Electromagnetic Waves: Characteristics, transverse nature, and velocity

📖 Topic Discussion — Read & Understand

Electromagnetic waves are waves produced by continuously changing electric and magnetic fields. They are one of the most important concepts in modern physics because light, radio waves, microwaves, infrared radiation, ultraviolet radiation, X-rays and gamma rays are all electromagnetic waves.

Unlike mechanical waves such as sound waves, electromagnetic waves do not require a material medium for their propagation. They can travel through empty space or vacuum. This is why sunlight can travel from the Sun to the Earth through the vacuum of space.

An electromagnetic wave consists of an oscillating electric field and an oscillating magnetic field. These two fields are continuously changing and are linked with each other. A changing electric field produces a magnetic field, while a changing magnetic field produces an electric field.

The most important characteristic of electromagnetic waves is their transverse nature. The electric field and magnetic field oscillate perpendicular to each other and also perpendicular to the direction in which the wave travels.

E ⟂ B ⟂ Direction of Propagation

Suppose an electromagnetic wave is travelling along the positive X-axis. The electric field can oscillate along the Y-axis and the magnetic field can oscillate along the Z-axis. Therefore, the electric field, magnetic field and direction of propagation are mutually perpendicular.

Electromagnetic waves carry energy from one place to another. They also carry momentum and can exert radiation pressure. Their energy transport is described using the Poynting vector.

According to Maxwell's electromagnetic theory, the speed of an electromagnetic wave in vacuum is equal to the speed of light.

c = 3 × 10⁸ m/s

The velocity of electromagnetic waves in vacuum can also be written as:

c = 1 / √(μ₀ε₀)

where μ₀ is the permeability of free space and ε₀ is the permittivity of free space. Maxwell's theory therefore established a deep connection between electromagnetic waves and light.

💡 Core Concept:

A changing electric field and changing magnetic field continuously support each other. This combination travels through space as an electromagnetic wave.

🖼️ Animated Electromagnetic Wave

The diagram below shows the electric field, magnetic field and direction of propagation simultaneously.

🔴 Electric Field E

🔵 Magnetic Field B

↑ E

↕ B

Wave Direction ➜

Observe: The red electric field and blue magnetic field are perpendicular to each other, while the wave travels forward.

🔄 Transverse Nature

In a transverse wave, the disturbance is perpendicular to the direction of propagation. Electromagnetic waves have this property.

Electric Field
E ↑↓

Magnetic Field
B ↑↓

Propagation
➜

E ⟂ B

E ⟂ Direction of Propagation

B ⟂ Direction of Propagation

⭐ Characteristics of Electromagnetic Waves

1. No Medium Required

Electromagnetic waves can travel through vacuum.

2. Transverse

Electric and magnetic fields oscillate perpendicular to propagation.

3. Carry Energy

They transfer energy from one point to another.

4. Carry Momentum

Electromagnetic radiation possesses momentum.

5. Travel at Light Speed

In vacuum their speed is 3 × 10⁸ m/s.

6. Produced by Accelerating Charges

Accelerating charges can produce electromagnetic radiation.

🚀 Velocity of Electromagnetic Waves

In vacuum, electromagnetic waves travel with the maximum possible speed in nature.

c = 3 × 10⁸ m/s

From Maxwell's theory:

c = 1 / √(μ₀ε₀)

In a medium:

v = 1 / √(με)

If the refractive index of the medium is n:

v = c / n

Important: When electromagnetic radiation enters another medium, its frequency remains unchanged, but its velocity and wavelength change.

📐 Relation Between Speed, Frequency and Wavelength

c = fλ

where:

c
Speed of electromagnetic wave

f
Frequency

λ
Wavelength

⚡ Relationship Between Electric and Magnetic Fields

For an electromagnetic wave travelling in vacuum:

E / B = c

Therefore:

E = cB

The ratio of electric field amplitude to magnetic field amplitude is equal to the speed of light in vacuum.

🔋 Energy and Poynting Vector

Electromagnetic waves transport energy through space. The direction of energy flow is represented by the Poynting vector.

S = (1/μ₀)(E × B)

The direction of E × B gives the direction in which electromagnetic energy is transported.

🌈 Electromagnetic Spectrum

Electromagnetic waves are classified according to their frequency or wavelength.

Radio
Longest λ

Microwave

Infrared

Visible
Light

Ultraviolet

X-Rays

Gamma
Highest f

Frequency ↑ → Energy ↑ → Wavelength ↓

📊 Important Properties at a Glance

Property	Electromagnetic Wave
Nature	Transverse
Medium Required	Not required
Electric Field	Present
Magnetic Field	Present
Relationship	E ⟂ B
Speed in Vacuum	3 × 10⁸ m/s
Wave Equation	c = fλ
Energy	Transported by the wave

📐 Important Formulae

c = 1 / √(μ₀ε₀)

c = fλ

v = 1 / √(με)

v = c/n

E/B = c

S = (1/μ₀)(E × B)

🧮 Solved Example

An electromagnetic wave has a frequency of 6 × 10¹⁴ Hz. Find its wavelength in vacuum.

