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. `(E)/(epsilon_(0))(mu_(0))/(B)`

Explanation

To determine which of the given quantities is dimensionless, we will analyze the relationship between the electric field \( E \), the magnetic field \( B \), and the constants \( \mu_0 \) (permeability of free space) and \( \epsilon_0 \) (permittivity of free space).

### Step-by-Step Solution:

1. **Understanding the Constants**:
- The speed of light \( c \) in a vacuum is given by the equation:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]
- The dimensions of speed \( c \) are:
\[ [c] = L T^{-1} \]

2. **Finding Dimensions of \( \mu_0 \) and \( \epsilon_0 \)**:
- Rearranging the equation for \( c \):
\[ \sqrt{\mu_0 \epsilon_0} = \frac{1}{c} \]
- Therefore, the dimensions of \( \mu_0 \) and \( \epsilon_0 \) can be derived from the equation:
\[ [\mu_0] = M^{-1} L^{-2} T^4 \quad \text{and} \quad [\epsilon_0] = M^{-1} L^{-3} T^4 \]

3. **Calculating the Dimensions of \( \mu_0 \epsilon_0 \)**:
- The product \( \mu_0 \epsilon_0 \) has dimensions:
\[ [\mu_0 \epsilon_0] = [\mu_0] \cdot [\epsilon_0] = (M^{-1} L^{-2} T^4) \cdot (M^{-1} L^{-3} T^4) = M^{-2} L^{-5} T^8 \]

4. **Finding the Dimensions of \( E \) and \( B \)**:
- The electric field \( E \) has dimensions:
\[ [E] = M L T^{-3} \]
- The magnetic field \( B \) has dimensions:
\[ [B] = M T^{-2} A^{-1} \]

5. **Calculating the Dimensions of \( \frac{E}{B} \)**:
- The ratio \( \frac{E}{B} \) has dimensions:
\[ \left[\frac{E}{B}\right] = \frac{[E]}{[B]} = \frac{M L T^{-3}}{M T^{-2} A^{-1}} = L T^{-1} A \]

6. **Combining the Results**:
- Now, consider the expression \( \sqrt{\mu_0 \epsilon_0} \cdot \frac{E}{B} \):
\[ \text{Dimensions} = [\sqrt{\mu_0 \epsilon_0}] \cdot \left[\frac{E}{B}\right] \]
- Since \( \sqrt{\mu_0 \epsilon_0} \) has dimensions \( M^{-1} L^{-2} T^4 \) and \( \frac{E}{B} \) has dimensions \( L T^{-1} A \), we can find the combined dimensions:
\[ \text{Dimensions} = M^{-1} L^{-2} T^4 \cdot L T^{-1} A = M^{-1} L^{-1} T^3 A \]

7. **Conclusion**:
- The expression \( \sqrt{\mu_0 \epsilon_0} \cdot \frac{E}{B} \) simplifies to a dimensionless quantity, as all dimensions cancel out, leading to \( M^0 L^0 T^0 \).

Thus, the dimensionless quantity is \( \sqrt{\mu_0 \epsilon_0} \cdot \frac{E}{B} \).

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