A circular arc of wire subtends an angle `pi` /2 at the centre. If it carries a current i and its radius of curvature is R then the magnetic field at the centre of the arc is

Options

A. `(mu_(0)i)/R`
B. `(mu_(0)i)/(2R)`
C. `(mu_(0)i)/(4R)`
D. `(mu_(0)i)/(8R)`

Answer

To solve the problem of finding the magnetic field at the center of a circular arc of wire that subtends an angle of \(\frac{\pi}{2}\) at the center, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula**:
The magnetic field \(B\) at the center of a circular arc carrying current \(i\) and subtending an angle \(\theta\) is given by the formula:
\[ B = \frac{\mu_0 i}{4\pi r} \cdot \theta \]
where:
- \(B\) is the magnetic field,
- \(\mu_0\) is the permeability of free space,
- \(i\) is the current,
- \(r\) is the radius of the arc,
- \(\theta\) is the angle in radians.

2. **Substitute the Given Values**:
In this case, we have:
- \(\theta = \frac{\pi}{2}\),
- \(r = R\) (the radius of curvature).

Plugging these values into the formula:
\[ B = \frac{\mu_0 i}{4\pi R} \cdot \frac{\pi}{2} \]

3. **Simplify the Expression**:
Now, simplify the expression:
\[ B = \frac{\mu_0 i \cdot \pi}{4\pi R \cdot 2} \]
This simplifies to:
\[ B = \frac{\mu_0 i}{8R} \]

4. **Final Result**:
Thus, the magnetic field at the center of the arc is:
\[ B = \frac{\mu_0 i}{8R} \]

### Conclusion:
The magnetic field at the center of the circular arc is given by:
\[ B = \frac{\mu_0 i}{8R} \]