c = fλ

λ = c/f

λ = (3 × 10⁸) / (6 × 10¹⁴)

λ = 5 × 10⁻⁷ m

Answer: 5 × 10⁻⁷ m

🎯 Practice MCQs

Q1. Electromagnetic waves are:

A. Longitudinal waves B. Transverse waves C. Mechanical waves only D. Stationary waves

✅ Correct Answer: B. Transverse waves

Q2. Electromagnetic waves can travel through:

A. Vacuum B. Only solids C. Only liquids D. Only gases

✅ Correct Answer: A. Vacuum

Q3. In an electromagnetic wave, E and B are:

A. Parallel B. Perpendicular C. Identical D. Zero

✅ Correct Answer: B. Perpendicular

Q4. Speed of electromagnetic waves in vacuum is:

A. 3 × 10⁶ m/s B. 3 × 10⁷ m/s C. 3 × 10⁸ m/s D. 3 × 10⁹ m/s

✅ Correct Answer: C. 3 × 10⁸ m/s

Q5. The correct relation for an EM wave in vacuum is:

A. E/B = c B. E/B = 1/c C. E/B = c² D. E/B = 0

✅ Correct Answer: A. E/B = c

🏆 Quick Revision

🌊 Electromagnetic waves consist of electric and magnetic fields.

↔️ They are transverse waves.

⚡ E ⟂ B ⟂ Direction of propagation.

🚀 Speed in vacuum = 3 × 10⁸ m/s.

📐 c = fλ

⚡ E/B = c

🌌 Electromagnetic waves do not require a material medium.

🌊 Electromagnetic Waves
Characteristics • Transverse Nature • Velocity

Interactive Physics Learning

Displacement Current | Maxwell's Modification of Ampere's Law

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1. Introduction

Ampere's circuital law relates the magnetic field around a closed path to the electric current enclosed by that path.

For a steady current, Ampere's law is written as:

∮ B · dl = μ₀ I

However, this form creates a problem when the current is changing with time, especially during the charging of a capacitor.

Maxwell solved this problem by introducing the concept of displacement current.

2. The Problem with Ampere's Law

Consider a capacitor being charged by a battery. Current flows through the connecting wires, but there is no ordinary conduction current through the insulating gap between the capacitor plates.

🔋 Battery

→

⚡ Conduction Current

→

🔵 Capacitor

The electric field between the capacitor plates changes with time. This changing electric field produces a magnetic field.

3. 🔄 Charging Capacitor Animation

Observe the changing electric field between the capacitor plates. This changing field is associated with displacement current.

BATTERY

↑
⚡
↓

⬆

Observation: There is no conduction current across the capacitor gap, but the changing electric field behaves like a current in producing a magnetic field.

4. What is Displacement Current?

Displacement current is the current associated with a time-varying electric field.

It is not an actual flow of electric charges through the dielectric gap like conduction current.

I<sub>d</sub> = ε₀ dΦ<sub>E</sub>/dt

Where:

I<sub>d</sub>
Displacement current

ε₀
Permittivity of free space

Φ<sub>E</sub>
Electric flux

dΦ<sub>E</sub>/dt
Rate of change of electric flux

5. Maxwell's Modification of Ampere's Law

Maxwell added the displacement current term to Ampere's law.

∮ B · dl = μ₀(I + I<sub>d</sub>)

Therefore:

∮ B · dl = μ₀I + μ₀ε₀ dΦ<sub>E</sub>/dt

6. General Maxwell-Ampere Law

∇ × B = μ₀J + μ₀ε₀ ∂E/∂t

This equation shows that magnetic fields can be produced by:

🔌 Conduction Current

Produced by actual movement of electric charges.

⚡ Changing Electric Field

Produces displacement current and therefore contributes to the magnetic field.

7. Displacement Current in a Charging Capacitor

For a parallel plate capacitor:

E = Q / (ε₀A)

Electric flux is:

Φ<sub>E</sub> = EA

Therefore:

Φ<sub>E</sub> = Q / ε₀

Differentiating with respect to time:

dΦ<sub>E</sub>/dt = 1/ε₀ × dQ/dt

Since:

dQ/dt = I

We obtain:

I<sub>d</sub> = I

Important Result:

During charging of an ideal capacitor, the displacement current between the plates is equal in magnitude to the conduction current in the wires.

8. Conduction Current vs Displacement Current

Property	Conduction Current	Displacement Current
Origin	Movement of charges	Changing electric field
Symbol	I	Id
Expression	I = dQ/dt	Id = ε₀dΦE/dt
Capacitor gap	No conduction current	Displacement current exists
Produces magnetic field?	Yes	Yes

9. ⭐ Key Concept

Changing Electric Field

→

Displacement Current

→

Magnetic Field

The major contribution of Maxwell was recognizing that a time-varying electric field can produce a magnetic field.

