Introduction to Electromagnetism
Dear student, you have already studied about electric charges, electric fields, and electric currents in earlier grades. Now we are going to explore a very important connection between electricity and magnetism. Have you ever wondered how an electric motor works? How does a generator produce electricity? How does a transformer change voltage? All of these depend on the relationship between electricity and magnetism — and that is what electromagnetism is all about.
In this unit, we will learn about magnets and magnetic fields, how electric currents create magnetic fields, how changing magnetic fields can produce electric currents (electromagnetic induction), and the practical devices that use these principles. This is one of the most important and most-tested units in the Grade 12 exam, so let us study it carefully!
4.1 Magnets and Magnetic Field
What is a Magnet?
A magnet is any material or object that can attract certain metals (like iron, nickel, and cobalt) and can exert a force on other magnets. The word “magnetism” comes from the ancient Greek region of Magnesia, where people first discovered a naturally occurring magnetic mineral called lodestone (magnetite).
Properties of Magnets
- Attractive property: Magnets attract materials containing iron, nickel, and cobalt. These materials are called ferromagnetic materials.
- Directive property: A freely suspended magnet always aligns itself in the north-south direction. This property is used in compasses.
- Poles exist in pairs: Every magnet has two poles — a north pole (N) and a south pole (S). You cannot isolate a single pole. If you break a magnet in half, each piece becomes a new magnet with both poles.
- Like poles repel, unlike poles attract: N repels N, S repels S, but N attracts S.
- Magnetic poles are strongest at the ends: The magnetic effect is strongest near the poles and weakest at the center (the neutral region).
Types of Magnets
| Type | Description | Examples |
|---|---|---|
| Natural magnets | Found in nature; weak magnetic strength | Lodestone (magnetite) |
| Artificial (permanent) magnets | Made by humans; strong and durable | Bar magnets, horseshoe magnets, disk magnets |
| Electromagnets | Temporary magnets made by passing current through a coil | Electromagnets in cranes, relays, motors |
| Temporary magnets | Act as magnets only when near a permanent magnet | Soft iron nails near a magnet |
What is a Magnetic Field?
A magnetic field is the region around a magnet where its magnetic influence can be felt. If you place a small compass near a magnet, the compass needle deflects — that tells you a magnetic field exists there. The magnetic field is a vector quantity — it has both magnitude and direction.
The magnetic field strength (also called magnetic flux density or magnetic induction) is denoted by the symbol B. Its SI unit is the Tesla (T). Another common unit is the Gauss (G), where 1 T = 10,000 G.
Magnetic Field of a Bar Magnet
The magnetic field of a bar magnet emerges from the north pole, curves through space, and enters the south pole. Inside the magnet, the field lines run from south to north, forming continuous closed loops.
- Like poles repel; unlike poles attract.
- Magnetic poles always exist in pairs (no magnetic monopoles).
- Magnetic field B is a vector; unit: Tesla (T).
- The Earth’s magnetic field protects us from harmful solar radiation.
- Ferromagnetic materials: iron, nickel, cobalt.
- A compass needle points north because Earth’s geographic North is actually a magnetic South.
Practice Question 1: What happens if you break a bar magnet into two pieces? Does each piece have one pole?
Practice Question 2: Why does a compass needle deflect when placed near a current-carrying wire?
4.2 Magnetic Field Lines
What are Magnetic Field Lines?
Dear student, we cannot see magnetic fields with our eyes. But we can represent them using magnetic field lines (also called lines of magnetic flux). These are imaginary lines drawn to show the direction and strength of the magnetic field at different points in space.
Properties of Magnetic Field Lines
- Direction: Magnetic field lines point from the north pole to the south pole outside the magnet, and from south to north inside the magnet.
- Continuous closed loops: Magnetic field lines always form closed loops. They never start or end in space (unlike electric field lines that start on positive charges and end on negative charges).
- Never intersect: Two magnetic field lines can never cross each other. If they did, it would mean two different directions of the magnetic field at the same point, which is impossible.
- Closeness indicates strength: Where field lines are close together, the magnetic field is strong. Where they are far apart, the field is weak. Near the poles, field lines are crowded (strong field); far from the magnet, they spread out (weak field).
- Tangent gives direction: At any point on a field line, the tangent to the line gives the direction of the magnetic field B at that point.
- Can pass through any medium: Magnetic field lines can pass through vacuum, air, and most materials.
Detecting Magnetic Field Lines
We can visualize magnetic field patterns using:
- Iron filings method: Sprinkle iron filings around a magnet on a sheet of paper. The filings align along the field lines, showing the pattern clearly.
- Compass method: Place several small compasses around a magnet. Each compass needle aligns with the local field direction, tracing out the field lines.
Magnetic Field Lines Between Unlike Poles (Attraction)
When north and south poles of two magnets face each other, the field lines go from the N pole of one magnet directly to the S pole of the other, showing attraction.
Magnetic Field Lines Between Like Poles (Repulsion)
When two north poles face each other, the field lines push away from each other, creating a neutral point between them where the net field is zero. This shows repulsion.
Magnetic Flux
The number of magnetic field lines passing through a given area is called magnetic flux, denoted by Φ (phi):
where B is the magnetic field strength, A is the area, and θ is the angle between B and the normal to the surface.
Unit: Weber (Wb). Note that 1 Wb = 1 T·m². So 1 T = 1 Wb/m².
Worked Example 1
A magnetic field of 0.5 T passes through a rectangular area of 0.04 m² at an angle of 60° to the normal of the surface. Find the magnetic flux through the area.
$$\Phi = BA\cos\theta = 0.5 \times 0.04 \times \cos 60°$$ $$\Phi = 0.02 \times 0.5 = 0.01 \text{ Wb}$$
- Field lines go from N to S outside, S to N inside the magnet.
- They form continuous closed loops — never break or end.
- They never cross each other.
- Closeness of lines = strength of field.
- Tangent at any point gives the direction of B.
- Magnetic flux Φ = BA cos θ; unit: Weber (Wb).
- Between unlike poles: field lines connect them (attraction).
- Between like poles: neutral point exists where B = 0 (repulsion).
Practice Question 3: A circular loop of radius 5 cm is placed perpendicular to a uniform magnetic field of 0.3 T. Find the magnetic flux through the loop.
