Question

A solenoid of length \[10\,{\text{cm}}\], diameter \[1\,{\text{cm}}\], number of turns \[500\] with relative permeability of the core \[2000\], is connected to an ac source of frequency \[50\,{\text{Hz}}\]. Then, the reactance is:
A. Zero
B. \[55\,\Omega \]
C. \[105\,\Omega \]
D. \[155\,\Omega \]

Answer 1

Hint: First of all, we will find the inductance of the solenoid using the conventional formula. Then we will find the inductive reactance by manipulating accordingly.

Complete step by step answer:
In the given problem, we are supplied with the following data:
The length of the solenoid is \[10\,{\text{cm}}\] .
The diameter of the solenoid is \[1\,{\text{cm}}\] .
The number of turns in the solenoid is \[500\] .
The frequency of the ac source is \[50\,{\text{Hz}}\] .
We are asked to find the reactance in the coil.
To begin with, first we need to find the inductance in the coil. Then only we will be able to find the reactance of the coil.
Since, the radius is half the diameter. So,
$ r = \dfrac{d}{2} \\$
$\implies r = \dfrac{1}{2}\,{\text{cm}} \\$
$\implies r = 0.5\,{\text{cm}} \\ $
We will convert the units from centimetre to metre.
So, now radius will be:
$r = 0.5\,{\text{cm}} \\
r = 0.5 \times {10^{ - 2}}\,{\text{m}} \\$
We will convert the length of the solenoid as well:
$l = 10\,{\text{cm}} \\$
$l = 10 \times {10^{ - 2}}\,{\text{m}} \\$
$l = 0.1\,{\text{m}} \\$
To calculate the inductance, we will use the formula as given below:
\[L = \dfrac{{\mu {N^2}A}}{l}\] …… (1)
Where,
\[L\] indicates inductance of the solenoid.
\[\mu \] indicates relative permeability.
\[N\] indicates the number of turns.
\[A\] indicates the cross-sectional area of the solenoid.
\[l\] indicates the length of the solenoid.
Now, substituting the required values in the equation (1), we get:
$L = \dfrac{{2000 \times {{\left( {500} \right)}^2} \times \pi \times {{\left( {0.5 \times {{10}^{ - 2}}} \right)}^2}}}{{0.10}}\,{\text{H}} \\$
$L = 0.493\,{\text{H}} \\$
Therefore, the value of the inductance is found to be \[0.493\,{\text{H}}\] .
Now, to calculate the reactance we use the formula:
\[{X_{\text{L}}} = L \times 2\pi f\] …… (2)
Where,
\[{X_{\text{L}}}\] indicates the reactance of the coil.
\[L\] indicates the inductance.
\[f\] indicates the frequency of the ac source.
Now, by substituting the required values in the equation (2), we get:
${X_{\text{L}}} = L \times 2\pi f \\$
$\implies {X_{\text{L}}} = 0.493 \times 2 \times 3.14 \times 50 \\$
$\implies {X_{\text{L}}} = 154.8\,\Omega \\$
$\therefore {X_{\text{L}}} \sim 155\,\Omega \\$
Hence, the reactance of the solenoid is \[155\,\Omega \] .

So, the correct answer is “Option D”.

Note:
This problem is based on the magnetic effects of current. The inductive reactance increases as the frequency of the ac source starts increasing. We can also say that inductive reactance is directly proportional to the frequency supplied.