10. 📐 Important Formulae

I<sub>d</sub> = ε₀ dΦ<sub>E</sub>/dt

∮ B · dl = μ₀(I + I<sub>d</sub>)

I<sub>d</sub> = I
For an ideal charging capacitor

∇ × B = μ₀J + μ₀ε₀ ∂E/∂t

11. 🧮 Worked Example

The electric flux through a surface changes at a rate of 2 × 10<sup>6</sup> N m²/C per second. Find the displacement current.

I<sub>d</sub> = ε₀ dΦ<sub>E</sub>/dt

ε₀ = 8.85 × 10<sup>−12</sup> C²/(N m²)

I<sub>d</sub> = (8.85 × 10<sup>−12</sup>) (2 × 10<sup>6</sup>)

I<sub>d</sub> = 1.77 × 10<sup>−5</sup> A

Answer: 1.77 × 10<sup>−5</sup> A

12. 🎯 Practice MCQs

Q1. Maxwell introduced displacement current to modify:

A. Faraday's law B. Ampere's law C. Ohm's law D. Coulomb's law

✅ Correct Answer: B. Ampere's law

Q2. Displacement current is associated with:

A. Constant electric field B. Time-varying electric field C. Constant magnetic field D. Static charge only

✅ Correct Answer: B. Time-varying electric field

Q3. The expression for displacement current is:

A. I = V/R B. I = dQ/dt C. I<sub>d</sub> = ε₀dΦ<sub>E</sub>/dt D. I = QR

✅ Correct Answer: C. I<sub>d</sub> = ε₀dΦ<sub>E</sub>/dt

Q4. During charging of an ideal capacitor:

A. I<sub>d</sub> = 0 B. I<sub>d</sub> = I C. I<sub>d</sub> = 2I D. I<sub>d</sub> = I/2

✅ Correct Answer: B. I<sub>d</sub> = I

Q5. A changing electric field produces:

A. Only electric charge B. Magnetic field C. Zero field D. Only heat

✅ Correct Answer: B. Magnetic field

13. 🏆 One-Minute Revision

Ampere's Law
Relates magnetic field to current.

Maxwell's Modification
Added displacement current.

Displacement Current
I<sub>d</sub> = ε₀dΦ<sub>E</sub>/dt

Charging Capacitor
I<sub>d</sub> = I

⭐ Remember

Changing Electric Field → Displacement Current → Magnetic Field

⚡ Displacement Current | Maxwell's Modification of Ampere's Law
Interactive Physics Learning • JEE / CBSE

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Displacement Current: Maxwell’s modification of Ampere's Law

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1. Introduction

Ampere's circuital law relates the magnetic field around a closed path to the electric current enclosed by that path.

For a steady current, Ampere's law is written as:

∮ B · dl = μ₀ I

However, this form creates a problem when the current is changing with time, especially during the charging of a capacitor.

Maxwell solved this problem by introducing the concept of displacement current.

2. The Problem with Ampere's Law

Consider a capacitor being charged by a battery. Current flows through the connecting wires, but there is no ordinary conduction current through the insulating gap between the capacitor plates.

🔋 Battery

→

⚡ Conduction Current

→

🔵 Capacitor

The electric field between the capacitor plates changes with time. This changing electric field produces a magnetic field.

3. 🔄 Charging Capacitor Animation

Observe the changing electric field between the capacitor plates. This changing field is associated with displacement current.

BATTERY

↑
⚡
↓

⬆

Observation: There is no conduction current across the capacitor gap, but the changing electric field behaves like a current in producing a magnetic field.

4. What is Displacement Current?

Displacement current is the current associated with a time-varying electric field.

It is not an actual flow of electric charges through the dielectric gap like conduction current.

I<sub>d</sub> = ε₀ dΦ<sub>E</sub>/dt

Where:

I<sub>d</sub>
Displacement current

ε₀
Permittivity of free space

Φ<sub>E</sub>
Electric flux

dΦ<sub>E</sub>/dt
Rate of change of electric flux

5. Maxwell's Modification of Ampere's Law

Maxwell added the displacement current term to Ampere's law.

∮ B · dl = μ₀(I + I<sub>d</sub>)

Therefore:

∮ B · dl = μ₀I + μ₀ε₀ dΦ<sub>E</sub>/dt

6. General Maxwell-Ampere Law

∇ × B = μ₀J + μ₀ε₀ ∂E/∂t

This equation shows that magnetic fields can be produced by:

🔌 Conduction Current

Produced by actual movement of electric charges.

⚡ Changing Electric Field

Produces displacement current and therefore contributes to the magnetic field.

7. Displacement Current in a Charging Capacitor

For a parallel plate capacitor:

E = Q / (ε₀A)

Electric flux is:

Φ<sub>E</sub> = EA

Therefore:

Φ<sub>E</sub> = Q / ε₀

Differentiating with respect to time:

dΦ<sub>E</sub>/dt = 1/ε₀ × dQ/dt

Since:

dQ/dt = I

We obtain:

I<sub>d</sub> = I

Important Result:

During charging of an ideal capacitor, the displacement current between the plates is equal in magnitude to the conduction current in the wires.