A = πr² = π(0.05)² = π × 0.0025 = 0.00785 m²
$$\Phi = BA\cos\theta = 0.3 \times 0.00785 \times 1 = 0.00236 \text{ Wb}$$
Practice Question 4: Why can magnetic field lines never cross each other?
4.3 Current and Magnetism
Oersted’s Discovery
Dear student, for a long time, people thought electricity and magnetism were completely separate phenomena. But in 1820, a Danish scientist named Hans Christian Oersted made a remarkable discovery during a lecture demonstration. He noticed that a compass needle placed near a current-carrying wire deflected from its usual north-south position. This proved that an electric current produces a magnetic field — electricity and magnetism are related!
Magnetic Field Due to a Straight Current-Carrying Wire
When current flows through a straight wire, the magnetic field forms concentric circles around the wire. The direction of the field is given by the right-hand grip rule:
Right-hand grip rule: Grasp the wire with your right hand such that your thumb points in the direction of the conventional current. Your curled fingers then point in the direction of the magnetic field lines around the wire.
Magnitude of the Magnetic Field
The strength of the magnetic field at a distance r from a long straight wire carrying current I is:
where μ₀ = 4π × 10⁻⁷ T·m/A is the permeability of free space (also called the magnetic constant).
Worked Example 2
A long straight wire carries a current of 5 A. What is the magnetic field at a point 10 cm from the wire?
I = 5 A, r = 10 cm = 0.1 m
$$B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.1}$$ $$B = \frac{4\pi \times 5 \times 10^{-7}}{0.2\pi} = \frac{20 \times 10^{-7}}{0.2} = 100 \times 10^{-7} = 10^{-5} \text{ T}$$ $$B = 10 \mu\text{T}$$
Magnetic Field at the Center of a Circular Loop
When current flows through a circular loop of radius R, the magnetic field at the center of the loop is:
The direction is given by another right-hand rule: curl your right-hand fingers in the direction of current around the loop, and your thumb points in the direction of the magnetic field at the center.
Magnetic Field of a Solenoid
A solenoid is a long coil of wire with many turns. When current flows through it, the magnetic field inside is:
where n = N/L is the number of turns per unit length (N = total turns, L = length of solenoid).
- Inside a long solenoid, the magnetic field is nearly uniform and parallel to the axis.
- Outside the solenoid, the field is very weak (similar to a bar magnet).
- A solenoid with a ferromagnetic core (like soft iron) becomes a powerful electromagnet.
- The field inside is independent of the solenoid’s cross-sectional area.
Worked Example 3
A solenoid of length 0.5 m has 500 turns and carries a current of 2 A. Find the magnetic field inside the solenoid.
n = N/L = 500/0.5 = 1000 turns/m
$$B = \mu_0 n I = 4\pi \times 10^{-7} \times 1000 \times 2$$ $$B = 8\pi \times 10^{-4} = 25.13 \times 10^{-4} = 2.513 \times 10^{-3} \text{ T} \approx 2.5 \text{ mT}$$
Worked Example 4
A circular loop of radius 0.1 m carries a current of 3 A. Find the magnetic field at the center of the loop.
$$B = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \times 3}{2 \times 0.1}$$ $$B = \frac{12\pi \times 10^{-7}}{0.2} = 60\pi \times 10^{-7} = 1.885 \times 10^{-5} \text{ T}$$
Force on a Current-Carrying Conductor in a Magnetic Field
When a current-carrying wire is placed in an external magnetic field, it experiences a force. This is the principle behind electric motors.
where B is the external magnetic field, I is the current, L is the length of wire in the field, and θ is the angle between the current direction and the magnetic field direction.
The direction of the force is given by Fleming’s Left-Hand Rule:
- When θ = 90° (wire perpendicular to B): F = BIL (maximum force)
- When θ = 0° (wire parallel to B): F = 0 (no force)
- When θ = 0° or 180°: F = 0
Worked Example 5
A wire of length 0.3 m carrying a current of 4 A is placed in a magnetic field of 0.5 T. If the wire makes an angle of 30° with the field, find the force on the wire.
$$F = BIL\sin\theta = 0.5 \times 4 \times 0.3 \times \sin 30°$$ $$F = 0.6 \times 0.5 = 0.3 \text{ N}$$
Worked Example 6
A straight wire carrying 10 A is placed perpendicular to a uniform magnetic field of 0.2 T. The force on the wire is 0.6 N. What is the length of the wire in the field?
θ = 90°, so sin 90° = 1
$$F = BIL$$ $$0.6 = 0.2 \times 10 \times L$$ $$L = \frac{0.6}{2} = 0.3 \text{ m}$$
Force on a Moving Charge in a Magnetic Field
A charged particle moving in a magnetic field also experiences a force:
where q is the charge, v is the velocity of the particle, B is the magnetic field, and θ is the angle between v and B.
Worked Example 7
An electron (q = 1.6 × 10⁻¹⁹ C) moves at 2 × 10⁶ m/s perpendicular to a magnetic field of 0.4 T. Find the force on the electron.
θ = 90°, sin 90° = 1
$$F = qvB = 1.6 \times 10^{-19} \times 2 \times 10^6 \times 0.4$$ $$F = 1.28 \times 10^{-13} \text{ N}$$
- Oersted discovered that current produces a magnetic field.
- Right-hand grip rule: thumb = current direction, fingers = B direction.
- B for straight wire: B = μ₀I/(2πr)
- B at center of circular loop: B = μ₀I/(2R)
- B inside solenoid: B = μ₀nI (n = turns per unit length)
- μ₀ = 4π × 10⁻⁷ T·m/A
- Force on wire: F = BIL sin θ (max when θ = 90°, zero when θ = 0°)
- Force on charge: F = qvB sin θ
- Fleming’s Left-Hand Rule: thumb = F, index = B, middle = I (for motors)
- Magnetic force does no work (always perpendicular to velocity).
Practice Question 5: A wire carrying 8 A is bent into a circular loop of radius 0.05 m. Find the magnetic field at the center of the loop.
Practice Question 6: A 20 cm wire carrying 3 A is placed in a 0.6 T magnetic field. Calculate the maximum and minimum force on the wire.
