
A copper wire is wound on a wooden frame , whose shape is that of an equilateral triangle . If the linear dimension of each side of the frame is increased by a factor of 3., keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil?
(A) Decrease by a factor of $9\sqrt 3 $
(B) Increase by a factor of $3$
(C) Decrease by a factor of $9$
(D) Increases by a factor of $27$
Answer
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Hint: Since the wire is wounded on the frame , this makes the case of a solenoid. We will first find the coefficient of self-inductance of solenoid for both the lengths and then compare both of the values.
Complete step by step answer:
Self induction is the process of inducing emf and thus inducing current in the coil whenever there is a change in current. The direction of induced emf and induced current is given by Lenz Law.
For a solenoid , the value of self –inductance is given by-
$L = {\mu _0}{n^2}\pi {R^2}l$
where $L$ = coefficient of self-inductance
$n$ = number of turns per unit length
$R$ = radius of the wire
$l$ = length of the solenoid
Case 1: When the length of one side of the triangle is $l$.
Length( perimeter) of the whole triangle = $3l$
Number of turns per unit length = $\dfrac{{{N_1}}}{{3l}}$ = ${n_1}$……(i)
Self- inductance= ${L_1} = {\mu _0}{n_1}^2\pi {R^2}(3l)$ ………….(ii)
Case 2: When the length is tripled .
Length of the whole triangle = $3(3l)$ = $9l$
Number of turns per unit length = $\dfrac{{{N_2}}}{{9l}} = {n_2}$ ……(iii)
Self- inductance = ${L_2} = {\mu _0}n{{}_2^2}\pi {R^2}(9l)$ ………(iv)
According to question: the number of turn per unit length remains constant i.e. ${n_1} = {n_2}$
In self- inductances , keep the values of ${n_1}$ and ${n_2}$ from eq(i) and eq (iii)
$
{L_1} = \dfrac{{{\mu _0}{N_1}^2\pi {R^2}}}{{3l}} \\
\therefore{L_2} = \dfrac{{{\mu _0}{N_2}^2\pi {R^2}}}{{9l}} \\
$
Comparing these two we get : ${L_2} = 3{L_1}$
Self –inductance increases by 3 times.
Hence, option B is correct.
Note:While performing calculations we need to keep in mind the area that is given in the formula refers to the area of the wire of which solenoid is made and not the area of the triangle. The area of wire will not change.The self-inductance provides work as magnetic resistance. It resists any change in the magnetic flux of the coil.
Complete step by step answer:
Self induction is the process of inducing emf and thus inducing current in the coil whenever there is a change in current. The direction of induced emf and induced current is given by Lenz Law.
For a solenoid , the value of self –inductance is given by-
$L = {\mu _0}{n^2}\pi {R^2}l$
where $L$ = coefficient of self-inductance
$n$ = number of turns per unit length
$R$ = radius of the wire
$l$ = length of the solenoid
Case 1: When the length of one side of the triangle is $l$.
Length( perimeter) of the whole triangle = $3l$
Number of turns per unit length = $\dfrac{{{N_1}}}{{3l}}$ = ${n_1}$……(i)
Self- inductance= ${L_1} = {\mu _0}{n_1}^2\pi {R^2}(3l)$ ………….(ii)
Case 2: When the length is tripled .
Length of the whole triangle = $3(3l)$ = $9l$
Number of turns per unit length = $\dfrac{{{N_2}}}{{9l}} = {n_2}$ ……(iii)
Self- inductance = ${L_2} = {\mu _0}n{{}_2^2}\pi {R^2}(9l)$ ………(iv)
According to question: the number of turn per unit length remains constant i.e. ${n_1} = {n_2}$
In self- inductances , keep the values of ${n_1}$ and ${n_2}$ from eq(i) and eq (iii)
$
{L_1} = \dfrac{{{\mu _0}{N_1}^2\pi {R^2}}}{{3l}} \\
\therefore{L_2} = \dfrac{{{\mu _0}{N_2}^2\pi {R^2}}}{{9l}} \\
$
Comparing these two we get : ${L_2} = 3{L_1}$
Self –inductance increases by 3 times.
Hence, option B is correct.
Note:While performing calculations we need to keep in mind the area that is given in the formula refers to the area of the wire of which solenoid is made and not the area of the triangle. The area of wire will not change.The self-inductance provides work as magnetic resistance. It resists any change in the magnetic flux of the coil.
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