
A coil of wire of a certain radius $(r)$ has \[600\] turns and a self- inductance of \[108mH\] . The self- inductance of coil with same radius and \[500\] turns will is:
A. $80mH$
B. $75mH$
C. $108mH$
D. $90mH$
Answer
497.1k+ views
Hint: To answer this question, we must first understand self-inductance in general before moving on to the query that is solely based on it. The property of a current-carrying coil that resists or opposes the change in current flowing through it is known as self-inductance. This is primarily owing to the self-induced e.m.f. generated by the coil.
Complete answer:
We know that a solenoid's or coil's self-inductance is determined by,
$L = \dfrac{{{\mu _0}{N^2}A}}{l}$
Where \[L\] is the solenoid's self-inductance, is the medium's permeability, \[A\] is the cross-sectional area, \[N\] is the number of turns, and \[l\] is the solenoid's or coil's length.
The self -inductance of a coil is clearly proportional to the square of the number of turns.
Therefore,
$\dfrac{{{L_1}}}{{{L_2}}} = {\left( {\dfrac{{{n_1}}}{{{n_2}}}} \right)^2}$
So, now we will put all the given values in the above equation and find the unknown.
$ \Rightarrow \dfrac{{108mH}}{{{L_2}}} = \dfrac{{{{600}^2}}}{{{{500}^2}}}$
Now proceeding into the equation, we will cross multiply and evaluate for ${L_2}$
$
\Rightarrow {L_2} = 108 \times \dfrac{{{{500}^2}}}{{{{600}^2}}} \\
\Rightarrow {L_2} = 108 \times \dfrac{{25}}{{36}} \\
$
$\therefore {L_2} = 75mH$
Therefore, the self- inductance of coil with same radius and \[500\] turns will be $75mH$
Hence, the correct option is: (B) $75mH$
Additional information:
The attribute of self-induction of a coil is that it tends to retain the magnetic flux associated with it and opposes any change in the flux by generating current in it. A coil's inertia is equivalent to mechanical inertia. Self-induction is known as the inertia of electricity for this reason.
Note:
Mutual inductance should not be confused with self-inductance. Mutual inductance involves two or more inductors, whereas the former uses a single coil or inductor. Mutual inductance is when an inductor's magnetic flux is connected into another inductor.
Complete answer:
We know that a solenoid's or coil's self-inductance is determined by,
$L = \dfrac{{{\mu _0}{N^2}A}}{l}$
Where \[L\] is the solenoid's self-inductance, is the medium's permeability, \[A\] is the cross-sectional area, \[N\] is the number of turns, and \[l\] is the solenoid's or coil's length.
The self -inductance of a coil is clearly proportional to the square of the number of turns.
Therefore,
$\dfrac{{{L_1}}}{{{L_2}}} = {\left( {\dfrac{{{n_1}}}{{{n_2}}}} \right)^2}$
So, now we will put all the given values in the above equation and find the unknown.
$ \Rightarrow \dfrac{{108mH}}{{{L_2}}} = \dfrac{{{{600}^2}}}{{{{500}^2}}}$
Now proceeding into the equation, we will cross multiply and evaluate for ${L_2}$
$
\Rightarrow {L_2} = 108 \times \dfrac{{{{500}^2}}}{{{{600}^2}}} \\
\Rightarrow {L_2} = 108 \times \dfrac{{25}}{{36}} \\
$
$\therefore {L_2} = 75mH$
Therefore, the self- inductance of coil with same radius and \[500\] turns will be $75mH$
Hence, the correct option is: (B) $75mH$
Additional information:
The attribute of self-induction of a coil is that it tends to retain the magnetic flux associated with it and opposes any change in the flux by generating current in it. A coil's inertia is equivalent to mechanical inertia. Self-induction is known as the inertia of electricity for this reason.
Note:
Mutual inductance should not be confused with self-inductance. Mutual inductance involves two or more inductors, whereas the former uses a single coil or inductor. Mutual inductance is when an inductor's magnetic flux is connected into another inductor.
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