A solenoid of length $L=0.5\text{ m}$ has $N=500$ turns and carries a current $I=2\text{ A}$. The magnetic field $B$ at the center of the solenoid is:

Solution

1. Identify the given values: length $L=0.5\text{ m}$, number of turns $N=500$, and current $I=2\text{ A}$.
2. Calculate the turn density $$\begin{array}{rl}n& =N/L\\ & =500/0.5\\ & =1000\text{ turns/m}\end{array}$$.
3. Use the formula for the magnetic field of a solenoid: $B={\mu }_{0}nI$.
4. Substitute the values: $$\begin{array}{rl}B& =(4\pi \times 10^{-7}\text{ T}\cdot \text{m/A})\times (1000{\text{ m}}^{-1})\times (2\text{ A})\\ & =8\pi \times 10^{-4}\text{ T}\end{array}$$.
5. Hence the answer is (A).