Dimension of resistance `R` in terms of `mu_0 & epsilon_(0)` are

Options

A. `sqrt((mu_(0))/(epsilon_(0)))`
B. `sqrt((epsilon_(0))/(mu_(0)))`
C. `(epsilon_(0))/(mu_(0))`
D. `(mu_(0))/(epsilon_(0))`

Answer

To find the dimension of resistance \( R \) in terms of \( \mu_0 \) (the permeability of free space) and \( \epsilon_0 \) (the permittivity of free space), we can follow these steps:

### Step 1: Start with the formula relating power, current, and resistance
We know from electrical theory that:
\[ I^2 R = P \]
where \( P \) is the power, \( I \) is the current, and \( R \) is the resistance.

### Step 2: Express power in terms of fundamental quantities
Power \( P \) can be expressed as:
\[ P = \frac{\text{Work}}{\text{Time}} = \frac{\text{Force} \times \text{Displacement}}{\text{Time}} \]
The unit of work is Joules (J), which can be expressed as:
\[ \text{J} = \text{N} \cdot \text{m} = \text{kg} \cdot \text{m}^2/\text{s}^2 \]
Thus, we can write:
\[ P = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^3} \]

### Step 3: Substitute for \( P \) in the power equation
Substituting this into the equation \( I^2 R = P \):
\[ I^2 R = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^3} \]

### Step 4: Express current \( I \) in terms of its dimensions
Current \( I \) has the dimension of:
\[ I = \text{A} = \text{C/s} \]
where \( \text{C} \) is the unit of charge (Coulombs).

### Step 5: Substitute the expression for \( I \) into the equation
Substituting \( I \) into the equation gives:
\[ (\text{C/s})^2 R = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^3} \]
This simplifies to:
\[ \frac{\text{C}^2}{\text{s}^2} R = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^3} \]

### Step 6: Solve for \( R \)
Rearranging gives:
\[ R = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^3} \cdot \frac{\text{s}^2}{\text{C}^2} = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^1 \cdot \text{C}^2} \]

### Step 7: Find the dimensions of \( \mu_0 \) and \( \epsilon_0 \)
The dimensions of \( \mu_0 \) and \( \epsilon_0 \) are:
- \( \mu_0 \) (permeability of free space) has the dimension:
\[ \mu_0 = \frac{\text{N}}{\text{A}^2} = \frac{\text{kg} \cdot \text{m/s}^2}{\text{C}^2/\text{s}^2} = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^2 \cdot \text{C}^2} \]
- \( \epsilon_0 \) (permittivity of free space) has the dimension:
\[ \epsilon_0 = \frac{\text{C}^2}{\text{N} \cdot \text{m}^2} = \frac{\text{C}^2}{\text{kg} \cdot \text{m/s}^2 \cdot \text{m}^2} = \frac{\text{C}^2 \cdot \text{s}^2}{\text{kg} \cdot \text{m}^3} \]

### Step 8: Find the ratio \( \frac{\epsilon_0}{\mu_0} \)
Now, we can find the ratio:
\[ \frac{\epsilon_0}{\mu_0} = \frac{\frac{\text{C}^2 \cdot \text{s}^2}{\text{kg} \cdot \text{m}^3}}{\frac{\text{kg} \cdot \text{m}^2}{\text{s}^2 \cdot \text{C}^2}} = \frac{\text{C}^4 \cdot \text{s}^4}{\text{kg}^2 \cdot \text{m}^5} \]

### Step 9: Relate \( R \) to \( \mu_0 \) and \( \epsilon_0 \)
From the derived dimensions, we can express resistance \( R \) in terms of \( \mu_0 \) and \( \epsilon_0 \):
\[ R = \sqrt{\frac{\mu_0}{\epsilon_0}} \]

### Conclusion
Thus, the dimension of resistance \( R \) in terms of \( \mu_0 \) and \( \epsilon_0 \) is:
\[ R = \sqrt{\frac{\mu_0}{\epsilon_0}} \]