If electronic charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental quantities, the permeability of vacuum `mu_(0)` can be expressed in units of :

Options

A. `((mc^(2))/(he^(2)))`
B. `((h)/(me^(2)))`
C. `((hc)/(me^(2)))`
D. `((h)/(ce^(2)))`

Answer

To find the expression for the permeability of vacuum \( \mu_0 \) in terms of the fundamental quantities: electronic charge \( e \), electron mass \( m \), speed of light in vacuum \( c \), and Planck's constant \( h \), we will follow these steps:

### Step 1: Write the dimensions of each fundamental quantity
1. **Electronic charge \( e \)**: The dimension of charge is given by \( [e] = [A][T] = [I][T] \) where \( [I] \) is current and \( [T] \) is time.
2. **Electron mass \( m \)**: The dimension of mass is \( [m] = [M] \).
3. **Speed of light \( c \)**: The dimension of speed is \( [c] = [L][T]^{-1} \).
4. **Planck's constant \( h \)**: The dimension of Planck's constant is \( [h] = [M][L]^2[T]^{-1} \).

### Step 2: Write the dimension of \( \mu_0 \)
The permeability of vacuum \( \mu_0 \) has the dimension:
\[ [\mu_0] = [M][L][T]^{-2}[A]^{-2} \]

### Step 3: Express \( \mu_0 \) in terms of \( e \), \( m \), \( c \), and \( h \)
Assume:
\[ [\mu_0] = K e^a m^b c^c h^d \]
where \( K \) is a dimensionless constant and \( a, b, c, d \) are the powers we need to find.

### Step 4: Write the dimensions of each term
Substituting the dimensions:
- \( e^a \) contributes \( [A]^a \) or \( [I]^a[T]^a \)
- \( m^b \) contributes \( [M]^b \)
- \( c^c \) contributes \( [L]^c[T]^{-c} \)
- \( h^d \) contributes \( [M]^d[L]^{2d}[T]^{-d} \)

### Step 5: Combine the dimensions
Combining these, we have:
\[ [\mu_0] = [M]^{b+d}[L]^{c+2d}[T]^{-2-c-d}[A]^a \]

### Step 6: Set up equations by comparing dimensions
From the dimensions of \( \mu_0 \):
\[ [M]^{1} = b + d \]
\[ [L]^{1} = c + 2d \]
\[ [T]^{-2} = -2 - c - d \]
\[ [A]^{-2} = a \]

### Step 7: Solve the equations
From the equations:
1. \( b + d = 1 \) (1)
2. \( c + 2d = 1 \) (2)
3. \( -2 - c - d = -2 \) (3)
4. \( a = -2 \) (4)

From (3), we can simplify to find:
\[ c + d = 0 \implies c = -d \]
Substituting \( c = -d \) into (2):
\[ -d + 2d = 1 \implies d = 1, c = -1 \]
Substituting \( d = 1 \) into (1):
\[ b + 1 = 1 \implies b = 0 \]

### Step 8: Final expression for \( \mu_0 \)
Substituting \( a = -2 \), \( b = 0 \), \( c = -1 \), and \( d = 1 \):
\[ \mu_0 = h^1 c^{-1} e^{-2} = \frac{h}{c e^2} \]

### Conclusion
Thus, the permeability of vacuum \( \mu_0 \) can be expressed in units as:
\[ \mu_0 = \frac{h}{c e^2} \]