The magnetic field inside a solenoid can be calculated using the formula:

$$B = \mu_0 n I$$

where:

• $B$ is the magnetic field in teslas (T),
• $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \mathrm{~Tm/A}$),
• $n$ is the number of turns per unit length of the solenoid (turns/m),
• $I$ is the current in amperes (A).

Given:

• The magnetic field $B = 6.28 \times 10^{-3} \mathrm{~T}$,
• The current $I = 5 \mathrm{~A}$,
• The length of the solenoid $L = 0.5 \mathrm{~m}$,

First, let's calculate the number of turns per unit length $n$, which is $n = \frac{m}{L}$ where $m$ is the total number of turns and $L$ is the length of the solenoid.

Rearrange the formula for $B$ to solve for $m$:

$$B = \mu_0 \frac{m}{L} I$$

Therefore,

$$m = \frac{B L}{\mu_0 I}$$

Substituting the values we have:

$$m = \frac{(6.28 \times 10^{-3} \mathrm{T}) (0.5 \mathrm{m})}{(4\pi \times 10^{-7} \mathrm{Tm/A}) (5 \mathrm{A})}$$

$$m = \frac{6.28 \times 10^{-3} \times 0.5}{4\pi \times 10^{-7} \times 5}$$

$$m = \frac{6.28 \times 0.5 \times 10^{-3}}{20\pi \times 10^{-7}}$$

$$m = \frac{3.14 \times 10^{-3}}{20 \pi \times 10^{-7}}$$

$$m = \frac{3.14 \times 10^{-3}}{20 \times 3.14 \times 10^{-7}}$$

$$m = \frac{1}{20 \times 10^{-4}}$$

$$m = \frac{1 \times 10^4}{20}$$

$$m = 500$$

Therefore, the value of $m$ is 500 turns.