F_max = BIL = 0.6 × 3 × 0.2 = 0.36 N
Minimum force (when θ = 0° or 180°):
F_min = BIL sin 0° = 0 N
Practice Question 7: A solenoid has 2000 turns per meter and carries a current of 5 A. What is the magnetic field inside it? If a soft iron core with relative permeability 200 is inserted, what is the new field?
B = μ₀nI = 4π × 10⁻⁷ × 2000 × 5 = 4π × 10⁻³ ≈ 0.01257 T = 12.57 mT
With iron core:
B = μ₀μᵣnI = 200 × 0.01257 = 2.514 T
The iron core increases the field by a factor of 200! This is why electromagnets use iron cores.
4.4 Electromagnetic Induction
Faraday’s Discovery
Dear student, after Oersted showed that electricity can produce magnetism, scientists wondered: can magnetism produce electricity? In 1831, Michael Faraday answered this question with a brilliant experiment. He showed that a changing magnetic field can produce an electric current. This process is called electromagnetic induction.
Faraday discovered that an electromotive force (EMF) is induced in a coil when:
- A magnet is moved toward or away from the coil
- The coil is moved toward or away from a magnet
- The current in a nearby coil is changed (switched on or off)
- A coil is rotated in a magnetic field
In all these cases, the key factor is that the magnetic flux through the coil changes.
Magnetic Flux
We already introduced magnetic flux. Let us recall:
The flux changes if B changes, A changes, or θ changes.
Faraday’s Law of Electromagnetic Induction
In words: the faster the flux changes, the greater the induced EMF. Also, more turns in the coil means a larger EMF.
Lenz’s Law
This is really a statement of conservation of energy. If the induced current did not oppose the change, it would amplify it, creating energy from nothing — which is impossible.
Worked Example 8
A coil of 200 turns is placed in a magnetic field that decreases from 0.5 T to 0.1 T in 0.02 seconds. The area of each turn is 0.01 m² and the field is perpendicular to the coil. Find the induced EMF.
ΔΦ = Φ₂ − Φ₁ = B₂A cos θ − B₁A cos θ
Since B is perpendicular to the coil: cos 0° = 1
ΔΦ = (0.1 × 0.01) − (0.5 × 0.01) = 0.001 − 0.005 = −0.004 Wb
$$|\varepsilon| = N\frac{|\Delta\Phi|}{\Delta t} = 200 \times \frac{0.004}{0.02} = 200 \times 0.2 = 40 \text{ V}$$
Worked Example 9
A coil of 50 turns and area 0.04 m² is placed in a magnetic field of 0.3 T. The coil is rotated from a position where the field is perpendicular to it to a position where the field is parallel to it in 0.1 s. Find the average induced EMF.
Initial flux: Φ₁ = BA cos 0° = 0.3 × 0.04 × 1 = 0.012 Wb
Final flux: Φ₂ = BA cos 90° = 0.3 × 0.04 × 0 = 0 Wb
ΔΦ = 0 − 0.012 = −0.012 Wb
$$|\varepsilon| = N\frac{|\Delta\Phi|}{\Delta t} = 50 \times \frac{0.012}{0.1} = 50 \times 0.12 = 6 \text{ V}$$
Worked Example 10
A rectangular coil of 100 turns, 10 cm wide and 20 cm long, rotates at 50 revolutions per second in a magnetic field of 0.5 T. The axis of rotation is perpendicular to the field. Find the maximum induced EMF.
A = 0.1 × 0.2 = 0.02 m²
Angular frequency: ω = 2πf = 2π × 50 = 100π rad/s
The EMF in a rotating coil: ε = NBAω sin ωt
Maximum EMF: ε_max = NBAω
$$\varepsilon_{\text{max}} = 100 \times 0.5 \times 0.02 \times 100\pi$$ $$\varepsilon_{\text{max}} = 100\pi = 314.16 \text{ V}$$
- EMF is induced ONLY when magnetic flux CHANGES (not when it is constant).
- Faraday’s law: ε = −NΔΦ/Δt (magnitude: |ε| = N|ΔΦ|/Δt)
- Lenz’s law: induced current opposes the change that caused it.
- More turns N → larger EMF.
- Faster change (smaller Δt) → larger EMF.
- Flux can change by: changing B, changing A, changing θ, or moving the coil.
- For a rotating coil: ε_max = NBAω (very common exam formula!)
- Lenz’s law is really conservation of energy in disguise.
Practice Question 8: A coil of 500 turns encloses an area of 0.05 m². A magnetic field of 0.4 T passes through it perpendicular to the plane of the coil. If the field is reduced to zero in 0.05 s, find the induced EMF.
$$|\varepsilon| = N\frac{|\Delta\Phi|}{\Delta t} = 500 \times \frac{0.02}{0.05} = 500 \times 0.4 = 200 \text{ V}$$
Practice Question 9: When a magnet is pushed into a coil, the induced current in the coil creates a magnetic field that repels the magnet. Which law explains this?
Practice Question 10: A coil of area 0.02 m² with 100 turns is in a 0.3 T field. It is removed from the field in 0.04 s. Find the average EMF induced.
$$|\varepsilon| = 100 \times \frac{0.006}{0.04} = 100 \times 0.15 = 15 \text{ V}$$
4.5 Faraday’s Law of Electromagnetic Induction (Detailed)
Factors Affecting the Induced EMF
Dear student, let us now look more carefully at what affects the size of the induced EMF. From Faraday’s law:
The induced EMF depends on:
- Number of turns (N): Doubling the turns doubles the EMF.
- Rate of change of flux (ΔΦ/Δt): Making the change faster increases the EMF.
- Area of the coil (A): Larger area captures more flux, so larger EMF for the same B change.
- Strength of the magnetic field (B): Stronger field means more flux, so larger EMF when it changes.
EMF Induced by a Moving Conductor
When a straight conductor of length L moves with velocity v perpendicular to a uniform magnetic field B, an EMF is induced across its ends:
This is sometimes called the motional EMF. The direction of the induced current is found by Fleming’s Right-Hand Rule:
• Left Hand = Motor (force on current in field) → F = BIL
• Right Hand = Generator (current from motion in field) → ε = BLv
Worked Example 11
An airplane with a wingspan of 30 m flies horizontally at 200 m/s through a region where the vertical component of Earth’s magnetic field is 5 × 10⁻⁵ T. Find the EMF induced between the wing tips.
