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

We need to find the number of turns $$m$$ in a solenoid, given its dimensions and the magnetic field it produces.

The magnetic field inside a long solenoid is given by $$B = \mu_0 n I = \mu_0 \frac{m}{L} I$$ where $$\mu_0 = 4\pi \times 10^{-7}$$ T m/A is the permeability of free space, $$n = m/L$$ is the number of turns per unit length, $$m$$ is the total number of turns, $$L$$ is the length of the solenoid, and $$I$$ is the current.

Substituting the given values $$B = 6.28 \times 10^{-3}$$ T, $$L = 0.5$$ m, and $$I = 5$$ A into this formula gives $$6.28 \times 10^{-3} = 4\pi \times 10^{-7} \times \frac{m}{0.5} \times 5$$.

This simplifies to $$6.28 \times 10^{-3} = 4\pi \times 10^{-7} \times 10m$$ and hence $$6.28 \times 10^{-3} = 4\pi \times 10^{-6} \times m$$. Using $$\pi \approx 3.14$$, we have $$6.28 \times 10^{-3} = 4 \times 3.14 \times 10^{-6} \times m = 12.56 \times 10^{-6} \times m$$. From this, $$m = \frac{6.28 \times 10^{-3}}{12.56 \times 10^{-6}} = \frac{6.28}{12.56} \times 10^{3} = 0.5 \times 10^{3} = 500$$.

The answer is 500.