8. Conduction Current vs Displacement Current

Property	Conduction Current	Displacement Current
Origin	Movement of charges	Changing electric field
Symbol	I	Id
Expression	I = dQ/dt	Id = ε₀dΦE/dt
Capacitor gap	No conduction current	Displacement current exists
Produces magnetic field?	Yes	Yes

9. ⭐ Key Concept

Changing Electric Field

→

Displacement Current

→

Magnetic Field

The major contribution of Maxwell was recognizing that a time-varying electric field can produce a magnetic field.

10. 📐 Important Formulae

I<sub>d</sub> = ε₀ dΦ<sub>E</sub>/dt

∮ B · dl = μ₀(I + I<sub>d</sub>)

I<sub>d</sub> = I
For an ideal charging capacitor

∇ × B = μ₀J + μ₀ε₀ ∂E/∂t

11. 🧮 Worked Example

The electric flux through a surface changes at a rate of 2 × 10<sup>6</sup> N m²/C per second. Find the displacement current.

I<sub>d</sub> = ε₀ dΦ<sub>E</sub>/dt

ε₀ = 8.85 × 10<sup>−12</sup> C²/(N m²)

I<sub>d</sub> = (8.85 × 10<sup>−12</sup>) (2 × 10<sup>6</sup>)

I<sub>d</sub> = 1.77 × 10<sup>−5</sup> A

Answer: 1.77 × 10<sup>−5</sup> A

12. 🎯 Practice MCQs

Q1. Maxwell introduced displacement current to modify:

A. Faraday's law B. Ampere's law C. Ohm's law D. Coulomb's law

✅ Correct Answer: B. Ampere's law

Q2. Displacement current is associated with:

A. Constant electric field B. Time-varying electric field C. Constant magnetic field D. Static charge only

✅ Correct Answer: B. Time-varying electric field

Q3. The expression for displacement current is:

A. I = V/R B. I = dQ/dt C. I<sub>d</sub> = ε₀dΦ<sub>E</sub>/dt D. I = QR

✅ Correct Answer: C. I<sub>d</sub> = ε₀dΦ<sub>E</sub>/dt

Q4. During charging of an ideal capacitor:

A. I<sub>d</sub> = 0 B. I<sub>d</sub> = I C. I<sub>d</sub> = 2I D. I<sub>d</sub> = I/2

✅ Correct Answer: B. I<sub>d</sub> = I

Q5. A changing electric field produces:

A. Only electric charge B. Magnetic field C. Zero field D. Only heat

✅ Correct Answer: B. Magnetic field

13. 🏆 One-Minute Revision

Ampere's Law
Relates magnetic field to current.

Maxwell's Modification
Added displacement current.

Displacement Current
I<sub>d</sub> = ε₀dΦ<sub>E</sub>/dt

Charging Capacitor
I<sub>d</sub> = I

⭐ Remember

Changing Electric Field → Displacement Current → Magnetic Field

⚡ Displacement Current | Maxwell's Modification of Ampere's Law
Interactive Physics Learning • JEE / CBSE

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Alternating Current (AC): Peak, RMS, and average values of AC voltage and current

📌 1. Question Description

⚡ Alternating Current

Alternating current (AC) is an electric current whose magnitude and direction change periodically with time.

For a sinusoidal AC voltage:

v = V₀ sin(ωt)

For sinusoidal AC current:

i = I₀ sin(ωt)

Here V₀ and I₀ are the peak values of voltage and current.

🔵 2. Important Terms

Peak Value

The maximum instantaneous value attained by AC voltage or current.

V₀ = Peak Voltage

I₀ = Peak Current

RMS Value

The RMS value of AC is the equivalent DC value producing the same heating effect in a resistor.

V_rms = V₀/√2

I_rms = I₀/√2

Average Value

For a complete sinusoidal cycle, positive and negative halves cancel. Therefore:

V_avg = 0

I_avg = 0

📈 3. Animated AC Waveform

Peak Voltage V₀ = 220 V

Frequency f = 50 Hz

🔵 Instantaneous AC voltage | 🟡 Peak value | 🟢 RMS level

📐 4. Peak Value of AC

For sinusoidal alternating voltage:

v = V₀ sin(ωt)

The maximum value occurs when:

sin(ωt) = 1

Therefore:

V_max = V₀

Similarly, for current:

I_max = I₀

🧮 5. RMS Value — Derivation

Consider a sinusoidal current:

i = I₀ sin(ωt)

The mean square current over one complete cycle is:

I²_rms = Average(i²)

Therefore:

I²_rms = I₀² Average(sin²ωt)

For a complete cycle:

Average(sin²ωt) = 1/2

Hence:

I²_rms = I₀²/2

I_rms = I₀/√2

Similarly:

V_rms = V₀/√2

⭐ For sinusoidal AC:

RMS value = 0.707 × Peak value

🟠 6. Average Value of AC

For a sinusoidal AC, the average value over one complete cycle is zero because the positive and negative halves cancel each other.

