This chapter covers the key formulas related to electromagnetic waves: displacement current, the Ampere–Maxwell law, the wave equations for electric and magnetic fields, the relation between their amplitudes, and the speed of electromagnetic waves in vacuum and in a medium. These are the equations you need to solve problems about wave propagation, energy, and the electromagnetic spectrum.
Each formula is grouped by topic below, with the meaning of every symbol, when to use it, and original worked examples. For the detailed derivations and concepts, see the Class 12 Physics notes for Electromagnetic Waves.
Formulas at a Glance
| Purpose (what you are finding) | Formula |
|---|---|
| Displacement current | \( i_d = \varepsilon_0 \dfrac{d\Phi_E}{dt} \) |
| Total current (Ampere–Maxwell law) | \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 i_c + \mu_0 \varepsilon_0 \dfrac{d\Phi_E}{dt} \) |
| Electric field in a plane EM wave | \( E_x = E_0 \sin(kz – \omega t) \) |
| Magnetic field in a plane EM wave | \( B_y = B_0 \sin(kz – \omega t) \) |
| Wave number from wavelength | \( k = \dfrac{2\pi}{\lambda} \) |
| Angular frequency from frequency | \( \omega = 2\pi \nu \) |
| Speed of EM wave in vacuum | \( c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}} \) |
| Frequency–wavelength relation | \( \nu \lambda = c \) |
| Relation between amplitude of E and B | \( B_0 = \dfrac{E_0}{c} \) |
| Speed of EM wave in a medium | \( v = \dfrac{1}{\sqrt{\mu \varepsilon}} \) |
All Formulas, Grouped by Topic
Displacement Current and Ampere–Maxwell Law
The displacement current is defined as (NCERT, p. 4):
\[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \]
The generalised Ampere’s circuital law (Ampere–Maxwell law) includes both conduction and displacement currents (NCERT, p. 4):
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 i_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} \]
Wave Equations for a Plane Electromagnetic Wave
For a wave propagating along the z-direction, the electric field is along the x-axis and the magnetic field along the y-axis (NCERT, p. 6):
\[ E_x = E_0 \sin(kz – \omega t) \]
\[ B_y = B_0 \sin(kz – \omega t) \]
These can also be written using wavelength and frequency:
\[ E = E_0 \sin\left[2\pi\left(\frac{z}{\lambda} – \nu t\right)\right] \]
\[ B = B_0 \sin\left[2\pi\left(\frac{z}{\lambda} – \nu t\right)\right] \]
Relation between Wave Number, Angular Frequency, and Speed
\[ k = \frac{2\pi}{\lambda} \qquad \omega = 2\pi \nu \]
\[ \omega = c k \quad \text{or} \quad \nu \lambda = c \tag{NCERT, p. 7} \]
Speed of Light in Vacuum
\[ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \tag{NCERT, p. 7} \]
Relation between Electric and Magnetic Field Amplitudes
\[ B_0 = \frac{E_0}{c} \qquad \text{or} \qquad E_0 = c B_0 \tag{NCERT, p. 7} \]
Speed of Electromagnetic Waves in a Material Medium
\[ v = \frac{1}{\sqrt{\mu \varepsilon}} \tag{NCERT, p. 7} \]
What Each Symbol Means
| Symbol | Meaning | SI Unit |
|---|---|---|
| \( i_d \) | Displacement current | A (ampere) |
| \( i_c \) | Conduction current | A |
| \( \Phi_E \) | Electric flux through a surface | \( \text{V}\cdot\text{m} \) |
| \( \varepsilon_0 \) | Permittivity of free space | \( \text{F/m} \) |
| \( \mu_0 \) | Permeability of free space | \( \text{H/m} \) |
| \( \mathbf{B} \) | Magnetic field | T (tesla) |
| \( \mathbf{E} \) | Electric field | V/m |
| \( E_0 \) | Amplitude of electric field | V/m |
| \( B_0 \) | Amplitude of magnetic field | T |
| \( k \) | Wave number (angular spatial frequency) | rad/m |
| \( \omega \) | Angular frequency | rad/s |
