Question

A wire of fixed length l and radius r. Its self- inductance is found to be L. Now, if same wire is wound on a solenoid of length \[\dfrac{1}{2}\] and radius \[\dfrac{r}{2}\], then self-inductance will be:
A) \[2L\]
B) \[\dfrac{L}{2}\]
C) \[3L\]
D) \[4L\]

Answer 1

Hint: The induction of voltage is called self-inductance of that coil. When the current carrying wire is wound on a solenoid, the change in the current inside the wire induces the voltage. This self-induction of voltage is a concept of self-inductance. The self-inductance depends upon the number of turns wounded on a solenoid, length of the wire etc…

Formula used:
\[L = {\mu _0}{n^2}V\]
Where,
L=self-inductance,
\[{\mu _0}\]=permeability
n= number of turns,
V= volume of a wire.

Complete step by step answer:
The wire is in the shape of a cylinder. Here the unit length l is wound on a solenoid. And the volume of the wire is equal to \[\pi {r^2}l\]
Applying these values in the known self-inductance formula gives,
\[L = {\mu _0} \times \dfrac{{{N^2}}}{l} \times \pi {r^2}l - - - - - {\text{ }}\left( 1 \right)\]

The number of turns wounded on a solenoid is N. Now the length of the wire is taken as \[{l_1}\]
\[{l_1} = 2\pi r \times N\]
\[ \Rightarrow N = \dfrac{{{l_1}}}{{2\pi r}}\]

Applying this value of N in (1),
\[L = \dfrac{{{\mu _0}\pi {r^2}}}{l} \times {N^2}\]
\[ \Rightarrow L = \dfrac{{{\mu _0}\pi {r^2}}}{l} \times \dfrac{{{l_1}^2}}{{4{\pi ^2}{r^2}}}\]
Cancelling the common terms,
\[ \Rightarrow L = \dfrac{{{\mu _0}{l_1}^2}}{{4\pi l}} - - - {\text{ }}\left( 2 \right)\]

From the equation (2), self-inductance L is directly proportional to \[\dfrac{{{l_1}}}{l}\]
\[ \Rightarrow L\alpha \dfrac{{{l_1}}}{l}\]
Applying the value of length of solenoid,
\[ \Rightarrow L = \dfrac{{{l_1}}}{{\tfrac{1}{2}}}\]
\[ \Rightarrow L = 2{l_1}\]
Now changing the length of the wire is changed as L.
The self-inductance of the wire L is equal to 2L.

Hence, the correct answer is option (A).

Additional information:
There is another concept called mutual induction. Transformer is an example of mutual induction. When the current is flowing in the primary coil induces the current in the secondary coil. The SI unit of mutual inductance is ‘henry’.

Note: The self-inductance of the solenoid is the amount of voltage induced when the current changes inside the wire. The unit of self-inductance is Henry (H). The self-inductance is the concept used in the choke coils etc… It is used to store the electrical energy. This electrical energy is in the form of a magnetic field.