I_{avg, complete cycle} = 0

V_{avg, complete cycle} = 0

For the positive half cycle:

I_avg = 2I₀/π

Similarly:

V_avg = 2V₀/π

✔ Half-cycle average = 0.637 × Peak value

📊 7. Peak, RMS & Average — Comparison

Quantity	Formula	For 220 V AC
Peak Voltage	V₀	≈ 311 V
RMS Voltage	V₀/√2	220 V
Half-cycle Average	2V₀/π	≈ 198 V
Complete-cycle Average	0	0 V

🧪 8. Practical Demonstration — AC Measurement

Experiment: Measuring AC Voltage

Apparatus:

• AC source/function generator
• AC voltmeter or multimeter
• Oscilloscope
• Connecting wires

Procedure:

1. Connect the AC source to the measuring instrument.
2. Observe the sinusoidal waveform on an oscilloscope.
3. Measure the RMS voltage using an AC voltmeter.
4. Determine the peak voltage from the waveform.

If the measured RMS voltage is 220 V:

V₀ = √2 × 220 ≈ 311 V

📚 9. Important Formulae

v = V₀ sin(ωt)

i = I₀ sin(ωt)

ω = 2πf

T = 1/f

V_rms = V₀/√2

I_rms = I₀/√2

V_{avg, half} = 2V₀/π

I_{avg, half} = 2I₀/π

V_{avg, complete} = 0

I_{avg, complete} = 0

📝 10. MCQ Practice — 15 Questions

1. The maximum value of an AC voltage is called:

A. RMS value

B. Average value

C. Peak value

D. Effective value

✔ Answer: C

2. RMS value of sinusoidal AC voltage is:

A. V₀

B. V₀/2

C. V₀/√2

D. 2V₀

✔ Answer: C

3. RMS value of 100 V peak AC is:

A. 50 V

B. 70.7 V

C. 100 V

D. 141.4 V

✔ Answer: B

4. Average value of sinusoidal AC over a complete cycle is:

A. V₀

B. V₀/√2

C. 2V₀/π

D. 0

✔ Answer: D

5. Average value over a positive half-cycle is:

A. V₀/2

B. 2V₀/π

C. V₀/√2

D. Zero

✔ Answer: B

6. For a sinusoidal AC, V₀/V_rms is:

A. 1

B. √2

C. 1/√2

D. 2

✔ Answer: B

7. The frequency of AC having time period 0.02 s is:

A. 20 Hz

B. 25 Hz

C. 50 Hz

D. 100 Hz

✔ Answer: C

8. The angular frequency of AC is:

A. ω = f/2π

B. ω = 2πf

C. ω = 1/f

D. ω = π/f

✔ Answer: B

9. 220 V AC refers to its:

A. Peak value

B. RMS value

C. Average value

D. Instantaneous value

✔ Answer: B

10. Peak value corresponding to 220 V RMS is approximately:

A. 155.6 V

B. 220 V

C. 311 V

D. 440 V

✔ Answer: C

11. RMS value is also known as:

A. Effective value

B. Peak value

C. Maximum value

D. Instantaneous value

✔ Answer: A

12. The average value of AC over a complete cycle becomes zero because:

A. Current is constant

B. Positive and negative halves cancel

C. Frequency becomes zero

D. Voltage becomes infinite

✔ Answer: B

13. For sinusoidal AC, I_rms equals approximately:

A. 0.5I₀

B. 0.637I₀

C. 0.707I₀

D. 1.414I₀

✔ Answer: C

14. Which instrument normally indicates the RMS value of AC?

A. AC voltmeter

B. Compass

C. Galvanometer only

D. Potentiometer

✔ Answer: A

15. If peak current is 10 A, RMS current is:

A. 5 A

B. 7.07 A

C. 10 A

D. 14.14 A

✔ Answer: B

🟣 11. Assertion–Reason — 5 Questions

Options:
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, but R is not the correct explanation of A.
C. A is true, but R is false.
D. A is false, but R is true.

Assertion: The average value of sinusoidal AC over a complete cycle is zero.

Reason: The positive and negative half-cycles cancel each other.

Answer: A

Assertion: RMS value of sinusoidal AC is less than its peak value.

Reason: V_rms = V₀/√2.

Answer: A

Assertion: 220 V AC has a peak voltage of approximately 311 V.

Reason: V₀ = √2 V_rms.

Answer: A

Assertion: The average value of AC is not always zero.

Reason: Average value over a half-cycle is 2V₀/π.

Answer: A

Assertion: RMS value is called the effective value of AC.

Reason: It gives the equivalent DC value for the same heating effect in a resistor.

Answer: A

🟢 12. 2 Marks — 6 Questions

Q1 Define alternating current. Write its mathematical expression.

Q2 Define peak value of AC voltage and current.

Q3 Define RMS value of alternating current.

Q4 Write the RMS value of sinusoidal AC voltage in terms of its peak value.

Q5 Why is the average value of sinusoidal AC over a complete cycle zero?

Q6 Write the relation between RMS value and peak value of AC current.

🟡 13. 3 Marks — 6 Questions

Q1 Explain peak value and RMS value of AC.

Q2 Derive the relation I_rms = I₀/√2.

Q3 Explain the average value of sinusoidal AC over a half-cycle.