L = 30 m, v = 200 m/s, B = 5 × 10⁻⁵ T (vertical), v is horizontal (perpendicular to B)
$$\varepsilon = BLv = 5 \times 10^{-5} \times 30 \times 200 = 5 \times 10^{-5} \times 6000 = 0.3 \text{ V}$$
Worked Example 12
A conductor of length 0.5 m moves at 4 m/s at an angle of 60° to a magnetic field of 0.8 T. Find the induced EMF.
The component of velocity perpendicular to B: v_perp = v sin 60°
$$\varepsilon = BLv\sin\theta = 0.8 \times 0.5 \times 4 \times \sin 60°$$ $$\varepsilon = 1.6 \times 0.866 = 1.386 \text{ V}$$
AC Generator
An AC generator converts mechanical energy into electrical energy using electromagnetic induction. A coil is rotated in a magnetic field, producing an alternating EMF:
where ω = 2πf is the angular velocity of rotation.
- Maximum EMF: ε₀ = NBAω
- The EMF varies sinusoidally between +ε₀ and −ε₀.
- Frequency f = ω/(2π) — the number of complete cycles per second.
- Armature: The rotating coil.
- Field magnets: Provide the magnetic field.
- Slip rings: Allow the coil to rotate while maintaining electrical connection.
- Brushes: Carbon contacts that press against the slip rings to collect the current.
DC Generator (Dynamo)
A DC generator is similar to an AC generator but uses a split ring commutator instead of slip rings. The commutator reverses the connections every half turn, so the output current always flows in the same direction (though it pulsates).
- Motional EMF: ε = BLv (conductor moving perpendicular to B).
- ε = BLv sin θ when motion is at angle θ to B.
- Fleming’s Right-Hand Rule: for generators (thumb = v, index = B, middle = I).
- AC generator EMF: ε = NBAω sin(ωt); maximum: ε₀ = NBAω.
- AC generator uses slip rings; DC generator uses split-ring commutator.
- The faster the change, the larger the EMF — this is the core of Faraday’s law.
Practice Question 11: A rod of length 0.4 m moves perpendicularly to a 0.5 T magnetic field at 10 m/s. What EMF is induced across its ends?
Practice Question 12: An AC generator has 200 turns, each of area 0.03 m², and rotates at 60 Hz in a 0.4 T field. Find the maximum EMF.
$$\varepsilon_{\text{max}} = NBA\omega = 200 \times 0.4 \times 0.03 \times 120\pi$$ $$= 200 \times 0.012 \times 120\pi = 2.4 \times 120\pi = 288\pi \approx 904.8 \text{ V}$$
Practice Question 13: What is the difference between a slip ring and a split-ring commutator?
Split-ring commutator: A single ring split into two halves, with each half connected to one end of the coil. Every half rotation, the brushes switch from one half to the other, reversing the external connections at the same time the coil current reverses. This produces unidirectional (DC) output — used in DC generators.
4.6 Transformers
What is a Transformer?
Dear student, have you noticed that the electricity coming to your house from the power line is at a very high voltage (like 220V in Ethiopia), but your phone charger needs only 5V? How does this voltage get reduced? The answer is a transformer.
A transformer is a device that changes (transforms) AC voltage from one level to another — either stepping it up (increasing) or stepping it down (decreasing) — using the principle of electromagnetic induction.
Construction
A transformer consists of:
- A primary coil (input side) with Nₚ turns
- A secondary coil (output side) with Nₛ turns
- A soft iron core that links the two coils magnetically
How It Works
When an alternating current flows through the primary coil, it creates a changing magnetic field in the iron core. This changing field passes through the secondary coil, inducing an EMF (by Faraday’s law). Since the current is alternating, the flux is always changing — so the EMF is continuously induced.
Transformer Equation
where Vₚ = primary voltage, Vₛ = secondary voltage, Nₚ = primary turns, Nₛ = secondary turns.
- Step-up transformer: Nₛ > Nₚ → Vₛ > Vₚ (increases voltage)
- Step-down transformer: Nₛ < Nₚ → Vₛ < Vₚ (decreases voltage)
Ideal Transformer (No Energy Loss)
In an ideal transformer (100% efficient), the power in equals the power out:
This means: when voltage is stepped UP, current is stepped DOWN (and vice versa). You do not get something for nothing!
Worked Example 13
A transformer has 500 turns in the primary and 50 turns in the secondary. If the primary voltage is 2200 V, find the secondary voltage. Is this a step-up or step-down transformer?
$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$ $$V_s = V_p \times \frac{N_s}{N_p} = 2200 \times \frac{50}{500} = 2200 \times 0.1 = 220 \text{ V}$$
This is a step-down transformer (voltage decreases from 2200 V to 220 V).
Worked Example 14
A step-up transformer converts 200 V to 2000 V. If the primary has 100 turns, how many turns does the secondary have? If the primary current is 10 A, what is the secondary current (assuming ideal transformer)?
$$N_s = N_p \times \frac{V_s}{V_p} = 100 \times \frac{2000}{200} = 100 \times 10 = 1000 \text{ turns}$$
For ideal transformer: VₚIₚ = VₛIₛ
$$I_s = \frac{V_p I_p}{V_s} = \frac{200 \times 10}{2000} = 1 \text{ A}$$
Voltage increased 10 times, current decreased 10 times.
Worked Example 15
A transformer is 90% efficient. The primary voltage is 220 V, secondary voltage is 1100 V, and primary current is 5 A. Find: (a) the power input, (b) the power output, (c) the secondary current.
(a) Power input:
$$P_{in} = V_p I_p = 220 \times 5 = 1100 \text{ W}$$
(b) Power output:
$$\eta = \frac{P_{out}}{P_{in}} \times 100\%$$ $$P_{out} = 0.90 \times 1100 = 990 \text{ W}$$
(c) Secondary current:
$$I_s = \frac{P_{out}}{V_s} = \frac{990}{1100} = 0.9 \text{ A}$$
(If ideal: Iₛ = 220 × 5/1100 = 1.0 A. The actual current is less due to losses.)
