An arc of a circle of raduis `R` subtends an angle `(pi)/2` at the centre. It carriers a current `i`. The magnetic field at the centre will be

Options

A. `(mu_(0)i)/(2R)`
B. `(mu_(0)i)/(8R)`
C. ``(mu_(0)i)/(4R)`
D. `(2mu_(0)i)/(5R)`

Answer

To find the magnetic field at the center of an arc of a circle that subtends an angle of \(\frac{\pi}{2}\) at the center and carries a current \(i\), we can follow these steps:

### Step 1: Understand the formula for the magnetic field due to a circular arc
The magnetic field \(B\) at the center of a circular arc carrying a current can be derived from the Biot-Savart Law. For a full circular loop, the magnetic field at the center is given by:
\[ B = \frac{\mu_0 I}{2R} \]
where \(\mu_0\) is the permeability of free space, \(I\) is the current, and \(R\) is the radius of the circular arc.

### Step 2: Adjust the formula for the angle subtended by the arc
Since the arc subtends an angle \(\theta\) at the center, we need to adjust the formula to account for the fraction of the full circle represented by this angle. The total angle of a full circle is \(2\pi\). Therefore, the magnetic field due to the arc can be expressed as:
\[ B = \frac{\mu_0 I}{2R} \cdot \frac{\theta}{2\pi} \]

### Step 3: Substitute the given angle into the formula
In this case, the angle \(\theta\) is given as \(\frac{\pi}{2}\). Substituting this value into the formula gives:
\[ B = \frac{\mu_0 I}{2R} \cdot \frac{\frac{\pi}{2}}{2\pi} \]

### Step 4: Simplify the expression
Now, we simplify the expression:
\[ B = \frac{\mu_0 I}{2R} \cdot \frac{1}{4} = \frac{\mu_0 I}{8R} \]

### Conclusion
Thus, the magnetic field at the center of the arc is:
\[ B = \frac{\mu_0 I}{8R} \]

### Final Answer
The magnetic field at the center of the arc is \(\frac{\mu_0 I}{8R}\).

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