A solenoid of length $0.5 \mathrm{~m}$ has a radius of $1 \mathrm{~cm}$ and is made up of '$\mathrm{m}$' number of turns. It carries a current of $5 \mathrm{~A}$. If the magnitude of the magnetic field inside the solenoid is $6.28 \times 10^{-3} \mathrm{~T}$ then the value of $\mathrm{m}$ is __________.

<p>The magnetic field inside a solenoid can be calculated using the formula:</p> <p>$B = \mu_0 n I$</p> <p>where:</p> <ul> <li>$B$ is the magnetic field in teslas (T),</li> <li>$\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \mathrm{~Tm/A}$),</li> <li>$n$ is the number of turns per unit length of the solenoid (turns/m),</li> <li>$I$ is the current in amperes (A).</li> </ul> <p>Given:</p> <ul> <li>The magnetic field $B = 6.28 \times 10^{-3} \mathrm{~T}$,</li> <li>The current $I = 5 \mathrm{~A}$,</li> <li>The length of the solenoid $L = 0.5 \mathrm{~m}$,</li> </ul> <p>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.</p> <p>Rearrange the formula for $B$ to solve for $m$:</p> <p>$B = \mu_0 \frac{m}{L} I$</p> <p>Therefore,</p> <p>$m = \frac{B L}{\mu_0 I}$</p> <p>Substituting the values we have:</p> <p>$$m = \frac{(6.28 \times 10^{-3} \mathrm{T}) (0.5 \mathrm{m})}{(4\pi \times 10^{-7} \mathrm{Tm/A}) (5 \mathrm{A})}$$</p> <p>$m = \frac{6.28 \times 10^{-3} \times 0.5}{4\pi \times 10^{-7} \times 5}$</p> <p>$m = \frac{6.28 \times 0.5 \times 10^{-3}}{20\pi \times 10^{-7}}$</p> <p>$m = \frac{3.14 \times 10^{-3}}{20 \pi \times 10^{-7}}$</p> <p>$m = \frac{3.14 \times 10^{-3}}{20 \times 3.14 \times 10^{-7}}$</p> <p>$m = \frac{1}{20 \times 10^{-4}}$</p> <p>$m = \frac{1 \times 10^4}{20}$</p> <p>$m = 500$</p> <p>Therefore, the value of $m$ is 500 turns.</p>