Energy Losses in Transformers
In real transformers, some energy is lost:
- Copper loss (I²R loss): Heat produced in the windings due to their resistance. Reduced by using thick copper wire.
- Eddy current loss: Circulating currents in the iron core cause heating. Reduced by using a laminated core (thin insulated sheets instead of a solid block).
- Hysteresis loss: The iron core is magnetized and demagnetized repeatedly, losing energy each cycle. Reduced by using soft iron (which has a narrow hysteresis loop).
- Flux leakage: Not all magnetic flux from the primary links the secondary. Reduced by winding the coils close together on the same core.
- Transformer equation: Vₛ/Vₚ = Nₛ/Nₚ
- Ideal: VₚIₚ = VₛIₛ → Iₛ/Iₚ = Nₚ/Nₛ
- Step-up: Nₛ > Nₚ (V increases, I decreases)
- Step-down: Nₛ < Nₚ (V decreases, I increases)
- Transformers work ONLY with AC, not DC.
- Efficiency η = P_out/P_in × 100%
- Losses: copper loss, eddy current loss, hysteresis loss, flux leakage.
- Eddy currents reduced by laminated core.
- Hysteresis loss reduced by soft iron core.
- Power transmission uses step-up at the station, step-down near consumers.
Practice Question 14: A transformer has 2000 turns on the primary and 200 turns on the secondary. If the input voltage is 2400 V, what is the output voltage? What type of transformer is this?
Practice Question 15: Why can a transformer not work with DC?
Practice Question 16: A power station generates 10,000 V and sends power through transmission lines. A step-up transformer raises it to 200,000 V for transmission. If the transmission current at the high voltage is 50 A, what would the current have been without the transformer?
P = 200,000 × 50 = 10,000,000 W
Without transformer: I = P/V = 10,000,000/10,000 = 1000 A
The transformer reduces the transmission current from 1000 A to 50 A — a 20 times reduction! This dramatically reduces power loss in the transmission lines (P_loss = I²R).
4.7 Application and Safety
Applications of Electromagnetism
1. Electric Motor
An electric motor converts electrical energy into mechanical energy. It works on the principle that a current-carrying conductor in a magnetic field experiences a force (F = BIL sin θ).
- A rectangular coil is placed between the poles of a magnet.
- When current flows through the coil, opposite sides experience forces in opposite directions, creating a torque that rotates the coil.
- A split-ring commutator reverses the current direction every half turn, so the coil keeps rotating in the same direction.
2. Electric Generator (Dynamo)
A generator converts mechanical energy into electrical energy. It works on the principle of electromagnetic induction (Faraday’s law).
- A coil is rotated in a magnetic field.
- The changing flux through the coil induces an EMF.
- With slip rings → AC output; with split-ring commutator → DC output.
• Motor: current in → force → rotation (electrical to mechanical)
• Generator: rotation → changing flux → current out (mechanical to electrical)
• They are essentially the same device used in opposite directions!
3. Electromagnets
Temporary magnets made by passing current through a coil wrapped around a soft iron core. Used in:
- Electric bells and buzzers
- Relays (electrically controlled switches)
- Cranes for lifting scrap metal
- Magnetic locks and doors
- Loudspeakers
4. Transformers in Power Transmission
Electrical power is transmitted over long distances at very high voltage to reduce energy loss in the lines:
By stepping up the voltage (and thus reducing the current), the power loss I²R is greatly reduced. This is why power stations use step-up transformers and cities use step-down transformers.
5. Induction Cooktops
An induction cooktop uses a rapidly changing magnetic field to induce eddy currents in the bottom of a metal pot. These eddy currents heat the pot directly — the cooktop itself does not get hot (only the pot does). This is very efficient and safe.
6. Magnetic Levitation (Maglev Trains)
Maglev trains use powerful electromagnets to levitate above the track, eliminating friction. This allows them to travel at very high speeds (over 500 km/h). Electromagnetic induction is also used for propulsion and braking.
Safety Considerations
- High voltage danger: Transformers and transmission lines carry very high voltages. Never touch power lines or open transformers.
- Short circuits: Can cause fires and explosions. Use proper fuses and circuit breakers.
- Electromagnetic radiation: High-current devices produce electromagnetic fields. Maintain safe distances from high-voltage equipment.
- Eddy current heating: In transformers and motors, eddy currents can cause overheating. Laminated cores reduce this.
- Proper insulation: All electrical connections must be properly insulated to prevent electric shock.
- Grounding (earthing): Electrical appliances must be grounded to prevent shock in case of insulation failure.
- Motor: electrical → mechanical (F = BIL on coil in B field).
- Generator: mechanical → electrical (Faraday’s law, rotating coil).
- Motor and generator are the same device used differently.
- Electromagnets: temporary, strength controlled by current and number of turns.
- Power transmission: step-up at station, step-down at destination, to reduce I²R losses.
- Induction heating: eddy currents in conductors from changing B fields.
- Safety: insulation, grounding, fuses, circuit breakers, safe distances.
Practice Question 17: Explain why power is transmitted at high voltage over long distances.
Practice Question 18: What is the role of the split-ring commutator in a DC motor?
Unit 4 Summary
In this unit, you learned seven major topics:
- Magnets and Magnetic Field: Properties of magnets, poles, magnetic field as a vector quantity, B measured in Tesla.
- Magnetic Field Lines: Properties (closed loops, never cross, N→S outside), magnetic flux Φ = BA cos θ.
- Current and Magnetism: Oersted’s discovery, B for straight wire (μ₀I/2πr), circular loop (μ₀I/2R), solenoid (μ₀nI), force on wire (F = BIL sin θ), force on charge (F = qvB sin θ), Fleming’s Left-Hand Rule.
- Electromagnetic Induction: Faraday’s law (ε = −NΔΦ/Δt), Lenz’s law, factors affecting EMF.
- Faraday’s Law (Detailed): Motional EMF (ε = BLv), Fleming’s Right-Hand Rule, AC generator (ε = NBAω sin ωt), DC generator.
- Transformers: Vₛ/Vₚ = Nₛ/Nₚ, ideal: VₚIₚ = VₛIₛ, losses, efficiency, only works with AC.
- Applications and Safety: Motors, generators, electromagnets, power transmission, safety measures.
