The maximum value of permeability of `mu`-metal `(77% Ni, 16% Fe, 5% Cu, 2% Cr)` is `0*126T-m//A`. Find the maximum relative permeability and susceptibility.

Explanation

To solve the problem, we need to find the maximum relative permeability and susceptibility of the given `mu`-metal with a maximum permeability of \( \mu = 0.126 \, \text{T-m/A} \).

### Step-by-Step Solution:

1. **Understanding Permeability**:
The permeability (\( \mu \)) of a material is defined as its ability to support the formation of a magnetic field within itself. The permeability of free space (\( \mu_0 \)) is given by:
\[ \mu_0 = 4\pi \times 10^{-7} \, \text{T-m/A} \]

2. **Calculating Relative Permeability (\( \mu_r \))**:
The relative permeability (\( \mu_r \)) is the ratio of the permeability of the material to the permeability of free space:
\[ \mu_r = \frac{\mu}{\mu_0} \]
Substituting the values:
\[ \mu_r = \frac{0.126}{4\pi \times 10^{-7}} \]

3. **Calculating \( \mu_r \)**:
First, calculate \( 4\pi \):
\[ 4\pi \approx 12.5664 \]
Now, calculate \( 4\pi \times 10^{-7} \):
\[ 4\pi \times 10^{-7} \approx 12.5664 \times 10^{-7} \approx 1.25664 \times 10^{-6} \]
Now, substitute this value into the equation for \( \mu_r \):
\[ \mu_r = \frac{0.126}{1.25664 \times 10^{-6}} \approx 1.003 \times 10^5 \]
Rounding off, we get:
\[ \mu_r \approx 1 \times 10^5 \]

4. **Calculating Susceptibility (\( \chi \))**:
The relationship between relative permeability and susceptibility is given by:
\[ \mu_r = 1 + \chi \]
Rearranging this gives:
\[ \chi = \mu_r - 1 \]
Substituting the value of \( \mu_r \):
\[ \chi = 1 \times 10^5 - 1 \approx 1 \times 10^5 \]

5. **Final Results**:
- Maximum Relative Permeability (\( \mu_r \)): \( 1 \times 10^5 \)
- Maximum Susceptibility (\( \chi \)): \( 1 \times 10^5 \)

### Summary:
- The maximum relative permeability of the `mu`-metal is \( 1 \times 10^5 \).
- The maximum susceptibility of the `mu`-metal is \( 1 \times 10^5 \).