Question

How many meters of a thin wire is required to design a solenoid of length $1\,m$ and $L = 1\,mH$? Assume the cross-sectional diameter is very small.
A. $10\,m$
B. $40\,m$
C. $70\,m$
D. $100\,m$
E. $140\,m$

Answer 1

Hint: We know that inductance of solenoid is given as,
$L = \dfrac{{{\mu _0}{N^2}A}}{l}$
Where $N$ is the number of turns in the solenoid. $A$ is the area of each turn $l$ is the length of the solenoid and ${\mu _0}$ is a constant whose value is $4\pi \times {10^{ - 7}}$ .
We can express the number of turns in terms of length of the wire as
$N = \dfrac{x}{{2\pi r}}$
Where $x$ denotes the length of wire and the circumference of each turn is $2\pi r$
Substituting this in the first equation we can find the length of wire.

Complete step by step answer:
The inductance of a solenoid is given by the formula
$L = \dfrac{{{\mu _0}{N^2}A}}{l}$ …………………..(1)
Where $N$ is the number of turns in the solenoid. $A$ is the area of each turn. And $l$ is the length of the solenoid and ${\mu _0}$ is a constant whose value is $4\pi \times {10^{ - 7}}$.
The length of the wire, $x$ of the solenoid is circumference of each turn multiplied by the number of turns. That is,
$x = 2\pi r \times N$
Since, circumference of a circle is $2\pi r$
Therefore, the value of $N$ is given by
$N = \dfrac{x}{{2\pi r}}$
Area of the circle is
$A = \pi {r^2}$
Substituting all these values in equation 1 we get
We get,
$L = \dfrac{{4\pi \times {{10}^{ - 7}}{{\left( {\dfrac{x}{{2\pi r}}} \right)}^2}\pi {r^2}}}{l}$
$\Rightarrow L = \dfrac{{4\pi \times {{10}^{ - 7}} \times {x^2} \times \pi {r^2}}}{{4{\pi ^2}{r^2}l}}$
$ \Rightarrow x = \sqrt {\dfrac{{l \times L}}{{10 - 7}}} $
Given value of $l = 1\,m$ and $L = 1mH$
On Substituting we get
\[
\Rightarrow x = \sqrt {\dfrac{{1\, \times 1 \times {{10}^{ - 3}}\,}}{{{{10}^{ - 7}}}}} \\
\Rightarrow x = 100\,m \\
 \]

Therefore, the correct option is (D).

Note:
Here, we took the turns of the solenoid to be circular because it is given that the cross-sectional diameter of the solenoid is very small. That is why in the equation for the number of turns we divided the total length of the wire by circumference of the circle, $2\pi r$. But for large cross-sectional diameter, we cannot assume each turn as a circle.