A rod with circular cross-section area `2cm^(2)` and length `40cm` is wound uniformly with `400` turns of an insulated wire.If a current of `0.4` A flows in the wire windings,the total magnetic flux produced inside windings is `4 pi times10^(-6)Wb`.The relative permeability of the rod is (Given: Permeability of vacuum `mu_(0)=4 pi times10^(-7)NA^(-2)`?

Options

A. `125`
B. `12.5`
C. `5/(16)`
D. `(32)/5`

Answer

To solve the problem, we need to find the relative permeability (μ_r) of the rod given the magnetic flux (Φ), the area (A), the number of turns (N), the current (I), and the permeability of vacuum (μ₀).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Cross-sectional area of the rod, A = 2 cm² = 2 × 10^(-4) m² (convert cm² to m²)
- Length of the rod, L = 40 cm = 0.4 m (convert cm to m)
- Number of turns, N = 400
- Current, I = 0.4 A
- Magnetic flux, Φ = 4π × 10^(-6) Wb
- Permeability of vacuum, μ₀ = 4π × 10^(-7) N/A²

2. **Calculate the Number of Turns per Meter (n):**
\[ n = \frac{N}{L} = \frac{400}{0.4} = 1000 \text{ turns/m} \]

3. **Use the Formula for Magnetic Flux:**
The magnetic flux (Φ) can be expressed as:
\[ Φ = μ_r \cdot μ₀ \cdot n \cdot I \cdot A \]
Rearranging this gives:
\[ μ_r = \frac{Φ}{μ₀ \cdot n \cdot I \cdot A} \]

4. **Substitute the Known Values:**
\[ μ_r = \frac{4π \times 10^{-6}}{(4π \times 10^{-7}) \cdot (1000) \cdot (0.4) \cdot (2 \times 10^{-4})} \]

5. **Simplify the Expression:**
- The \(4π\) terms cancel out:
\[ μ_r = \frac{10^{-6}}{(10^{-7}) \cdot (1000) \cdot (0.4) \cdot (2 \times 10^{-4})} \]
- Calculate the denominator:
\[ = 10^{-7} \cdot 1000 \cdot 0.4 \cdot 2 \times 10^{-4} = 10^{-7} \cdot 800 \times 10^{-4} = 800 \times 10^{-11} = 8 \times 10^{-9} \]
- Thus,
\[ μ_r = \frac{10^{-6}}{8 \times 10^{-9}} = \frac{10^{-6}}{8} \times 10^{9} = \frac{10^{3}}{8} = 125 \]

6. **Final Answer:**
The relative permeability of the rod is:
\[ μ_r = 125 \]