Quick Revision Notes — Unit 4: Electromagnetism
All Key Formulas
| Topic | Formula | Notes |
|---|---|---|
| Magnetic flux | Φ = BA cos θ | Unit: Weber (Wb) |
| B of straight wire | B = μ₀I/(2πr) | μ₀ = 4π × 10⁻⁷ T·m/A |
| B at center of loop | B = μ₀I/(2R) | R = radius of loop |
| B inside solenoid | B = μ₀nI | n = N/L (turns per meter) |
| Force on wire in B | F = BIL sin θ | Max when θ = 90° |
| Force on charge in B | F = qvB sin θ | Always ⊥ to v and B |
| Faraday’s law | |ε| = N|ΔΦ|/Δt | N = number of turns |
| Motional EMF | ε = BLv | v ⊥ to B |
| Motional EMF (angle) | ε = BLv sin θ | θ between v and B |
| AC generator EMF | ε = NBAω sin(ωt) | ω = 2πf |
| Max EMF (generator) | ε₀ = NBAω | Very common exam formula |
| Transformer voltage | Vₛ/Vₚ = Nₛ/Nₚ | Turns ratio |
| Transformer power (ideal) | VₚIₚ = VₛIₛ | Conservation of energy |
| Transformer current | Iₛ/Iₚ = Nₚ/Nₛ | Inverse of voltage ratio |
| Efficiency | η = P_out/P_in × 100% | Always less than 100% |
| Power loss in lines | P_loss = I²R | Reason for high-voltage transmission |
Hand Rules Summary
| Rule | Used For | Thumb | Index | Middle |
|---|---|---|---|---|
| Right-hand grip | B direction around wire | Current (I) | — | Fingers curl in B direction |
| Fleming’s Left Hand | Motor (force on wire) | Force (F) | Field (B) | Current (I) |
| Fleming’s Right Hand | Generator (induced current) | Velocity (v) | Field (B) | Induced current (I) |
Important Definitions
- Magnet: An object that attracts ferromagnetic materials and has north and south poles.
- Magnetic field (B): The region around a magnet where magnetic force is experienced; measured in Tesla.
- Magnetic flux (Φ): The number of field lines passing through an area; Φ = BA cos θ; unit: Weber.
- Magnetic field lines: Imaginary lines showing the direction and relative strength of the magnetic field.
- Solenoid: A long coil of wire; produces a uniform magnetic field inside when current flows.
- Electromagnet: A temporary magnet made by passing current through a coil with an iron core.
- Electromagnetic induction: The production of EMF by changing magnetic flux through a circuit.
- Faraday’s law: The induced EMF equals the negative rate of change of magnetic flux.
- Lenz’s law: The induced current opposes the change in flux that caused it.
- Motional EMF: EMF induced in a conductor moving through a magnetic field.
- Transformer: A device that changes AC voltage using electromagnetic induction.
- Step-up transformer: Nₛ > Nₚ, increases voltage, decreases current.
- Step-down transformer: Nₛ < Nₚ, decreases voltage, increases current.
- Eddy currents: Circular currents induced in conducting materials by changing magnetic fields.
Common Mistakes to Avoid
- Mixing up Left and Right Hand rules: Left Hand = Motor (force ON wire), Right Hand = Generator (current FROM motion). Memorize: “Left = Laa (force), Right = Right (induced current).”
- Forgetting the negative sign in Faraday’s law: The negative sign represents Lenz’s law. For magnitude calculations, use |ε| = N|ΔΦ|/Δt.
- Using wrong angle: In F = BIL sin θ, θ is between I and B. In Φ = BA cos θ, θ is between B and the NORMAL to the surface (NOT between B and the surface itself).
- Confusing Φ = BA cos θ angle: If B is parallel to the surface, θ = 90°, cos 90° = 0, so Φ = 0. If B is perpendicular to the surface, θ = 0°, cos 0° = 1, so Φ = BA (maximum).
- Forgetting that transformers need AC: Transformers CANNOT work with DC. DC gives constant flux → no induced EMF.
- Wrong current ratio in transformers: Iₛ/Iₚ = Nₚ/Nₛ (inverted, not same as voltage ratio!).
- Not converting units: Convert cm to m, cm² to m² before calculating.
- Confusing motor and generator: Motor = electrical in, mechanical out. Generator = mechanical in, electrical out.
- Forgetting N in Faraday’s law: Always multiply by the number of turns!
- Sign of ΔΦ: ΔΦ = Φ_final − Φ_initial. Be careful with the order!
Important Constants
- μ₀ (permeability of free space) = 4π × 10⁻⁷ T·m/A
- Electron charge: e = 1.6 × 10⁻¹⁹ C
- 1 Tesla = 10⁴ Gauss
- 1 Weber = 1 Tesla·m²
- Ethiopian mains voltage: 220 V, 50 Hz
Challenge Exam Questions — Unit 4
Section A: Multiple Choice Questions
Question 1: The magnetic field at the center of a circular loop of radius R carrying current I is B. If the radius is doubled and the current is halved, the new magnetic field is:
B = μ₀I/(2R). New values: I’ = I/2, R’ = 2R
B’ = μ₀(I/2)/(2 × 2R) = μ₀I/(8R) = (1/4) × μ₀I/(2R) = B/4
Question 2: A wire of length L carrying current I is placed in a magnetic field B. The maximum force on the wire occurs when the angle between the wire and the field is:
F = BIL sin θ. sin θ is maximum (= 1) when θ = 90°. At 0° and 180°, sin θ = 0, so F = 0.
Question 3: A transformer has 100 turns in the primary and 500 turns in the secondary. If the primary current is 10 A, the secondary current (ideal) is:
Iₛ/Iₚ = Nₚ/Nₛ = 100/500 = 1/5
Iₛ = 10/5 = 2 A
(Note: this is a step-up transformer for voltage, so current steps down.)
Question 4: Which of the following is NOT a property of magnetic field lines?
Magnetic field lines do NOT start or end on charges. They form continuous closed loops from N to S outside the magnet and S to N inside. Electric field lines start on positive charges and end on negative charges — this is a key difference.
Question 5: Lenz’s law is a consequence of the conservation of:
Lenz’s law ensures that the induced current opposes the change causing it. If it did not oppose the change (i.e., if it aided it), energy would be created from nothing. By opposing the change, the induced current ensures energy is conserved.
