IB Physics How to Solve: IB Physics – Electric and Magnetic Fields (Coulomb, Ampere, Induction, Transformers)

How to Solve: IB Physics – Electric and Magnetic Fields (Coulomb, Ampere, Induction, Transformers)

Complete Guide

Introduction

"Mastering electric and magnetic fields unlocks 15–20% of your IB Physics Paper 2—questions on Coulomb’s Law, Ampere’s Law, electromagnetic induction, and transformers appear every year. Get this right, and you’re one step closer to a 7."

WHAT YOU NEED TO KNOW FIRST

1. Newton’s Laws of Motion – Forces cause acceleration.
2. Basic Circuit Theory – Current, voltage, resistance, and Ohm’s Law.
3. Vector Addition – Forces and fields have direction and magnitude.

KEY TERMS & FORMULAS

Electric Fields

1. Coulomb’s Law (MEMORISE THIS)
   [
   F = k \frac{q_1 q_2}{r^2}
   ]
2. (F) = electrostatic force (N)
3. (k) = Coulomb’s constant ((8.99 \times 10^9 \, \text{N m}^2 \text{C}^{-2}))
4. (q_1, q_2) = charges (C)

5. (r) = distance between charges (m)

6. Electric Field Strength (MEMORISE THIS)
   [
   E = \frac{F}{q} = k \frac{Q}{r^2}
   ]

7. (E) = electric field strength (N/C or V/m)

8. (Q) = source charge (C)

9. Electric Potential Energy (MEMORISE THIS)
   [
   U = k \frac{q_1 q_2}{r}
   ]

10. (U) = potential energy (J)

11. Electric Potential (Voltage) (MEMORISE THIS)
    [
    V = k \frac{Q}{r}
    ]

12. (V) = electric potential (V)

Magnetic Fields

1. Magnetic Force on a Moving Charge (MEMORISE THIS)
   [
   F = qvB \sin \theta
   ]
2. (F) = magnetic force (N)
3. (q) = charge (C)
4. (v) = velocity (m/s)
5. (B) = magnetic field strength (T)

6. (\theta) = angle between (v) and (B)

7. Magnetic Force on a Current-Carrying Wire (MEMORISE THIS)
   [
   F = ILB \sin \theta
   ]

8. (I) = current (A)

9. (L) = length of wire (m)

10. Ampere’s Law (GIVEN ON EXAM SHEET)
    [
    \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}
    ]

11. (\mu_0) = permeability of free space ((4\pi \times 10^{-7} \, \text{T m A}^{-1}))

12. (I_{\text{enc}}) = current enclosed by loop (A)

13. Magnetic Field Inside a Solenoid (MEMORISE THIS)
    [
    B = \mu_0 n I
    ]

14. (n) = number of turns per unit length (m⁻¹)

Electromagnetic Induction

1. Faraday’s Law (GIVEN ON EXAM SHEET)
   [
   \mathcal{E} = -N \frac{d\Phi_B}{dt}
   ]
2. (\mathcal{E}) = induced EMF (V)
3. (N) = number of turns

4. (\Phi_B) = magnetic flux (Wb)

5. Magnetic Flux (MEMORISE THIS)
   [
   \Phi_B = BA \cos \theta
   ]

   • (A) = area (m²)
   • (\theta) = angle between (B) and normal to area

6. Lenz’s Law (MEMORISE THIS)

   • The induced current opposes the change in flux.

Transformers

1. Transformer Equation (MEMORISE THIS)
   [
   \frac{V_p}{V_s} = \frac{N_p}{N_s}
   ]

   • (V_p, V_s) = primary/secondary voltage (V)
   • (N_p, N_s) = primary/secondary turns

2. Power in Transformers (MEMORISE THIS)
   [
   V_p I_p = V_s I_s \quad \text{(assuming 100% efficiency)}
   ]

STEP-BY-STEP METHOD

Step 1: Identify the Topic

• Is it electric fields (Coulomb, potential)?
• Is it magnetic fields (Ampere, force on wire)?
• Is it induction (Faraday, Lenz)?
• Is it transformers (voltage ratio, power)?

