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 solve the problem of finding the magnetic induction at a point 5 cm from either end of a straight wire carrying a current of 50 A, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Current (I) = 50 A
- Length of the wire (L) = 6 cm = 0.06 m
- Distance from either end to the point (r) = 5 cm = 0.05 m
- Conversion factor: 1 Gauss = \(10^{-4}\) T

2. **Understand the Geometry:**
- The wire is of finite length, and we need to find the magnetic field at a point P which is 5 cm from either end of the wire.
- The wire can be considered as extending from point A to point B, where A and B are the ends of the wire.

3. **Determine the Angles:**
- The distance from the midpoint of the wire to point P is 3 cm (half of the wire length) vertically and 5 cm horizontally.
- Using the Pythagorean theorem, we can find the distances:
- \(OP^2 = OA^2 + AP^2\)
- \(OP^2 = 3^2 + 5^2\)
- \(OP = \sqrt{9 + 25} = \sqrt{34} \approx 5.83 \text{ cm}\)

4. **Calculate Sine of Angles:**
- For the angles θ1 and θ2 at points A and B:
- \(\sin(\theta_1) = \frac{3}{5}\) and \(\sin(\theta_2) = \frac{3}{5}\)

5. **Use the Formula for Magnetic Field:**
- The formula for the magnetic field (B) due to a finite straight wire is given by:
\[ B = \frac{\mu_0 I}{4\pi r} (\sin \theta_1 + \sin \theta_2) \]
- Where:
- \(\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}\)

6. **Substituting Values:**
- Substitute the known values into the formula:
\[ B = \frac{(4\pi \times 10^{-7}) \times 50}{4\pi \times 0.05} \left(\frac{3}{5} + \frac{3}{5}\right) \]
- Simplifying:
\[ B = \frac{(10^{-7} \times 50)}{0.05} \times \frac{6}{5} \]

7. **Calculating the Magnetic Field:**
- Calculate:
\[ B = 10^{-6} \times 6 = 6 \times 10^{-6} \text{ T} \]

8. **Convert to Gauss:**
- Since \(1 \text{ Gauss} = 10^{-4} \text{ T}\):
\[ B = \frac{6 \times 10^{-6}}{10^{-4}} = 60 \text{ Gauss} \]

### Final Answer:
The magnetic induction at the point 5 cm from either end of the wire is **60 Gauss**.

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