A current of 10 A flow around a closed path in a circuit which is in the horizontal plane as shown in the figure. The circuit consists of eight alternative arcs of radii `r_(1)=0.08m` and `r_(2)=0.12m` . Each subtends the same angle at the centre. If magnetic field at centre produced by this circuit is `etaxx10^(-5)T`, then the value of `eta` is ?

Explanation

To solve the problem, we need to calculate the magnetic field at the center of a closed circuit consisting of eight alternating arcs of two different radii, \( r_1 = 0.08 \, \text{m} \) and \( r_2 = 0.12 \, \text{m} \), with a current of \( I = 10 \, \text{A} \) flowing through it.

### Step-by-Step Solution:

1. **Understanding the Circuit Configuration**:
The circuit consists of eight arcs, alternating between radius \( r_1 \) and \( r_2 \). Each arc subtends the same angle at the center.

2. **Magnetic Field Contribution from Each Arc**:
The magnetic field \( B \) at the center due to a circular arc of radius \( R \) carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I \theta}{4 \pi R} \]
where \( \theta \) is the angle subtended by the arc in radians and \( \mu_0 \) is the permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Calculating the Total Angle**:
Since there are eight arcs, and each arc subtends the same angle, we can denote the angle subtended by each arc as \( \theta \). The total angle for all arcs is \( 8\theta \). For simplicity, we can assume each arc subtends \( \frac{\pi}{4} \) radians (which is \( 45^\circ \)), leading to:
\[ 8\theta = 2\pi \, \text{radians} \quad \text{(which is a full circle)} \]

4. **Magnetic Field Contribution from \( r_1 \) and \( r_2 \)**:
- For arcs of radius \( r_1 \):
\[ B_1 = \frac{\mu_0 I \cdot \frac{\pi}{4}}{4 \pi r_1} = \frac{\mu_0 I}{16 r_1} \]
- For arcs of radius \( r_2 \):
\[ B_2 = \frac{\mu_0 I \cdot \frac{\pi}{4}}{4 \pi r_2} = \frac{\mu_0 I}{16 r_2} \]

5. **Total Magnetic Field at the Center**:
Since there are four arcs of each radius, the total magnetic field \( B \) at the center is:
\[ B = 4B_1 + 4B_2 = 4\left(\frac{\mu_0 I}{16 r_1}\right) + 4\left(\frac{\mu_0 I}{16 r_2}\right) \]
Simplifying this gives:
\[ B = \frac{\mu_0 I}{4} \left(\frac{1}{r_1} + \frac{1}{r_2}\right) \]

6. **Substituting the Values**:
Plugging in the values:
\[ B = \frac{(4\pi \times 10^{-7}) \cdot 10}{4} \left(\frac{1}{0.08} + \frac{1}{0.12}\right) \]

7. **Calculating the Terms**:
- Calculate \( \frac{1}{0.08} = 12.5 \) and \( \frac{1}{0.12} \approx 8.33 \).
- Therefore, \( \frac{1}{0.08} + \frac{1}{0.12} = 12.5 + 8.33 = 20.83 \).

8. **Final Calculation**:
\[ B = (10^{-7} \cdot 10 \cdot \pi) \cdot 20.83 = 6.54 \times 10^{-5} \, \text{T} \]

9. **Finding \( \eta \)**:
According to the question, the magnetic field at the center is given as \( \eta \times 10^{-5} \, \text{T} \). Thus, we have:
\[ \eta = 6.54 \]

### Final Answer:
\[ \eta = 6.54 \]