In SI units, the dimensions of `sqrt((epsilon_(0))/(mu_(0)))` is

Options

A. `AT^(-3)ML^(3//2)`
B. `A^(-1)TML^(3)`
C. `AT^(2)M^(-1)L^(-1)`
D. `A^(2)T^(3)M^(-1)L^(-2)`

Explanation

To find the dimensions of \(\sqrt{\frac{\epsilon_0}{\mu_0}}\), we will start by determining the dimensions of \(\epsilon_0\) (the permittivity of free space) and \(\mu_0\) (the permeability of free space).

### Step 1: Determine the dimensions of \(\epsilon_0\)

The permittivity \(\epsilon_0\) is defined in terms of the force between two point charges. The formula for the force \(F\) between two charges \(Q_1\) and \(Q_2\) separated by a distance \(r\) is given by:

\[ F = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_1 Q_2}{r^2} \]

Rearranging this gives:

\[ \epsilon_0 = \frac{Q_1 Q_2}{4\pi F r^2} \]

### Step 2: Identify the dimensions of the quantities involved

- The dimension of force \(F\) is \([F] = [M L T^{-2}]\).
- The dimension of charge \(Q\) is \([Q] = [I T]\), where \(I\) is the current in amperes.
- The dimension of distance \(r\) is \([r] = [L]\).

### Step 3: Substitute the dimensions into the equation for \(\epsilon_0\)

Substituting the dimensions into the equation for \(\epsilon_0\):

\[ \epsilon_0 = \frac{(I T)(I T)}{4\pi (M L T^{-2}) (L^2)} = \frac{I^2 T^2}{M L^2 T^{-2}} = \frac{I^2 T^4}{M L^2} \]

Thus, the dimensions of \(\epsilon_0\) are:

\[ [\epsilon_0] = \frac{I^2 T^4}{M L^2} \]

### Step 4: Determine the dimensions of \(\mu_0\)

The permeability \(\mu_0\) is related to the magnetic field produced by a current. The relationship can be expressed as:

\[ B = \mu_0 \cdot \frac{I}{2\pi r} \]

Where \(B\) is the magnetic field. Rearranging gives:

\[ \mu_0 = \frac{B \cdot 2\pi r}{I} \]

### Step 5: Identify the dimensions of the quantities involved for \(\mu_0\)

- The dimension of the magnetic field \(B\) is \([B] = [M T^{-2} I^{-1}]\).
- The dimension of current \(I\) is \([I] = [I]\).
- The dimension of distance \(r\) is \([L]\).

### Step 6: Substitute the dimensions into the equation for \(\mu_0\)

Substituting the dimensions into the equation for \(\mu_0\):

\[ \mu_0 = \frac{(M T^{-2} I^{-1}) (L)}{I} = \frac{M L}{T^2 I^2} \]

Thus, the dimensions of \(\mu_0\) are:

\[ [\mu_0] = \frac{M L}{T^2 I^2} \]

### Step 7: Find the dimensions of \(\sqrt{\frac{\epsilon_0}{\mu_0}}\)

Now we can find the dimensions of \(\sqrt{\frac{\epsilon_0}{\mu_0}}\):

\[ \frac{\epsilon_0}{\mu_0} = \frac{\frac{I^2 T^4}{M L^2}}{\frac{M L}{T^2 I^2}} = \frac{I^2 T^4 \cdot T^2 I^2}{M^2 L^3} = \frac{I^4 T^6}{M^2 L^3} \]

Taking the square root:

\[ \sqrt{\frac{\epsilon_0}{\mu_0}} = \sqrt{\frac{I^4 T^6}{M^2 L^3}} = \frac{I^2 T^3}{M L^{3/2}} \]

### Final Answer

Thus, the dimensions of \(\sqrt{\frac{\epsilon_0}{\mu_0}}\) are:

\[ \sqrt{\frac{\epsilon_0}{\mu_0}} = \frac{I^2 T^3}{M L^{3/2}} \]