A solenoid of length $l=0.5\text{ m}$ has $N=1000$ turns and across-sectional area $A=10\times 10^{-4}{\text{ m}}^2$. What is the self-inductance $L$ of the solenoid? (Take ${\mu }_0=4\pi \times 10^{-7}\text{ T m/A}$)

Solution

1. Identify the given values: length $l=0.5\text{ m}$, number of turns $N=1000$, area $A=10\times 10^{-4}{\text{ m}}^2$, and ${\mu }_0=4\pi \times 10^{-7}\text{ T m/A}$.
2. Use the formula for the self-inductance of a solenoid: $L=\frac{{\mu }_0{N}^{2}A}{l}$.
3. Substitute the values into the equation: $L=\frac{(4\pi \times 10^{-7})\times (1000{)}^2\times (10\times 10^{-4})}{0.5}$.
4. Simplify the expression: $$\begin{array}{rl}L& =\frac{4\pi \times 10^{-7}\times 10^6\times 10^{-3}}{0.5}\\ & =\frac{4\pi \times 10^{-4}}{0.5}\\ & =8\pi \times 10^{-4}\text{ H}\end{array}$$.
5. Calculate the final value: $L\approx 2.51\times 10^{-3}\text{ H}=2.51\text{ mH}$.
6. Hence the answer is (A).