A current of `50 A` is placed through a straight wire of length `6 cm` then the magnetic induction at a point `5cm` from the either end of the wire is (`1gauss=10^-4T`)

Options

A. `2.5gauss`
B. `1.25gauss`
C. `1.5gauss`
D. `3.0gauss`

Explanation

To find the magnetic induction (magnetic field) at a point 5 cm from either end of a straight wire carrying a current of 50 A, we can use the formula for the magnetic field around a long straight conductor.

### Step-by-Step Solution:

1. **Identify the Parameters:**
- Current (I) = 50 A
- Length of the wire (L) = 6 cm = 0.06 m
- Distance from the wire (r) = 5 cm = 0.05 m

2. **Determine the Angles:**
- The point where we want to find the magnetic field is 5 cm from either end of the 6 cm wire. This means that the distance from the center of the wire to the point is:
\[ d = \sqrt{(3 \text{ cm})^2 + (5 \text{ cm})^2} = \sqrt{(0.03 \text{ m})^2 + (0.05 \text{ m})^2} = \sqrt{0.0009 + 0.0025} = \sqrt{0.0034} \approx 0.0583 \text{ m} \]
- The angles (α and β) from the wire to the point can be calculated using trigonometry. Since the point is symmetric with respect to the wire, both angles are equal.

3. **Use the Formula for Magnetic Field:**
- The magnetic field (B) at a distance r from a long straight conductor is given by:
\[ B = \frac{\mu_0 I}{4 \pi r} \cdot (\sin \alpha + \sin \beta) \]
- Since α = β, we can write:
\[ B = \frac{\mu_0 I}{4 \pi r} \cdot 2 \sin \alpha \]

4. **Calculate the Sine Values:**
- For α (which is the angle opposite to the distance r), we can use:
\[ \sin \alpha = \frac{5 \text{ cm}}{d} = \frac{0.05}{0.0583} \approx 0.858 \]
- Thus, \( \sin \alpha + \sin \beta = 2 \cdot 0.858 \approx 1.716 \).

5. **Substitute Values into the Formula:**
- The permeability of free space (μ₀) = \( 4 \pi \times 10^{-7} \text{ T m/A} \).
- Now substituting the values:
\[ B = \frac{(4 \pi \times 10^{-7}) \cdot 50}{4 \pi \cdot 0.05} \cdot 1.716 \]
- Simplifying:
\[ B = \frac{(10^{-7}) \cdot 50}{0.05} \cdot 1.716 = \frac{5 \times 10^{-6}}{0.05} \cdot 1.716 = 1.5 \times 10^{-4} \text{ T} \]

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

### Final Answer:
The magnetic induction at a point 5 cm from either end of the wire is **1.5 Gauss**.