Question

A long solenoid has $500$ turns. When a current of $2\,A$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4\, \times \,{10^{ - 3}}\,Web$. The self-inductance of solenoid is:
A. $1.0\,Henry$
B. $4.0\,Henry$
C. $2.5\,Henry$
D. $2.0\,Henry$

Answer 1

The characteristic of the current-carrying coil that opposes the change of current that flows through it is called Self-inductance. The self-induced EMF produced in the coil is mainly caused due to this self-inductance . When current is passed through a solenoid, it develops some flux linked with the solenoid.

Complete step by step answer:
Self-inductance of a solenoid is explained as the flux associated with the solenoid when a unit amount of current is passed through the solenoid. The current in a coil induces an EMF in itself and opposes the change in current in the coil. Let us take a solenoid having $N$ number of turns, a length $l$ and area of cross-section $a$. When $I\,A$ current is flowing through it. There will be a magnetic field$B$at any given point in the solenoid.
We know that, the magnetic flux per turn is given by $B\, \times \,{\text{area}}\,{\text{of}}\,{\text{each}}\,{\text{turn}}$.
Magnetic field inside the solenoid is,
$\Rightarrow B\, = \,\dfrac{{{\mu _o}NI}}{l}$
Hence, magnetic flux through each turn is given by,
$\Rightarrow B\, = \,\dfrac{{{\mu _o}NIA}}{l}$

Now, the total magnetic flux associated with the solenoid is given
By,
$\Rightarrow B\, = \,\dfrac{{{\mu _o}{N^2}IA}}{l}$
Therefore self-inductance of solenoid is given by,
$\Rightarrow L = \dfrac{\phi }{I}$
$\Rightarrow L = \dfrac{{{\mu _o}{N^2}A}}{l}$
S.I. Unit of Self-inductance is $Henry$. Self-inductance of a solenoid only depends on the quantities of the solenoid, such as the number of turns per unit length of the solenoid and the cross-sectional area of each turn. If we increase the number of turns of the solenoid or the diameter of the solenoid, it will result in an increase of self-inductance of the solenoid.It always opposes the change in current in the circuit, whether the change in the current increases or decreases.

Now, as given in the question: the number of turns of the solenoid is $500$ turns.Current passing through the solenoid is$2\,A$. Magnetic flux associated with the solenoid per turn is$4 \times {10^{ - 3}}\,$.
Total magnetic flux associated with the solenoid is,
$\Rightarrow \phi \, = \,N\, \times \,4 \times {10^{ - 3}}\,Wb$
$\Rightarrow \phi \, = \,500\, \times 4 \times {10^{ - 3}}Wb$
$\Rightarrow \,\phi \, = \,2\,Wb$
Since, we know that Self-inductance is given by,
$\Rightarrow L\, = \dfrac{\phi }{I}\,Henry$
Substitute,
$\Rightarrow \,L\, = \,\dfrac{2}{2}\,Henry$
Simplfy,
$\therefore \,L\, = \,1.0\,Henry$
Hence, Self-Inductance of solenoid is $1.0{\text{ }}Henry$.

Therefore, option (A) is correct.

Note: When there is opposition to the change of current in one coil due to the presence of a second coil, then the opposition property is known as mutual induction. Here, in this, the current in one coil induces EMF in another coil and opposes the change in current.