In terms of basic units of mass (M), length (L), time (T), and charge (Q), the dimensions of magnetic permeability of vacuum `(mu_0)` would be

Options

A. `[MLQ^(-2)]`
B. `[LT^(-1)Q^(-1)]`
C. `[ML^(2)T^(-1)Q^(-2)]`
D. `[LTQ^(-1)]`

Answer

To find the dimensions of magnetic permeability of vacuum \((\mu_0)\), we can start with the Biot-Savart law, which relates the magnetic field \((B)\) to the current \((I)\) and the distance \((R)\).

### Step-by-Step Solution:

1. **Start with the Biot-Savart Law**:
\[ B = \frac{\mu_0 I \, dl}{R^2} \]
Here, \(B\) is the magnetic field, \(\mu_0\) is the magnetic permeability, \(I\) is the current, \(dl\) is an infinitesimal length, and \(R\) is the distance.

2. **Rearranging for \(\mu_0\)**:
Rearranging the equation gives:
\[ \mu_0 = \frac{B R^2}{I \, dl} \]

3. **Identify the dimensions**:
- The dimension of \(R\) (length) is \([L]\).
- The dimension of \(dl\) (length) is also \([L]\).
- The dimension of current \(I\) can be expressed in terms of charge \((Q)\) and time \((T)\) as:
\[ I = \frac{Q}{T} \]

4. **Substituting dimensions into the equation**:
The dimensions of \(\mu_0\) can now be expressed as:
\[ [\mu_0] = \frac{[B] \cdot [L^2]}{[I] \cdot [L]} \]

5. **Finding the dimensions of the magnetic field \(B\)**:
From the Lorentz force law:
\[ F = Q \cdot V \cdot B \]
Rearranging gives:
\[ B = \frac{F}{Q \cdot V} \]
The dimensions of force \(F\) are \([M L T^{-2}]\), velocity \(V\) is \([L T^{-1}]\), and charge \(Q\) is \([Q]\). Therefore:
\[ [B] = \frac{[M L T^{-2}]}{[Q] \cdot [L T^{-1}]} = \frac{[M L T^{-2}]}{[Q] \cdot [L] \cdot [T^{-1}]} = \frac{[M]}{[Q] \cdot [T]} \]

6. **Substituting back to find \(\mu_0\)**:
Now substituting the dimension of \(B\) back into the equation for \(\mu_0\):
\[ [\mu_0] = \frac{\left(\frac{[M]}{[Q] \cdot [T]}\right) \cdot [L^2]}{\left(\frac{Q}{T}\right) \cdot [L]} = \frac{[M] \cdot [L^2]}{[Q] \cdot [T] \cdot \frac{Q}{T} \cdot [L]} \]
Simplifying this gives:
\[ [\mu_0] = \frac{[M] \cdot [L^2]}{[Q^2] \cdot [T^{-2}]} \]
Which can be rearranged to:
\[ [\mu_0] = [M] \cdot [L] \cdot [T^2] \cdot [Q^{-2}] \]

7. **Final Dimension**:
Therefore, the final dimension of magnetic permeability of vacuum \((\mu_0)\) is:
\[ [\mu_0] = [M L T^{-2} Q^{-2}] \]