If `E` and `B` denote electric and magnetic fields respectively, which of the following is dimensionless?

Options

A. `sqrt(mu_(0)epsilon_(0))(E )/(B)`
B. `mu_(0)epsilon_(0)(E )/(B)`
C. `mu_(0)epsilon_(0)((B)/(E ))^(2)`
D. `(mu_(0)E)/(Be_(0))`

Answer

To determine which of the given quantities involving electric field \( E \) and magnetic field \( B \) is dimensionless, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Quantities**:
- The electric field \( E \) has the dimension of \( [E] = \text{ML}^2\text{T}^{-3}\text{I}^{-1} \).
- The magnetic field \( B \) has the dimension of \( [B] = \text{ML}^1\text{T}^{-2}\text{I}^{-1} \).

2. **Finding the Ratio \( \frac{E}{B} \)**:
- To find \( \frac{E}{B} \), we can substitute the dimensions:
\[ \frac{E}{B} = \frac{\text{ML}^2\text{T}^{-3}\text{I}^{-1}}{\text{ML}^1\text{T}^{-2}\text{I}^{-1}} = \frac{\text{L}^2\text{T}^{-3}}{\text{L}\text{T}^{-2}} = \text{L}^{2-1}\text{T}^{-3+2} = \text{L}^1\text{T}^{-1} \]
- This shows that \( \frac{E}{B} \) has dimensions of velocity, specifically \( \text{LT}^{-1} \).

3. **Using the Relationship of Speed of Light**:
- The speed of light \( c \) is given by the relationship:
\[ c = \frac{E}{B} \]
- Since \( c \) is a speed, it has dimensions of \( \text{LT}^{-1} \).

4. **Considering \( \mu_0 \) and \( \epsilon_0 \)**:
- The permeability \( \mu_0 \) and permittivity \( \epsilon_0 \) are related to the speed of light by:
\[ c^2 = \frac{1}{\mu_0 \epsilon_0} \]
- This implies:
\[ \mu_0 \epsilon_0 = \frac{1}{c^2} \]
- The dimensions of \( \mu_0 \) are \( [\mu_0] = \text{MLT}^{-2}\text{I}^{-2} \) and for \( \epsilon_0 \) are \( [\epsilon_0] = \text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{I}^{2} \).

5. **Finding the Quantity \( \sqrt{\mu_0 \epsilon_0} \)**:
- The dimensions of \( \sqrt{\mu_0 \epsilon_0} \) can be calculated as follows:
\[ [\sqrt{\mu_0 \epsilon_0}] = \sqrt{[\mu_0][\epsilon_0]} = \sqrt{(\text{MLT}^{-2}\text{I}^{-2})(\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{I}^{2})} \]
- This simplifies to:
\[ = \sqrt{\text{M}^{0}\text{L}^{-1}\text{T}^{2}} = \text{L}^{-\frac{1}{2}}\text{T}^{1} \]

6. **Final Calculation**:
- Now, we can analyze the quantity \( \frac{E}{B} \sqrt{\mu_0 \epsilon_0} \):
\[ \frac{E}{B} \cdot \sqrt{\mu_0 \epsilon_0} = \text{LT}^{-1} \cdot \text{L}^{-\frac{1}{2}}\text{T}^{1} = \text{L}^{1 - \frac{1}{2}}\text{T}^{-1 + 1} = \text{L}^{\frac{1}{2}} \cdot 1 \]
- This indicates that it is not dimensionless.

7. **Conclusion**:
- The only dimensionless quantity derived from the relationships is the constant \( 1 \) when we equate \( \frac{E}{B} \) to \( c \) and use the relationships of \( \mu_0 \) and \( \epsilon_0 \).

### Final Answer:
The dimensionless quantity is \( \frac{E}{B} \sqrt{\mu_0 \epsilon_0} = 1 \).