The relation between electric field E and magnetic field induction B in an electromagnetic waves

Options

A. `E= sqrt((mu_(0))/varepsilon_(0)) B`
B. E = cB
C. `E= B/c`
D. `E= B/c^(2)`

Answer

To find the relation between the electric field \( E \) and the magnetic field induction \( B \) in electromagnetic waves, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Electromagnetic Waves**:
Electromagnetic (EM) waves are solutions to Maxwell's equations. They consist of oscillating electric fields \( E \) and magnetic fields \( B \) that propagate through space.

2. **Maxwell's Equations**:
The relationship between \( E \) and \( B \) can be derived from Maxwell's equations. In particular, we focus on the wave equations derived from these equations.

3. **Wave Function Representation**:
For a traveling electromagnetic wave, we can express the electric field \( E \) and magnetic field \( B \) as:
\[ E = E_0 \cos(kx - \omega t) \]
\[ B = B_0 \cos(kx - \omega t) \]
where \( E_0 \) and \( B_0 \) are the amplitudes, \( k \) is the wave number, and \( \omega \) is the angular frequency.

4. **Differentiating the Wave Equations**:
To find the relationship between \( E \) and \( B \), we differentiate the electric field with respect to \( x \) and the magnetic field with respect to \( t \):
\[ \frac{\partial E}{\partial x} = -E_0 k \sin(kx - \omega t) \]
\[ \frac{\partial B}{\partial t} = -B_0 \omega \sin(kx - \omega t) \]

5. **Setting Up the Relationship**:
From Maxwell's equations, we know that:
\[ \frac{\partial E}{\partial x} = -\frac{\partial B}{\partial t} \]
Substituting the derivatives we found:
\[ -E_0 k \sin(kx - \omega t) = -B_0 \omega \sin(kx - \omega t) \]

6. **Cancelling the Sin Terms**:
Since \( \sin(kx - \omega t) \) cannot be zero for a wave, we can cancel it from both sides:
\[ E_0 k = B_0 \omega \]

7. **Relating \( E \) and \( B \)**:
Rearranging gives:
\[ \frac{E_0}{B_0} = \frac{\omega}{k} \]
The ratio \( \frac{\omega}{k} \) is the phase velocity \( v \) of the wave. For electromagnetic waves in a vacuum, this velocity is equal to the speed of light \( c \):
\[ c = \frac{\omega}{k} \]

8. **Final Relation**:
Thus, we arrive at the relation:
\[ \frac{E}{B} = c \]
or equivalently:
\[ E = cB \]

9. **Expressing in Terms of Constants**:
The speed of light \( c \) can also be expressed in terms of the permittivity \( \epsilon_0 \) and permeability \( \mu_0 \) of free space:
\[ c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} \]
Therefore, we can also write:
\[ E = B \cdot \frac{1}{\sqrt{\epsilon_0 \mu_0}} \]

### Conclusion:
The relationship between the electric field \( E \) and the magnetic field induction \( B \) in electromagnetic waves is given by:
\[ E = cB \]
where \( c \) is the speed of light.