Question 6: The magnetic flux through a coil of 50 turns changes from 0.01 Wb to 0.04 Wb in 0.1 s. The induced EMF is:
|ε| = N|ΔΦ|/Δt = 50 × |0.04 − 0.01|/0.1 = 50 × 0.03/0.1 = 50 × 0.3 = 15 V
Section B: Fill in the Blanks
Question 7: The SI unit of magnetic flux density is ________.
Question 8: In a solenoid, the magnetic field inside is given by B = ________.
Question 9: Fleming’s ________ hand rule is used to find the direction of induced current in a generator.
Right Hand Rule for generators: thumb = velocity, index = field, middle = induced current.
Question 10: A transformer works only with ________ current because it requires a ________ magnetic flux.
DC produces a steady flux (no change), so no EMF is induced in the secondary.
Question 11: In a step-down transformer, the number of turns in the secondary is ________ than in the primary.
Nₛ < Nₚ for a step-down transformer, so Vₛ < Vₚ.
Question 12: The phenomenon of ________ currents is used in induction cooktops for heating.
Eddy currents are circular currents induced in conductors by changing magnetic fields. They cause heating, which is used in induction cooking.
Section C: Short Answer Questions
Question 13: State the difference between a bar magnet and an electromagnet. Give two advantages of an electromagnet over a permanent magnet.
Advantages of electromagnet:
(1) The strength can be controlled by changing the current — you can make it as strong or as weak as needed.
(2) It can be turned on and off by switching the current on and off.
(3) The polarity (which end is N and which is S) can be reversed by reversing the current direction.
Question 14: Explain why an AC generator produces a sinusoidal EMF.
ε = −dΦ/dt = −NBA d[cos(ωt)]/dt = NBAω sin(ωt)
Since sin(ωt) is a sinusoidal function, the induced EMF varies sinusoidally between +NBAω and −NBAω.
Question 15: Explain how eddy current losses are minimized in transformer cores.
The iron core is made of laminations — thin, flat sheets of iron that are insulated from each other by a coating of varnish or oxide. Each lamination has a small cross-sectional area, which increases the resistance to eddy current flow. Since the eddy currents are proportional to the area, using thin laminations instead of a solid block greatly reduces the eddy current magnitude and thus the energy loss.
Question 16: Distinguish between an electric motor and an electric generator based on: (a) energy conversion, (b) governing principle, (c) hand rule used.
• Motor: Electrical energy → Mechanical energy
• Generator: Mechanical energy → Electrical energy
(b) Governing principle:
• Motor: Force on a current-carrying conductor in a magnetic field (F = BIL sin θ)
• Generator: Electromagnetic induction (Faraday’s law: ε = −NΔΦ/Δt)
(c) Hand rule:
• Motor: Fleming’s Left-Hand Rule (thumb = F, index = B, middle = I)
• Generator: Fleming’s Right-Hand Rule (thumb = v, index = B, middle = I)
Section D: Step-by-Step Calculation Questions
Question 17: Two parallel wires, each 2 m long, are 10 cm apart. Wire A carries 8 A upward and wire B carries 5 A downward. Find: (a) the magnetic field at wire B due to wire A, (b) the force on wire B due to wire A, (c) whether the wires attract or repel. (μ₀ = 4π × 10⁻⁷ T·m/A)
(a) Magnetic field at wire B due to wire A:
$$B = \frac{\mu_0 I_A}{2\pi r} = \frac{4\pi \times 10^{-7} \times 8}{2\pi \times 0.1} = \frac{32\pi \times 10^{-7}}{0.2\pi} = 1.6 \times 10^{-5} \text{ T}$$
Direction: By right-hand grip rule (current up in wire A), B at wire B points INTO the page.
(b) Force on wire B:
Wire B carries current downward. B field at B is into the page.
By Fleming’s Left-Hand Rule: F is to the RIGHT (toward wire A).
$$F = B I_B L = 1.6 \times 10^{-5} \times 5 \times 2 = 1.6 \times 10^{-4} \text{ N}$$
(c) Attraction or repulsion?
The currents are in OPPOSITE directions (one up, one down). Parallel wires with opposite currents repel each other. The force on wire B is directed AWAY from wire A. (Note: I need to correct the direction — with opposite currents, the force is repulsive, so wire B is pushed away from wire A, meaning the force is to the LEFT if wire A is to the left of wire B.)
Question 18: A rectangular coil of 200 turns, 10 cm × 5 cm, is placed with its plane perpendicular to a uniform magnetic field of 0.4 T. The coil is rotated 90° about an axis perpendicular to the field in 0.05 s. Find the average EMF induced.
Initial: Φ₁ = BA cos 0° = 0.4 × 0.005 × 1 = 0.002 Wb (perpendicular to field → maximum flux)
After 90° rotation: Φ₂ = BA cos 90° = 0.4 × 0.005 × 0 = 0 Wb
ΔΦ = 0 − 0.002 = −0.002 Wb
$$|\varepsilon| = N\frac{|\Delta\Phi|}{\Delta t} = 200 \times \frac{0.002}{0.05} = 200 \times 0.04 = 8 \text{ V}$$
Question 19: A power station generates power at 5000 V. A step-up transformer with 100 primary turns and 10,000 secondary turns is used. The power is transmitted through lines of total resistance 20 Ω to a town 50 km away. A step-down transformer at the town supplies 220 V to consumers. If the town draws 100 A at 220 V: (a) What is the transmission voltage? (b) What is the transmission current? (c) What is the power loss in the lines? (d) What is the efficiency of the transmission system?
$$V_s = V_p \times \frac{N_s}{N_p} = 5000 \times \frac{10000}{100} = 5000 \times 100 = 500000 \text{ V} = 500 \text{ kV}$$
(b) Transmission current:
Power at generation = Power in transmission (ideal step-up)
P = VₚIₚ = VₛIₛ
We need to find P first:
Power to consumers = V_town × I_town = 220 × 100 = 22000 W
Assuming ideal step-down: P to town = P from transmission = 22000 W
$$I_{trans} = \frac{P}{V_{trans}} = \frac{22000}{500000} = 0.044 \text{ A}$$
(c) Power loss in lines:
$$P_{loss} = I_{trans}^2 \times R = (0.044)^2 \times 20 = 0.001936 \times 20 = 0.0387 \text{ W}$$
(d) Efficiency:
$$\eta = \frac{P_{out}}{P_{in}} \times 100 = \frac{22000}{22000 + 0.0387} \times 100 \approx 99.9998\%$$
This demonstrates the incredible advantage of high-voltage transmission — the power loss is negligible!
