Relation Between E and B: Comprehensive NEET Physics Notes

1. Relation Between Electric Field (E) and Magnetic Field (B)

Electromagnetic waves consist of oscillating electric and magnetic fields that propagate through space. These fields are perpendicular to each other and to the direction of wave propagation. Let's explore the detailed relationship between the electric field $E$ and the magnetic field $B$ in an electromagnetic wave.

1.1 Basic Relationship Between E and B

Maxwell's equations show that in an electromagnetic wave propagating through a vacuum or free space, the electric field $E$ and magnetic field $B$ are related by the equation: $E=c\cdot B$ where:

• $E$ is the magnitude of the electric field
• $B$ is the magnitude of the magnetic field
• $c$ is the speed of light in a vacuum, approximately $3\times 10^8\text{m/s}$

This equation indicates that the ratio of the magnitudes of electric and magnetic fields is equal to the speed of light.

Explanation:

1. The electric field ($E$) and the magnetic field ($B$) are always in phase, meaning they reach their maximum and minimum values simultaneously.
2. These fields are perpendicular to each other and to the direction of wave propagation, making electromagnetic waves transverse in nature.

Did You Know? The relationship $E=c\cdot B$ signifies that electromagnetic waves can travel through a vacuum without any medium, unlike sound waves which require a material medium.

Mnemonic: Remember "E-B-C" (Electric field, Magnetic field, Speed of light): In electromagnetic waves, the electric field (E) is always equal to magnetic field (B) multiplied by the speed of light (C).

1.2 Wave Propagation and E-B Relationship

In a plane electromagnetic wave traveling in the z-direction, the electric and magnetic fields vary sinusoidally. The fields can be represented as:

• Electric field: $E_x=E_0\sin (kz-\omega t)$
• Magnetic field: $B_y=B_0\sin (kz-\omega t)$

Where:

• $E_0$ and $B_0$ are the maximum amplitudes of the electric and magnetic fields.
• $k$ is the wave number, $k=\frac{2\pi }{\lambda }$, with $\lambda$ as the wavelength.
• $\omega$ is the angular frequency, $\omega =2\pi f$.

Using Maxwell's equations, it can be derived that: $\frac{E_0}{B_0}=c$ Thus, the amplitude of the electric field is always larger than that of the magnetic field by a factor equal to the speed of light.

Real-life Application: Electromagnetic waves, such as light waves, follow this exact relationship between $E$ and $B$, which allows them to travel vast distances through space, enabling us to receive light from stars millions of light-years away.

1.3 Power and Intensity in Electromagnetic Waves

The intensity or power per unit area carried by an electromagnetic wave is given by the Poynting vector: $S=E\times B$ The average intensity of the wave is: $⟨S⟩=\frac{E_0{B}_0}{2{\mu }_0}$ Where:

• ${\mu }_0$ is the permeability of free space, $4\pi \times 10^{-7}\text{H/m}$.

1.4 Visualizing the E and B Relationship

Consider a diagram where an electromagnetic wave travels along the z-axis. The electric field (E) oscillates in the x-direction, and the magnetic field (B) oscillates in the y-direction. This perpendicular arrangement (right-hand rule) is fundamental for understanding how electromagnetic waves propagate. Visual aids like this can greatly enhance understanding.

Illustration Suggestion:

• Include a diagram showing a wave traveling in the z-direction, with arrows depicting electric field (E) oscillating in the x-direction and magnetic field (B) oscillating in the y-direction.

Common Misconception: Many students think that electric and magnetic fields are independent in electromagnetic waves. In reality, they are interdependent, constantly generating each other as the wave propagates.

1.5 NEET Problem-Solving Strategy

When solving problems involving electromagnetic waves:

• Use $E=c\cdot B$ to find one field if the other is given.
• Remember that in vacuum, $c=3\times 10^8\text{m/s}$, but in other media, $v=\frac{c}{n}$ where $n$ is the refractive index.

NEET Tip: For questions asking you to find the ratio of electric and magnetic fields, remember that the speed of light $c$ always serves as the proportionality constant. This is a quick way to eliminate incorrect answers in multiple-choice questions.

Quick Recap

• The electric field $E$ and magnetic field $B$ are perpendicular to each other in an electromagnetic wave.
• The relationship between their magnitudes is $E=c\cdot B$.
• Both fields oscillate sinusoidally and are in phase with each other.
• The Poynting vector represents the power per unit area carried by the wave.

Practice Questions

Question 1

An electromagnetic wave traveling through a vacuum has an electric field amplitude of $120\text{N/C}$. Calculate the amplitude of the magnetic field.

Solution: Using $E=c\cdot B$: $B=\frac{E}{c}=\frac{120}{3\times 10^8}=4\times 10^{-7},\text{T}$

Question 2

The magnetic field in an electromagnetic wave is given as $B=2\times 10^{-7}\text{T}$. Find the electric field associated with this wave.

Solution: Using $E=c\cdot B$: $E=3\times 10^8\times 2\times 10^{-7}=60\text{N/C}$

Question 3

An electromagnetic wave propagates in the z-direction with a frequency of $6\times 10^{14}\text{Hz}$. Determine its wavelength.

Solution: Using $\lambda =\frac{c}{f}$: $\lambda =\frac{3\times 10^8}{6\times 10^{14}}=5\times 10^{-7}\text{m}\text{or}500\text{nm}$

Question 4

If the electric field amplitude is $10\text{V/m}$, calculate the intensity of the electromagnetic wave.

Solution: $I=\frac{E^2}{2{\mu }_{0}c}=\frac{}{2\times 4\pi \times 10^{-7}\times 3\times 10^8}$