Step 2: List Given Quantities

• Write down every number and unit from the question.
• Convert units if needed (e.g., cm → m, mA → A).

Step 3: Choose the Right Formula

• Match the question to the correct formula.
• If unsure, check the KEY TERMS & FORMULAS section.

Step 4: Solve for the Unknown

• Rearrange the formula if needed.
• Plug in numbers with units.
• Calculate.

Step 5: Check Direction (If Applicable)

• For forces: Use right-hand rule (positive charges) or left-hand rule (negative charges).
• For induction: Apply Lenz’s Law to determine direction of induced current.

Step 6: Verify Units and Significant Figures

• Final answer must have correct units.
• Round to 3 significant figures (IB standard).

WORKED EXAMPLES

Example 1 – Basic: Coulomb’s Law

Question: Two point charges, (q_1 = +3.0 \, \mu\text{C}) and (q_2 = -5.0 \, \mu\text{C}), are 0.20 m apart. Calculate the electrostatic force between them.

Solution: 1. Identify: Electric force → Coulomb’s Law. 2. Given:
- (q_1 = +3.0 \times 10^{-6} \, \text{C})
- (q_2 = -5.0 \times 10^{-6} \, \text{C})
- (r = 0.20 \, \text{m}) 3. Formula: (F = k \frac{q_1 q_2}{r^2}) 4. Plug in:
[
F = (8.99 \times 10^9) \frac{(3.0 \times 10^{-6})(-5.0 \times 10^{-6})}{(0.20)^2}
] 5. Calculate:
[
F = (8.99 \times 10^9) \frac{-15 \times 10^{-12}}{0.04} = -3.37 \, \text{N}
] 6. Interpret: Negative sign = attractive force (opposite charges). 7. Final Answer: (3.37 \, \text{N}) (attractive).

What we did and why: - Used Coulomb’s Law because we had two point charges. - Negative sign indicates attraction (opposite charges). - Converted (\mu\text{C}) to C for correct units.

Example 2 – Medium: Magnetic Force on a Wire

Question: A wire of length 0.50 m carries a current of 4.0 A at an angle of 30° to a uniform magnetic field of 0.20 T. Calculate the magnetic force on the wire.

Solution: 1. Identify: Magnetic force on current-carrying wire. 2. Given:
- (I = 4.0 \, \text{A})
- (L = 0.50 \, \text{m})
- (B = 0.20 \, \text{T})
- (\theta = 30°) 3. Formula: (F = ILB \sin \theta) 4. Plug in:
[
F = (4.0)(0.50)(0.20) \sin 30°
] 5. Calculate:
[
F = 0.40 \times 0.5 = 0.20 \, \text{N}
] 6. Direction: Use right-hand rule (thumb = current, fingers = field, palm = force). 7. Final Answer: (0.20 \, \text{N}) perpendicular to both wire and field.

What we did and why: - Used (F = ILB \sin \theta) because the wire is at an angle to the field. - (\sin 30° = 0.5) reduces the force. - Direction matters—always state it!

Example 3 – Exam-Style: Transformer Efficiency

Question: A transformer steps down 240 V to 12 V. The primary coil has 1200 turns. The secondary current is 2.0 A. If the transformer is 90% efficient, calculate the primary current.

