potential at a distance of $25\text{ cm}$ from the centre of this sphere.
• Describe the relationship between the electric field lines and the equipotential surfaces around this sphere.

Worked Solution:

1. Definition:

Electric potential at a point in an electric field is the work done per unit positive charge in bringing a small test charge from infinity to that specific point.

2. Point-Charge Field Calculations:

• Charge source ($Q$) = $+5.0\mu \text{C}=5.0\times {10}^{-6}\text{ C}$
• Radius distance ($r$) = $25\text{ cm}=0.25\text{ m}$
• Permittivity constant parameter ($\frac{1}{4\pi {\epsilon }_0}$) = $8.99\times {10}^9\text{ N m}{2}^{}\text{C}{-}^2$

Electric Potential (V) = 1.8 × 10⁵ V

3. Geometric Spatial Relationships:

For a radial charge distribution, electric field lines diverge symmetrically outwards in all directions. Equipotential surfaces form concentric spherical shells centered around the electrode core. Consequently, **electric field lines always intersect equipotential surfaces at right angles ($90\circ$)** everywhere in space.

Part 2: Capacitors (6B)

Question 3: Capacitor Bank Energy Storage

The power for the electron pulse is stored in a $470\mu \text{F}$ capacitor charged to a p.d. of $250\text{ V}$.

Calculate the total charge stored on the capacitor and the total energy it contains.

Worked Solution:

1. Identify Variables:

• Capacitance ($C$) = $470\mu \text{F}=470\times {10}^{-6}\text{ F}$
• Charging Voltage ($V$) = $250\text{ V}$

2. Compute Accumulated Charge ($Q$):

$$Q=CV=470\times {10}^{-6}\times 250=0.1175\text{ C}$$

3. Compute Potential Energy Stored ($W$):

$$W=\frac{1}{2}C{V}^2=\frac{1}{2}\times 470\times {10}^{-6}\times {250}^2=14.6875\text{ J}$$

Total Charge Stored (Q) = 0.12 C
Total Energy Stored (W) = 15 J

Question 4: Discharge Mathematics & Transient Curve

To create the pulse, the capacitor is discharged through a $2.0\text{ k}\Omega$ resistor.

• Calculate the time constant ($\tau$) for this circuit.
• Determine the percentage of the initial charge remaining on the capacitor after $1.5\text{ seconds}$ of discharge.
• Sketch a graph of current ($I$) against time ($t$) for this discharge, labeling the initial current ($I_0$).

Worked Solution:

1. Time Constant ($\tau$) Calculation:

$$\tau =RC=(2.0\times {10}^3)\times (470\times {10}^{-6})=0.94\text{ s}$$

2. Charge Retention Evaluation:

The standard model for exponential capacitive decay tracking remaining charge over time is:

$$Q=Q_0{e}^{-\frac{t}{RC}}⟹\frac{Q}{Q_0}=e^{-\frac{1.5}{0.94}}$$

$$\frac{Q}{Q_0}=e^{-1.5957}\approx 0.2027$$

Expressing this structural ratio as a percentage: $0.2027\times 100=20.27\%$.

3. Initial Current Value ($I_0$):

$$I_0=\frac{V_0}{R}=\frac{250}{2000}=0.125\text{ A}$$

Time Constant (τ) = 0.94 s
Remaining Charge Percentage = 20%

4. Circuit Discharge Characteristic Graph:

Part 3: Electromagnetic Effects (6C)

Question 5: Magnetic Deflection & Path Derivation

After acceleration, the electron beam enters a region with a uniform magnetic field of flux density $0.15\text{ T}$ acting at right angles to the beam's velocity.

• Explain why the magnetic field causes the electrons to follow a circular path and derive the expression $r=\frac{p}{BQ}$ for the radius of this path.
• Apply Fleming’s Left Hand Rule to determine the direction of the force on the electrons.

Worked Solution:

1. Circular Path Explanation:

The magnetic force acting on a moving charged particle is always perpendicular to its velocity vector ($F=BQv\cdot \sin 90\circ$). Because this force acts perpendicular to the direction of motion, it changes the direction of the particle's velocity without altering its speed. This perpendicular force provides a constant centripetal acceleration, causing the electron beam to follow a circular path.

2. Expression Derivation:

Equating centripetal force to the magnetic Lorentz force:

$$\frac{m{v}^2}{r}=BQv$$

Cancel one factor of linear velocity ($v$) from both sides:

$$\frac{mv}{r}=BQ⟹r=\frac{mv}{BQ}$$

Since linear momentum is defined as mass times velocity ($p=mv$), substituting $p$ into the equation gives:

$$r=\frac{p}{BQ}$$

3. Vector Alignment (Fleming's Left Hand Rule):

• First Finger: Points in the direction of the Magnetic Field ($B$).
• Second Finger: Points in the direction of Conventional Current ($I$). Since electrons carry a negative charge, conventional current flows in the opposite direction of the electron beam's velocity vector.
• Thumb: The resulting orientation of your thumb gives the direction of the magnetic force, which directs the beam into its circular arc.

Question 6: Electromagnetic Induction Monitor

To monitor the beam frequency, a small coil of $500\text{ turns}$ and cross-sectional area $2.0\text{ cm}{2}^{}$ is placed near the pulse generator. The magnetic flux through the coil changes from $0.5\text{ mWb}$ to zero in $0.01\text{ s}$ during a pulse.

• Define magnetic flux linkage.
• Calculate the average e.m.f. induced in the coil during this time interval using Faraday’s Law.
• State Lenz’s Law and explain how it determines the polarity of the induced e.m.f.

Worked Solution:

1. Definition:

Magnetic flux linkage ($N\Phi$) is the product of the magnetic flux passing through a surface and the number of turns in the coil cutting those magnetic field lines.

2. Faraday's Law Calculation Steps:

• Number of turns ($N$) = $500$
• Initial flux linkage (${\Phi }_1$) = $0.5\text{ mWb}=0.5\times {10}^{-3}\text{ Wb}$
• Final flux linkage (${\Phi }_2$) = $0\text{ Wb}$
• Time delta interval ($\Delta t$) = $0.01\text{ s}$

Applying Faraday's Law, which states that the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage:

Induced electromotive force (ε) = 25 V

3. Directionality Analysis via Lenz's Law:

Lenz’s Law states that the direction of an induced current or e.m.f. is always such that it opposes the change in magnetic flux that produced it.

In this system, the local magnetic flux is collapsing toward zero. According to Lenz's Law, the induced e.m.f. sets up a current with a polarity that creates its own magnetic field pointing in the same direction as the original field, attempting to sustain the collapsing flux. This opposition to changes in flux explains the negative sign in the mathematical expression of Faraday's Law ($\epsilon =-\frac{\Delta \left(N\Phi \right)}{\Delta t}$).