Question 20: A coil of 100 turns and radius 5 cm is placed in a magnetic field that increases uniformly from 0.2 T to 0.8 T in 0.04 s. The coil is positioned with its plane perpendicular to the field. Find: (a) the change in flux through each turn, (b) the induced EMF, (c) the induced current if the coil has a total resistance of 10 Ω.
(a) Change in flux per turn:
ΔΦ = B₂A − B₁A = (0.8 − 0.2) × 0.00785 = 0.6 × 0.00785 = 0.00471 Wb
(b) Induced EMF:
$$|\varepsilon| = N\frac{|\Delta\Phi|}{\Delta t} = 100 \times \frac{0.00471}{0.04} = 100 \times 0.11775 = 11.775 \text{ V}$$
(c) Induced current:
$$I = \frac{\varepsilon}{R} = \frac{11.775}{10} = 1.178 \text{ A}$$
By Lenz’s law, since B is increasing (flux increasing), the induced current opposes the increase, so it creates a field opposing the external field. Using the right-hand grip rule, the current flows clockwise when viewed from the direction of B.
Question 21: A proton (q = 1.6 × 10⁻¹⁹ C, m = 1.67 × 10⁻²⁷ kg) enters a uniform magnetic field of 0.5 T with a velocity of 4 × 10⁶ m/s perpendicular to the field. Find: (a) the force on the proton, (b) the radius of its circular path, (c) the period of revolution.
F = qvB = 1.6 × 10⁻¹⁹ × 4 × 10⁶ × 0.5 = 3.2 × 10⁻¹³ N
(b) Radius of circular path:
The magnetic force provides centripetal force: qvB = mv²/r
$$r = \frac{mv}{qB} = \frac{1.67 \times 10^{-27} \times 4 \times 10^6}{1.6 \times 10^{-19} \times 0.5}$$ $$r = \frac{6.68 \times 10^{-21}}{8 \times 10^{-20}} = 0.0835 \text{ m} = 8.35 \text{ cm}$$
(c) Period of revolution:
$$T = \frac{2\pi r}{v} = \frac{2\pi \times 0.0835}{4 \times 10^6} = \frac{0.5247}{4 \times 10^6} = 1.312 \times 10^{-7} \text{ s}$$
Or using: T = 2πm/(qB) = 2π × 1.67 × 10⁻²⁷/(1.6 × 10⁻¹⁹ × 0.5) = 1.312 × 10⁻⁷ s ✓
Note: The period is independent of velocity!
Question 22: A transformer is 85% efficient. The primary has 800 turns and the secondary has 40 turns. If the primary is connected to a 2400 V AC supply and the secondary delivers 5 A to a load, find: (a) the secondary voltage, (b) the power output, (c) the primary current.
$$V_s = V_p \times \frac{N_s}{N_p} = 2400 \times \frac{40}{800} = 2400 \times 0.05 = 120 \text{ V}$$
(b) Power output:
$$P_{out} = V_s \times I_s = 120 \times 5 = 600 \text{ W}$$
(c) Primary current:
$$\eta = \frac{P_{out}}{P_{in}} = 0.85$$ $$P_{in} = \frac{P_{out}}{0.85} = \frac{600}{0.85} = 705.88 \text{ W}$$ $$I_p = \frac{P_{in}}{V_p} = \frac{705.88}{2400} = 0.294 \text{ A}$$
(If ideal: Iₚ = VₛIₛ/Vₚ = 600/2400 = 0.25 A. The actual primary current is higher due to losses.)
Question 23: A metal rod of length 0.6 m is pivoted at one end and rotates at 300 rpm in a plane perpendicular to a magnetic field of 0.4 T. Find the EMF induced between the center of the rod and its end.
For a rod rotating about one end, the average velocity of each point between center and end:
v_avg = ω × r_avg, where r_avg = (0 + 0.3)/2 = 0.15 m (from pivot)
But more precisely, for rotation about the pivot, the average velocity of the outer half:
v at distance r from pivot = ωr
Average v for the section from 0.3 m to 0.6 m: v_avg = ω × (0.3 + 0.6)/2 = ω × 0.45 m
ω = 2πf = 2π × 300/60 = 10π rad/s
v_avg = 10π × 0.45 = 4.5π m/s
$$\varepsilon = B \times L \times v_{avg} = 0.4 \times 0.3 \times 4.5\pi = 0.54\pi \approx 1.70 \text{ V}$$
Question 24: An AC generator produces an EMF given by ε = 170 sin(314t) V. Find: (a) the peak voltage, (b) the RMS voltage, (c) the frequency, (d) the time period.
(a) Peak voltage: ε₀ = 170 V
(b) RMS voltage:
$$V_{rms} = \frac{\varepsilon_0}{\sqrt{2}} = \frac{170}{1.414} = 120.2 \text{ V}$$
(c) Frequency:
ω = 314 rad/s = 2πf
$$f = \frac{314}{2\pi} = \frac{314}{6.283} = 49.97 \approx 50 \text{ Hz}$$ (This is the standard Ethiopian power frequency!)
(d) Time period:
$$T = \frac{1}{f} = \frac{1}{50} = 0.02 \text{ s} = 20 \text{ ms}$$
End of Challenge Questions
Dear student, you have completed all the challenge questions for Unit 4! Final tips for your exam:
- Always identify which hand rule to use: Left Hand for motors (force), Right Hand for generators (induced current).
- In Faraday’s law problems, always find ΔΦ first, then divide by Δt, then multiply by N.
- For transformer problems, always check whether it is step-up or step-down — this tells you whether V increases or decreases.
- Remember: F = BIL sin θ uses the angle between I and B; Φ = BA cos θ uses the angle between B and the normal.
- Practice writing explanations for Lenz’s law — it appears almost every year!
- Know the difference between peak voltage and RMS voltage: V_rms = V_peak/√2.