Solution: 1. Identify: Transformer voltage ratio and efficiency. 2. Given:
- (V_p = 240 \, \text{V})
- (V_s = 12 \, \text{V})
- (N_p = 1200)
- (I_s = 2.0 \, \text{A})
- Efficiency = 90% = 0.90 3. Step 1: Find (N_s) (turns ratio)
[
\frac{V_p}{V_s} = \frac{N_p}{N_s} \implies \frac{240}{12} = \frac{1200}{N_s} \implies N_s = 60
] 4. Step 2: Find ideal (I_p) (100% efficiency)
[
V_p I_p = V_s I_s \implies 240 I_p = 12 \times 2.0 \implies I_p = 0.10 \, \text{A}
] 5. Step 3: Adjust for efficiency
[
\text{Power input} = \frac{\text{Power output}}{\text{Efficiency}} \implies V_p I_p = \frac{V_s I_s}{0.90}
]
[
240 I_p = \frac{12 \times 2.0}{0.90} \implies I_p = \frac{24}{240 \times 0.90} = 0.111 \, \text{A}
] 6. Final Answer: (0.111 \, \text{A}) (3 s.f.).

What we did and why: - First used voltage ratio to find secondary turns. - Then used power conservation (assuming 100% efficiency). - Finally, adjusted for real-world efficiency (90%).

COMMON MISTAKES

1. MISTAKE: Forgetting to convert (\mu\text{C}) to C in Coulomb’s Law.
   WHY IT HAPPENS: Students see (\mu\text{C}) and plug it in directly.
   CORRECT APPROACH: Always convert to base units ((1 \, \mu\text{C} = 10^{-6} \, \text{C})).

2. MISTAKE: Ignoring the (\sin \theta) in (F = ILB \sin \theta).
   WHY IT HAPPENS: Students assume (\theta = 90°) (max force) when it’s not given.
   CORRECT APPROACH: Check the angle—if not 90°, multiply by (\sin \theta).

3. MISTAKE: Mixing up right-hand and left-hand rules.
   WHY IT HAPPENS: Confusion between positive and negative charges.
   CORRECT APPROACH:

4. Right-hand rule for positive charges (or conventional current).

5. Left-hand rule for negative charges (or electron flow).

6. MISTAKE: Forgetting Lenz’s Law in induction problems.
   WHY IT HAPPENS: Students calculate magnitude but ignore direction.
   CORRECT APPROACH: Always state: "The induced current opposes the change in flux."

7. MISTAKE: Assuming transformers are 100% efficient.
   WHY IT HAPPENS: Questions often give efficiency but students ignore it.
   CORRECT APPROACH: If efficiency is given, use:
   [
   \text{Power input} = \frac{\text{Power output}}{\text{Efficiency}}
   ]

EXAM TRAPS

1. TRAP: Giving a force without direction.
   HOW TO SPOT IT: Question says "calculate the force" but doesn’t specify if direction is needed.
   HOW TO AVOID IT: Always state direction (e.g., "attractive force" or "perpendicular to the wire").

2. TRAP: Using (F = qvB) when the charge is stationary.
   HOW TO SPOT IT: Question mentions a charge in a magnetic field but doesn’t say it’s moving.
   HOW TO AVOID IT: Magnetic force only acts on moving charges. If (v = 0), (F = 0).

3. TRAP: Misapplying transformer power equations.
   HOW TO SPOT IT: Question gives both voltage and current but asks for power.
   HOW TO AVOID IT:

4. If efficiency is not given, use (V_p I_p = V_s I_s).
5. If efficiency is given, use (V_p I_p = \frac{V_s I_s}{\text{Efficiency}}).

1-MINUTE RECAP

"Alright, last-minute cram? Here’s what you need: 1. Coulomb’s Law: (F = k \frac{q_1 q_2}{r^2}). Opposite charges attract, same charges repel. Convert (\mu\text{C}) to C! 2. Magnetic Force: (F = qvB \sin \theta) for charges, (F = ILB \sin \theta) for wires. Use the right-hand rule for direction. 3. Induction: Faraday’s Law (\mathcal{E} = -N \frac{d\Phi_B}{dt}). Lenz’s Law says the induced current opposes the change. 4. Transformers: (\frac{V_p}{V_s} = \frac{N_p}{N_s}). If efficiency is given, adjust power input. 5. Units matter! Always check: C, T, A, V. Round to 3 s.f. You’ve got this—go ace that exam!