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

Answer

To solve the problem of finding the magnetic induction at a point distant \( \frac{L}{4} \) from the center of a straight conductor carrying current \( i \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Geometry**:
- We have a straight conductor of length \( L \) with a current \( i \) flowing through it. We need to find the magnetic induction at a point \( P \) which is located at a distance \( \frac{L}{4} \) from the center of the conductor.

2. **Use the Formula for Magnetic Induction**:
- The magnetic induction \( B \) at a distance \( R \) from a straight conductor can be calculated using the formula:
\[ B = \frac{\mu_0 I}{4\pi R} \left( \sin \theta_1 + \sin \theta_2 \right) \]
- Here, \( \theta_1 \) and \( \theta_2 \) are the angles subtended by the ends of the conductor at point \( P \).

3. **Determine the Angles**:
- Since the point \( P \) is at a distance \( \frac{L}{4} \) from the center, we can denote the distances from the center of the conductor to its ends as \( \frac{L}{2} \) (half the length of the conductor).
- Using the geometry of the situation, we can find the angles \( \theta_1 \) and \( \theta_2 \).

4. **Calculate the Hypotenuse**:
- The distance from point \( P \) to either end of the conductor can be calculated using the Pythagorean theorem:
\[ AP = \sqrt{\left(\frac{L}{2}\right)^2 + \left(\frac{L}{4}\right)^2} = \sqrt{\frac{L^2}{4} + \frac{L^2}{16}} = \sqrt{\frac{4L^2 + L^2}{16}} = \sqrt{\frac{5L^2}{16}} = \frac{L\sqrt{5}}{4} \]

5. **Calculate \( \sin \theta_1 \) and \( \sin \theta_2 \)**:
- Since \( \theta_1 \) and \( \theta_2 \) are equal due to symmetry, we can calculate:
\[ \sin \theta_1 = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\frac{L}{2}}{\frac{L\sqrt{5}}{4}} = \frac{2}{\sqrt{5}} \]
- Thus, \( \sin \theta_1 = \sin \theta_2 = \frac{2}{\sqrt{5}} \).

6. **Substitute Values into the Formula**:
- Now, substitute \( R = \frac{L}{4} \) and \( \sin \theta_1 + \sin \theta_2 = \frac{2}{\sqrt{5}} + \frac{2}{\sqrt{5}} = \frac{4}{\sqrt{5}} \) into the magnetic induction formula:
\[ B = \frac{\mu_0 I}{4\pi \left(\frac{L}{4}\right)} \left( \frac{4}{\sqrt{5}} \right) \]
- Simplifying this gives:
\[ B = \frac{\mu_0 I}{\pi L} \cdot \frac{4}{\sqrt{5}} = \frac{4 \mu_0 I}{\sqrt{5} \pi L} \]

7. **Final Result**:
- The magnetic induction at the point \( P \) is:
\[ B = \frac{4 \mu_0 I}{\sqrt{5} \pi L} \]