| \( \lambda \) | Wavelength | m |
| \( \nu \) | Frequency | Hz (s⁻¹) |
| \( T \) | Time period | s |
| \( c \) | Speed of light in vacuum | m/s |
| \( v \) | Speed of EM wave in a medium | m/s |
| \( \varepsilon \) | Permittivity of the medium | F/m |
| \( \mu \) | Permeability of the medium | H/m |
When to Use Each Formula
| Formula | Use it when… |
|---|---|
| \( i_d = \varepsilon_0 \dfrac{d\Phi_E}{dt} \) | You need to find the displacement current inside a capacitor or in any region where the electric flux is changing with time. |
| \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 i_c + \mu_0 \varepsilon_0 \dfrac{d\Phi_E}{dt} \) | You want the magnetic field produced by a combination of conduction current and changing electric field (e.g., around a charging capacitor). |
| \( E_x = E_0 \sin(kz – \omega t) \) and \( B_y = B_0 \sin(kz – \omega t) \) | You are given the parameters of a plane electromagnetic wave and need to write the expressions for the fields, or find the fields at a given point and time. |
| \( k = \dfrac{2\pi}{\lambda} \) | You know the wavelength of the wave and need the wave number, or vice versa. |
| \( \omega = 2\pi \nu \) | You have the frequency and need the angular frequency. |
| \( c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}} \) | You need to compute the speed of light in vacuum from the constants, or to verify the value of \( c \). |
| \( \nu \lambda = c \) | You know frequency and need wavelength, or vice versa, for an electromagnetic wave in vacuum. Also used to find the speed if both are given. |
| \( B_0 = \dfrac{E_0}{c} \) | You know the amplitude of the electric field and need the amplitude of the magnetic field (or the reverse). |
| \( v = \dfrac{1}{\sqrt{\mu \varepsilon}} \) | You need the speed of an EM wave in a dielectric or magnetic medium, given its permittivity and permeability. |
Worked Examples
Example 1: Finding the magnetic field amplitude from the electric field amplitude
A plane electromagnetic wave in vacuum has an electric field amplitude \( E_0 = 90 \text{ V/m} \). What is the corresponding magnetic field amplitude? Take \( c = 3.0 \times 10^8 \text{ m/s} \).
- Step 1: Identify the relation: \( B_0 = \dfrac{E_0}{c} \).
- Step 2: Substitute the values:
\[ B_0 = \frac{90 \text{ V/m}}{3.0 \times 10^8 \text{ m/s}} \]
\[ = 3.0 \times 10^{-7} \text{ T} \]
Final answer: \( B_0 = 3.0 \times 10^{-7} \text{ T} \).
Example 2: Determining wavelength and frequency from a given wave equation
The magnetic field of a plane electromagnetic wave in vacuum is given by \( B_y = (4.0 \times 10^{-7} \text{ T}) \sin(1.0 \times 10^3 z + 3.0 \times 10^{11} t) \). Find the wavelength and frequency of the wave.
Step 1: Compare with the standard form \( B_y = B_0 \sin(kz + \omega t) \).
Here \( k = 1.0 \times 10^3 \text{ rad/m} \), \( \omega = 3.0 \times 10^{11} \text{ rad/s} \).
Step 2: Wavelength: \( \lambda = \dfrac{2\pi}{k} \).
\[ \lambda = \frac{2\pi}{1.0 \times 10^3} \approx 6.28 \times 10^{-3} \text{ m} = 6.28 \text{ mm} \]
Step 3: Frequency: \( \nu = \dfrac{\omega}{2\pi} \).
\[ \nu = \frac{3.0 \times 10^{11}}{2\pi} \approx 4.77 \times 10^{10} \text{ Hz} = 47.7 \text{ GHz} \]
Final answer: \( \lambda \approx 6.28 \text{ mm} \), \( \nu \approx 47.7 \text{ GHz} \).
Example 3: Finding the electric field equation from the magnetic field equation
For the wave in Example 2, write the expression for the electric field \( E_x \). The wave propagates along the \( z \)-axis; \( \mathbf{B} \) is along \( y \)-axis and \( \mathbf{E} \) along \( x \)-axis.
