A current i is flowing in a straight conductor of length L. The magnetic induction at a point distant `L/4` from its centre will be-

Options

A. `(4mu_(0)i)/(sqrt(5)piL)`
B. `(mu_(0)i)/(2piL)`
C. `(mu_(0)i)/(sqrt(2)L)`
D. zero

Explanation

To find the magnetic induction (magnetic field) at a point located at a distance \( \frac{L}{4} \) from the center of a straight conductor of length \( L \) carrying a current \( I \), we can follow these steps:

### Step 1: Understanding the Geometry
We have a straight conductor of length \( L \), and we want to find the magnetic field at a point \( P \) which is \( \frac{L}{4} \) away from the center of the conductor. The distance from the ends of the conductor to point \( P \) is \( \frac{L}{2} \) (half the length of the conductor) on one side and \( \frac{L}{4} \) on the other side.

### Step 2: Use the Biot-Savart Law
The magnetic field \( B \) at a point due to a current-carrying conductor can be calculated using the Biot-Savart Law:
\[ B = \frac{\mu_0 I}{4\pi r} \sin \theta \]
where \( r \) is the distance from the current element to the point where the field is being calculated, and \( \theta \) is the angle between the current element and the line connecting the current element to the point.

### Step 3: Calculate the Angles
For the point \( P \):
- The distance from the center of the conductor to point \( P \) is \( \frac{L}{4} \).
- The distances from the ends of the conductor to point \( P \) are \( \frac{L}{2} \) and \( \frac{L}{4} \).

Using the geometry of the situation, we can find:
- \( \sin \theta_1 = \frac{\frac{L}{2}}{\sqrt{\left(\frac{L}{2}\right)^2 + \left(\frac{L}{4}\right)^2}} \)
- \( \sin \theta_2 = \sin \theta_1 \) (due to symmetry)

Calculating \( \sin \theta_1 \):
\[ \sin \theta_1 = \frac{\frac{L}{2}}{\sqrt{\left(\frac{L}{2}\right)^2 + \left(\frac{L}{4}\right)^2}} = \frac{\frac{L}{2}}{\sqrt{\frac{L^2}{4} + \frac{L^2}{16}}} = \frac{\frac{L}{2}}{\sqrt{\frac{4L^2 + L^2}{16}}} = \frac{\frac{L}{2}}{\sqrt{\frac{5L^2}{16}}} = \frac{L}{2} \cdot \frac{4}{L\sqrt{5}} = \frac{2}{\sqrt{5}} \]

### Step 4: Substitute into the Biot-Savart Law
Now substituting \( r = \frac{L}{4} \) and \( \sin \theta_1 = \sin \theta_2 = \frac{2}{\sqrt{5}} \) into the Biot-Savart Law:
\[ B = \frac{\mu_0 I}{4\pi \left(\frac{L}{4}\right)} \left(\sin \theta_1 + \sin \theta_2\right) = \frac{\mu_0 I}{4\pi \left(\frac{L}{4}\right)} \left(2 \cdot \frac{2}{\sqrt{5}}\right) \]
\[ B = \frac{\mu_0 I}{\pi L} \cdot \frac{4}{\sqrt{5}} \]

### Step 5: Final Result
Thus, the magnetic induction at the point \( P \) is:
\[ B = \frac{4 \mu_0 I}{\pi L \sqrt{5}} \]

### Conclusion
The magnetic induction at a point \( \frac{L}{4} \) from the center of the conductor is given by:
\[ B = \frac{4 \mu_0 I}{\pi L \sqrt{5}} \]