To find the dimensional formula for the permeability of free space, denoted as \( \mu_0 \), we can start from the Biot-Savart law. The law states that the magnetic field \( dB \) produced by a current-carrying conductor is given by:
\[
dB = \frac{\mu_0}{4\pi} \frac{I \, dL \, \sin \theta}{r^2}
\]
Where:
- \( dB \) is the magnetic field,
- \( I \) is the current,
- \( dL \) is the length element,
- \( r \) is the distance from the current element to the point where the magnetic field is being calculated,
- \( \theta \) is the angle between the current element and the line connecting the current element to the point.
### Step 1: Rearranging the equation to solve for \( \mu_0 \)
From the equation, we can rearrange it to isolate \( \mu_0 \):
\[
\mu_0 = \frac{4\pi \, dB \, r^2}{I \, dL \, \sin \theta}
\]
### Step 2: Identifying the dimensions of each term
1. **Dimensions of \( dB \)** (Magnetic Field):
The magnetic field \( B \) has the dimensional formula:
\[
[B] = [M][T^{-2}][A^{-1}] \quad \text{(where M = mass, T = time, A = current)}
\]
Therefore, \( [dB] = [M][T^{-2}][A^{-1}] \).
2. **Dimensions of \( r^2 \)**:
Since \( r \) is a length, we have:
\[
[r^2] = [L^2]
\]
3. **Dimensions of \( I \)** (Current):
The dimension of current \( I \) is:
\[
[I] = [A]
\]
4. **Dimensions of \( dL \)** (Length Element):
The dimension of length \( dL \) is:
\[
[dL] = [L]
\]
5. **Dimensions of \( \sin \theta \)**:
The sine function is dimensionless, so:
\[
[\sin \theta] = 1
\]
### Step 3: Substituting dimensions into the formula for \( \mu_0 \)
Now we can substitute the dimensions into the rearranged equation for \( \mu_0 \):
\[
[\mu_0] = \frac{[M][T^{-2}][A^{-1}] \cdot [L^2]}{[A] \cdot [L] \cdot 1}
\]
### Step 4: Simplifying the expression
Now, simplifying the dimensions:
\[
[\mu_0] = \frac{[M][T^{-2}][A^{-1}] \cdot [L^2]}{[A][L]} = \frac{[M][L^2][T^{-2}]}{[A^2]} = [M][L^2][T^{-2}][A^{-2}]
\]
### Final Result
Thus, the dimensional formula for the permeability of free space \( \mu_0 \) is:
\[
[\mu_0] = [M][L^2][T^{-2}][A^{-2}]
\]