Step 1: Use the relation \( E_0 = c B_0 \).
\[ E_0 = (3.0 \times 10^8 \text{ m/s})(4.0 \times 10^{-7} \text{ T}) = 120 \text{ V/m} \]
Step 2: The wave equation for \( E_x \) has the same \( k \), \( \omega \), and sign.
Since \( B_y \) has \( \sin(kz + \omega t) \), the same form applies to \( E_x \).
\[ E_x = 120 \sin(1.0 \times 10^3 z + 3.0 \times 10^{11} t) \text{ V/m} \]
Final answer: \( E_x = 120 \sin(1.0 \times 10^3 z + 3.0 \times 10^{11} t) \text{ V/m} \).
Common Mistakes to Avoid
| Mistake | Correct Rule | How to Check Your Answer |
|---|---|---|
| Confusing displacement current with conduction current; treating them as identical in all regions. | Displacement current exists wherever the electric flux changes with time; conduction current flows only in conductors. Inside a capacitor, only displacement current exists. | Verify that the total current \( i = i_c + i_d \) is the same through any surface bounded by the same loop. |
| Forgetting that \( \mathbf{E} \) and \( \mathbf{B} \) are perpendicular to each other and to the direction of propagation. | For a plane wave propagating along \( z \), \( \mathbf{E} \) is along \( x \) and \( \mathbf{B} \) along \( y \). The cross product \( \mathbf{E} \times \mathbf{B} \) gives the direction of propagation. | Use the right-hand rule: if \( \mathbf{E} \) is along \( \hat{i} \) and \( \mathbf{B} \) along \( \hat{j} \), then \( \hat{i} \times \hat{j} = \hat{k} \) (direction of propagation). |
| Using \( c = 3 \times 10^8 \text{ m/s} \) without checking if the wave is in vacuum or in a medium. | In a medium, the speed is \( v = 1/\sqrt{\mu\varepsilon} \), which is less than \( c \). Only use \( c \) for vacuum. | If the problem mentions a dielectric or a medium with given \( \varepsilon_r \) and \( \mu_r \), compute \( v \) accordingly. |
| Mixing up the sign in the wave equation: \( \sin(kz – \omega t) \) vs \( \sin(kz + \omega t) \). | \( \sin(kz – \omega t) \) represents a wave travelling in the +\( z \) direction; \( \sin(kz + \omega t) \) travels in the –\( z \) direction. | Check the sign of the \( \omega t \) term: minus means forward, plus means backward. |
Frequently Asked Questions
What is the displacement current and why is it needed?
The displacement current is defined as \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). It was introduced by Maxwell to fix the inconsistency in Ampere’s law when applied to a charging capacitor. It ensures that the magnetic field around a capacitor is the same whether we calculate it using the conduction current or the changing electric flux.
How are the electric and magnetic fields related in an electromagnetic wave?
In a plane electromagnetic wave, the amplitudes are related by \( E_0 = c B_0 \). The fields are perpendicular to each other and to the direction of propagation, and they oscillate in phase.
What is the speed of electromagnetic waves in a medium?
The speed in a medium is \( v = 1/\sqrt{\mu\varepsilon} \), where \( \mu \) is the permeability and \( \varepsilon \) the permittivity of the medium. It is always less than the speed in vacuum \( c \).
Can an accelerated charge produce electromagnetic waves of any frequency?
The frequency of the electromagnetic wave equals the frequency of oscillation of the charge. Accelerated charges produce waves across a spectrum, but practical generation (e.g., by an antenna) is efficient only when the antenna size is comparable to the wavelength.
Reference: NCERT Class 12 Physics textbook, chapter Electromagnetic Waves.
Explore Class 12 Physics Formulas
- Class 12 Physics Formulas
- Physics Formulas for Classes 1 to 12
- Previous: Alternating Current
- Next: Ray Optics and Optical Instruments
Related chapters:
- Electric Charges and Fields notes
- Electrostatic Potential and Capacitance notes
- Current Electricity notes
Official source: download the NCERT textbook free from ncert.nic.in.