A current of `1A` is flowing through a straight conductor of length `16cm`. The magnetic induction `(` in tesla `)` at a point `10cm` from the either end the wire is `:`

Options

A. `8/3 xx10^(-6)`
B. `1/(6sqrt(2))xx10^(-5)`
C. `1/(6sqrt(3))xx10^(-5)`
D. `sqrt3/6xx10^(-6)`

Answer

To solve the problem of finding the magnetic induction (magnetic field) at a point 10 cm from either end of a straight conductor carrying a current of 1 A, we can follow these steps:

### Step 1: Understand the setup
The straight conductor has a length of 16 cm. The point where we want to calculate the magnetic induction is located 10 cm from either end of the wire. This means that the point is effectively at the center of the wire when considering the distances from both ends.

### Step 2: Calculate the perpendicular distance (r)
The point of interest is 10 cm from either end. The distance from the center of the wire to the point where we want to calculate the magnetic field can be found using the Pythagorean theorem.

Given:
- Total length of the wire, L = 16 cm
- Distance from either end to the point, d = 10 cm

The half-length of the wire is:
\[ \text{Half-length} = \frac{L}{2} = \frac{16 \text{ cm}}{2} = 8 \text{ cm} \]

Using the Pythagorean theorem:
\[ r^2 = d^2 - \left(\frac{L}{2}\right)^2 \]
\[ r^2 = 10^2 - 8^2 = 100 - 64 = 36 \]
\[ r = \sqrt{36} = 6 \text{ cm} = 0.06 \text{ m} \]

### Step 3: Calculate angles (α and β)
Since the point is symmetrically located, the angles α and β (the angles subtended by the wire at the point) can be calculated.

Using:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8 \text{ cm}}{10 \text{ cm}} = \frac{4}{5} \]
Thus,
\[ \sin \alpha = \sin \beta = \frac{4}{5} \]

### Step 4: Use the formula for magnetic field due to a finite wire
The formula for the magnetic field (B) at a distance r from a finite wire carrying current I is given by:
\[ B = \frac{\mu_0 I}{4 \pi r} (\sin \alpha + \sin \beta) \]
Substituting the known values:
- Current, I = 1 A
- Permeability of free space, \(\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}\)
- \(r = 0.06 \text{ m}\)
- \(\sin \alpha + \sin \beta = \frac{4}{5} + \frac{4}{5} = \frac{8}{5}\)

Now substituting these values into the formula:
\[ B = \frac{(4\pi \times 10^{-7}) \times 1}{4 \pi \times 0.06} \times \frac{8}{5} \]
\[ B = \frac{10^{-7}}{0.06} \times \frac{8}{5} \]
\[ B = \frac{8 \times 10^{-7}}{0.3} = \frac{8 \times 10^{-7}}{3} \text{ T} \]
\[ B = \frac{8 \times 10^{-6}}{30} = \frac{4 \times 10^{-6}}{15} \text{ T} \]

### Final Answer
The magnetic induction at the point 10 cm from either end of the wire is:
\[ B = \frac{4}{15} \times 10^{-5} \text{ T} \approx 2.67 \times 10^{-6} \text{ T} \]