A very long solenoid has 800 turns per unit length. А current of 1.6 A flows through it. Then the magnetic induction in the middle of the solenoid on its axis is approximately:

The correct option is:
$16\times {10}^{-4}\text{T}$

Understanding the Physical Principles
The magnetic induction (or magnetic field strength) $B$ inside a very long solenoid on its axis is given by the formula:
$B={\mu }_0·n·I$
where:
• ${\mu }_0$ is the permeability of free space, which is approximately $4\pi \times {10}^{-7}\text{T·m/A}$ (or about $1.26\times {10}^{-6}\text{T·m/A}$),
• $n$ is the number of turns per unit length (turns/meter),
• $I$ is the current flowing through the solenoid.

Step-by-Step Calculation
1. Identify the given values from the problem:
• Number of turns per unit length, $n=800{\text{m}}^{-1}$
• Current, $I=1.6\text{A}$
• Permeability of free space constant, ${\mu }_0\approx 4·3.14\times {10}^{-7}\text{T·m/A}\approx 12.56\times {10}^{-7}\text{T·m/A}$

2. Substitute these values into the magnetic induction formula:
$B=(4\pi \times {10}^{-7})\times 800\times 1.6$

3. Simplify the terms:
$B=4\times \pi \times 1280\times {10}^{-7}$
$B=5120\times \pi \times {10}^{-7}$
Using $\pi \approx 3.1416$:
$B\approx 5120\times 3.1416\times {10}^{-7}$
$B\approx 16085\times {10}^{-7}\text{T}$

4. Express the value in scientific notation:
$B\approx 16.1\times {10}^{-4}\text{T}$

Therefore, the magnetic induction on the axis in the middle of the solenoid is approximately $16\times {10}^{-4}\text{T}$.