A current of `5` flowing is passed through a straight wire of length `6 cm`, then the magnetic induction at a point `5 cm` from the either end of the wire is :

Options

A. `0.25` gauss
B. `0.125` gauss
C. `0.15` gauss
D. `0.30` gauss

Explanation

To solve the problem of finding the magnetic induction at a point 5 cm from either end of a straight wire carrying a current of 5 A and having a length of 6 cm, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Current (I) = 5 A
- Length of the wire (L) = 6 cm = 0.06 m
- Distance from either end (d) = 5 cm = 0.05 m
- The distance from the center of the wire to the point where we want to find the magnetic field is calculated as follows:
- The midpoint of the wire is 3 cm from either end (half of 6 cm).
- Therefore, the distance from the midpoint to the point of interest (5 cm from either end) is:
\[ x = 5 \, \text{cm} - 3 \, \text{cm} = 2 \, \text{cm} = 0.02 \, \text{m} \]

2. **Use the Formula for Magnetic Field**:
The magnetic induction (B) at a distance \( r \) from a straight current-carrying conductor is given by:
\[ B = \frac{\mu_0 I}{4 \pi r} \left( \sin \theta_1 + \sin \theta_2 \right) \]
where:
- \( \mu_0 \) = permeability of free space = \( 4 \pi \times 10^{-7} \, \text{T m/A} \)
- \( \theta_1 \) and \( \theta_2 \) are the angles subtended by the wire at the point of interest.

3. **Calculate the Angles**:
- From the geometry of the situation:
- The distance from the center of the wire to the point of interest is 2 cm.
- The length from the center to either end of the wire is 3 cm.
- Using the sine function:
\[ \sin \theta_1 = \frac{3 \, \text{cm}}{5 \, \text{cm}} = \frac{3}{5} \]
\[ \sin \theta_2 = \frac{3 \, \text{cm}}{5 \, \text{cm}} = \frac{3}{5} \]

4. **Substitute Values into the Formula**:
- Substitute \( \sin \theta_1 \) and \( \sin \theta_2 \) into the magnetic field formula:
\[ B = \frac{\mu_0 I}{4 \pi r} \left( \frac{3}{5} + \frac{3}{5} \right) = \frac{\mu_0 I}{4 \pi r} \left( \frac{6}{5} \right) \]
- Now substituting \( I = 5 \, \text{A} \) and \( r = 0.02 \, \text{m} \):
\[ B = \frac{(4 \pi \times 10^{-7}) \times 5}{4 \pi \times 0.02} \times \frac{6}{5} \]

5. **Simplify the Expression**:
- The \( 4 \pi \) terms cancel out:
\[ B = \frac{10^{-7} \times 5 \times 6}{0.02} \]
- Calculate:
\[ B = \frac{30 \times 10^{-7}}{0.02} = 1.5 \times 10^{-5} \, \text{T} \]

6. **Convert to Gauss**:
- Since \( 1 \, \text{T} = 10^4 \, \text{G} \):
\[ B = 1.5 \times 10^{-5} \, \text{T} = 0.15 \, \text{G} \]

### Final Answer:
The magnetic induction at a point 5 cm from either end of the wire is **0.15 Gauss**.