Q4 Distinguish between peak value, RMS value and average value.

Q5 An AC current has a peak value of 10 A. Calculate its RMS value.

Q6 An AC voltage is represented by v = 311 sin(100πt). Find peak voltage, RMS voltage and frequency.

🟠 14. 4 Marks — 6 Questions

Q1 Derive the expression for RMS value of a sinusoidal AC current.

Q2 Derive the average value of sinusoidal AC over a positive half-cycle.

Q3 Explain peak, RMS and average values of AC with suitable equations.

Q4 For v = 200 sin(100πt), calculate peak voltage, RMS voltage, frequency and time period.

Q5 An AC current has RMS value 5 A. Find its peak current and average current over a positive half-cycle.

Q6 Explain why an AC voltmeter measures RMS value rather than the peak value.

🔴 15. 5 Marks — 6 Questions

Q1 Derive the RMS value of sinusoidal AC voltage from its instantaneous value.

Q2 Derive the average value of sinusoidal AC over a half-cycle.

Q3 Explain the physical significance of RMS value of AC.

Q4 Compare peak, RMS and average values of sinusoidal AC and derive their mutual relations.

Q5 For a 220 V AC supply, calculate peak voltage and average voltage over a positive half-cycle.

Q6 An alternating current is given by i = 20 sin(100πt). Determine peak current, RMS current, frequency, time period and half-cycle average current.

🔵 16. 6 Marks — 6 Questions

Q1 Starting from i = I₀ sin(ωt), derive the RMS value of sinusoidal alternating current in detail.

Q2 Derive the average value of sinusoidal AC over a positive half-cycle and explain why the complete-cycle average is zero.

Q3 Explain peak value, RMS value and average value of AC voltage and current with equations, graphs and physical significance.

Q4 For v = 311 sin(100πt), calculate peak voltage, RMS voltage, frequency, time period and positive half-cycle average voltage.

Q5 A 220 V, 50 Hz AC source is connected to a resistor. Explain the meaning of 220 V and calculate its peak voltage.

Q6 Derive and compare the relations V_rms = V₀/√2 and V_avg = 2V₀/π for sinusoidal AC.

🧮 17. Numerical Practice — 6 Questions

Q1. An AC voltage has peak value 311 V. Calculate its RMS value.

Q2. An AC current has RMS value 5 A. Calculate its peak value.

Q3. The instantaneous voltage is v = 100 sin(200πt). Find peak voltage, frequency and time period.

Q4. Calculate the average voltage over a positive half-cycle for an AC having peak voltage 200 V.

Q5. A 220 V, 50 Hz AC supply is given to a circuit. Find its peak voltage and angular frequency.

Q6. A sinusoidal AC current has peak value 14.14 A. Calculate its RMS value and positive half-cycle average value.

🚀 18. Quick Revision

Instantaneous Voltage

v = V₀ sin(ωt)

Instantaneous Current

i = I₀ sin(ωt)

Peak → RMS

V_rms = 0.707V₀

Peak → Half-Cycle Average

V_avg = 0.637V₀

Complete Cycle

V_avg = 0

Angular Frequency

ω = 2πf

Inductance: Self-induction and mutual induction coefficients (solenoid derivations)

📌 1. Question Description

⚡ Inductance

Inductance is the property of an electrical circuit by virtue of which it opposes any change in the current flowing through it.

Φ ∝ I

The induced EMF produced due to changing current is related to the rate of change of current.

ε = −L dI/dt

⭐ The negative sign represents Lenz's law: the induced EMF opposes the change in current.

🔵 2. Self-Induction

When the current flowing through a coil changes, the magnetic flux linked with the same coil changes. This produces an induced EMF in the same coil. This phenomenon is called self-induction.

ε = −L dI/dt

Here L is called the coefficient of self-inductance.

📐 3. Derivation of Self-Inductance of a Long Solenoid

Consider a long solenoid having:

• Number of turns = N
• Length = l
• Area of cross-section = A
• Current = I

Magnetic field inside a long solenoid is:

B = μ₀nI

where:

n = N/l

Therefore:

B = μ₀NI/l

Magnetic flux through each turn:

Φ = BA

Hence total flux linkage:

NΦ = N(BA)

Substituting B:

NΦ = N(μ₀NI/l)A

By definition:

NΦ = LI

Therefore:

L = μ₀N²A/l

✔ Self-inductance of a long air-core solenoid:

L = μ₀N²A/l

🧲 4. Solenoid with Magnetic Core

If a magnetic material having relative permeability μᵣ is inserted inside the solenoid:

L = μ₀μᵣN²A/l

Therefore increasing the permeability of the core increases the self-inductance.

🔬 5. Animated Practical — Self-Induction

Current I 5

Rate of Change dI/dt 5

Inductance L 5

↑

Changing current → Induced EMF produced

🧪 6. Practical Experiment — Self-Induction

Experiment: Coil and Battery

Apparatus:

• Inductor/coil
• Battery
• Switch
• Galvanometer or LED
• Connecting wires

Procedure:

1. Connect the coil with a battery and switch.
2. Close the switch and observe the current.
3. Open the switch suddenly.
4. Observe the induced EMF in the coil.

✔ Changing current produces changing magnetic flux.
✔ Changing flux induces EMF in the same coil.
✔ The induced EMF opposes the change in current.

🟣 7. Mutual Induction

When a changing current in one coil produces a changing magnetic flux through a nearby second coil, an induced EMF is produced in the second coil. This phenomenon is called mutual induction.

ε₂ = −M dI₁/dt

Here M is the coefficient of mutual inductance between the two coils.

🔬 8. Animated Practical — Mutual Induction

→→→

🟡 Primary coil current changes → 🔵 Magnetic flux changes → 🟢 EMF induced in secondary coil

📐 9. Derivation of Mutual Inductance of Two Long Coaxial Solenoids

Consider two long coaxial solenoids wound over the same region. Let:

• Primary turns = N₁
• Secondary turns = N₂
• Length = l
• Common area = A
• Primary current = I₁

Magnetic field produced by the primary solenoid:

B₁ = μ₀N₁I₁/l

Magnetic flux through each turn of the secondary:

Φ₂ = B₁A

Therefore total flux linkage with secondary:

N₂Φ₂ = N₂B₁A

Substituting B₁:

N₂Φ₂ = μ₀N₁N₂AI₁/l

By definition:

N₂Φ₂ = MI₁

Hence:

M = μ₀N₁N₂A/l

✔ Mutual inductance of two long coaxial solenoids:

M = μ₀N₁N₂A/l

🔗 10. Relation Between L, M and Coupling Coefficient

If two coils have self-inductances L₁ and L₂ and coefficient of coupling k, then:

M = k√(L₁L₂)

For perfect coupling:

k = 1

Therefore maximum mutual inductance is:

M = √(L₁L₂)

⚡ 11. Energy Stored in an Inductor

An inductor stores energy in its magnetic field.

U = ½LI²

Energy density in a magnetic field is:

u = B²/(2μ₀)

⚙️ 12. Applications of Inductance

🔌 Transformers

Mutual induction is the basic working principle of transformers.

📻 Radio Tuning

Inductors are used in LC circuits for selecting desired frequencies.

⚡ Choke Coil

Inductors can limit AC current with comparatively small power loss.

🔋 Energy Storage

Inductors store energy in their magnetic fields.

🔄 Oscillators

Inductors are used with capacitors in LC oscillatory circuits.

🛡️ Filters

Inductors are used in electrical and electronic filtering circuits.

📊 13. Self-Induction vs Mutual Induction

Feature	Self-Induction	Mutual Induction
Number of coils	One coil	Two coils
Induced EMF	In the same coil	In another coil
Formula	ε = −L dI/dt	ε₂ = −M dI₁/dt
Coefficient	L	M
Main application	Choke coils, filters	Transformers

📚 14. Important Formulae

ε = −L dI/dt

L = NΦ/I

L = μ₀N²A/l

L = μ₀μᵣN²A/l

ε₂ = −M dI₁/dt

M = N₂Φ₂/I₁

M = μ₀N₁N₂A/l

M = k√(L₁L₂)

U = ½LI²

📝 15. MCQ Practice — 15 Questions

1. The SI unit of inductance is:

A. Tesla

B. Weber

C. Henry

D. Farad

✔ Answer: C

2. Self-induced EMF is given by:

A. ε = LI

B. ε = −L dI/dt

C. ε = IR

D. ε = BLv

✔ Answer: B

3. The self-inductance of a solenoid is proportional to:

A. N

B. N²

C. 1/N

D. √N

✔ Answer: B

4. The self-inductance of a long solenoid is:

A. μ₀NA/l

B. μ₀N²A/l

C. μ₀Nl/A

D. μ₀A/N²l

✔ Answer: B

5. Mutual induction occurs between:

A. Two capacitors

B. Two coils

C. Two resistors

D. Two batteries

✔ Answer: B

6. The SI unit of mutual inductance is:

A. Henry

B. Ohm

C. Volt

D. Tesla

✔ Answer: A

7. Mutual inductance of two coaxial solenoids is proportional to:

A. N₁N₂

B. N₁/N₂

C. N₁+N₂

D. 1/N₁N₂

✔ Answer: A

8. The coefficient of coupling k lies between:

A. 0 and 1

B. 1 and 2

C. −1 and 0

D. 2 and 10

✔ Answer: A

9. For perfect coupling:

A. k = 0

B. k = 0.5

C. k = 1

D. k = 2

✔ Answer: C

10. Energy stored in an inductor is:

A. LI²

B. ½LI²

C. 2LI²

D. L/I²

✔ Answer: B

11. Mutual induction is the basic principle of:

A. Transformer

B. Ammeter

C. Voltmeter

D. Galvanometer only

✔ Answer: A

12. Increasing the number of turns of a solenoid increases its:

A. Resistance only

B. Self-inductance

C. Capacitance only

D. Temperature only

✔ Answer: B

13. The negative sign in ε = −L dI/dt represents:

A. Ohm's law

B. Lenz's law

C. Coulomb's law

D. Ampere's law

✔ Answer: B

14. If current through an inductor is constant, induced EMF is:

A. Maximum

B. Zero

C. Infinite

D. Negative infinity

✔ Answer: B

15. The dimension of inductance is:

A. ML²T⁻²A⁻²

B. ML²T⁻²A⁻¹

C. MLT⁻¹A⁻¹

D. ML⁻¹T⁻²

✔ Answer: A

🟣 16. Assertion–Reason — 5 Questions

Options:
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, but R is not the correct explanation of A.
C. A is true, but R is false.
D. A is false, but R is true.

Assertion: A coil opposes a change in current through it.

Reason: Self-induced EMF opposes the change in current.

Answer: A

Assertion: Self-inductance of a solenoid is proportional to N².

Reason: Both the magnetic field and flux linkage increase with the number of turns.

Answer: A

Assertion: A steady current produces no self-induced EMF.

Reason: Self-induced EMF depends on dI/dt.

Answer: A

Assertion: Mutual inductance is measured in henry.

Reason: Mutual inductance is the ratio of flux linkage in one coil to current in the other coil.

Answer: A

Assertion: Mutual inductance is maximum when coupling between two coils is perfect.

Reason: For perfect coupling, k = 1.

Answer: A

🟢 17. 2 Marks — 6 Questions

Q1 Define self-induction and write its mathematical expression.

Q2 Define mutual induction and write the expression for induced EMF.

Q3 Define coefficient of self-inductance. Give its SI unit.

Q4 What is the SI unit of mutual inductance?

Q5 What is the physical significance of the negative sign in ε = −L dI/dt?

Q6 Write the expression for energy stored in an inductor.

🟡 18. 3 Marks — 6 Questions

Q1 Explain the phenomenon of self-induction.

Q2 Explain mutual induction between two coils.

Q3 Write three factors affecting the self-inductance of a solenoid.

Q4 What is coefficient of coupling? State its range.

Q5 Explain why an inductor opposes sudden changes in current.

Q6 Write the expression for energy stored in an inductor and explain each term.

🟠 19. 4 Marks — 6 Questions

Q1 Derive the expression for self-inductance of a long solenoid.

Q2 Derive the expression for mutual inductance of two long coaxial solenoids.

Q3 Explain the factors affecting the self-inductance of a solenoid.

Q4 Explain the relation M = k√(L₁L₂).

Q5 Explain the working principle of a transformer using mutual induction.

Q6 Explain the energy stored in an inductor.

🔴 20. 5 Marks — 6 Questions

Q1 Derive the expression L = μ₀N²A/l for a long solenoid.

Q2 Derive the expression M = μ₀N₁N₂A/l for two long coaxial solenoids.

Q3 Explain self-induction, induced EMF and the significance of self-inductance.

Q4 Explain mutual induction and discuss its applications.

Q5 Derive the expression for energy stored in an inductor.

Q6 Explain coefficient of coupling and derive M = k√(L₁L₂).

🔵 21. 6 Marks — 6 Questions

Q1 Derive the coefficient of self-inductance of a long solenoid step-by-step and discuss the factors affecting it.

Q2 Derive the coefficient of mutual inductance of two long coaxial solenoids and explain the physical meaning of mutual inductance.

Q3 Explain self-induction using Lenz's law and derive ε = −L dI/dt.

Q4 Explain mutual induction, coefficient of coupling and the relation M = k√(L₁L₂).

Q5 Derive the expression for energy stored in an inductor and obtain the magnetic energy density.

Q6 Compare self-induction and mutual induction with their mathematical expressions, solenoid derivations and practical applications.

🧮 22. Numerical Practice — 5 Questions

Q1. A solenoid has 1000 turns, length 0.5 m and cross-sectional area 4 × 10⁻⁴ m². Calculate its self-inductance.

Q2. The current through a 2 H inductor changes at a rate of 3 A/s. Find the magnitude of induced EMF.

Q3. A 4 H inductor carries a current of 2 A. Calculate the energy stored in it.

Q4. Two coils have self-inductances 4 H and 9 H. If their coefficient of coupling is 0.5, calculate their mutual inductance.

Q5. Two coaxial solenoids have 500 and 1000 turns respectively, common area 2 × 10⁻⁴ m² and length 0.4 m. Calculate their mutual inductance assuming an air core.

🧪 23. Practical Applications

🔌 Transformer

Works on mutual induction between primary and secondary coils.

⚡ Choke Coil

Uses inductive reactance to control alternating current.

📻 Tuning Circuit

Inductor and capacitor form an LC circuit used for frequency selection.

🔋 Energy Storage

An inductor stores energy in its magnetic field.

🚀 24. Quick Revision

Self-Induction

Induced EMF is produced in the same coil.

ε = −L dI/dt

Mutual Induction

EMF is induced in a neighbouring coil.

ε₂ = −M dI₁/dt

Solenoid

L = μ₀N²A/l

Two Solenoids

M = μ₀N₁N₂A/l

Coupling

M = k√(L₁L₂)

Energy

U = ½LI²