If all linear dimensions of an inductor are tripled, then self inductance will become (keeping the total number of turns per unit length as constant)
(A) 3 times
(B) 9 times
(C) 27 times
(D) \[\dfrac{1}{3}\]times
Answer
299.7k+ views
Hint Understand the given question and write down the formula for self inductance. It is given that the length or linear dimension of the inductor are tripled and by the self induction formula, it is derived that self inductance depends on the length of the coil. Assume the frame to be rectangular and solve for it.
Complete Step By Step Solution
Self inductance is a property of any conducting coil that opposes the change of current flow through the coil. It attains an induction due to the self induced emf produced by coil itself when there is a change in current flow. For our given case, let us assume that the given inductor is shaped rectangular and the coils are wound rectangular in shape.
For a inductor of N turns and length l and area of cross section A, having a magnetic field , the magnetic flux is given as,
\[\phi = \dfrac{{({\mu _0}NIA) \times N}}{l}\], where \[{\mu _0}\]is the magnetic permeability of the material , N is the number of turns and l is the linear dimension of the inductor. Magnetic flux is the product of the magnetic field and the number of turns. Now, Magnetic flux can also be written as a product of the self inductance of the coil and the current through the coil.
\[\phi = LI\]
Equating both, we get
\[ \Rightarrow L = \dfrac{{{\mu _0}{N^2}A}}{l}\]
In our given case, the shape of the inductor is squared having length l as its sides. We know that the total number of turns N is given as the product of the number of turns n and length l. Substituting these in the above equation we get,
\[ \Rightarrow L = \dfrac{{{\mu _0}{{(n \times l)}^2}(l \times l)}}{l}\]
\[ \Rightarrow L = {\mu _0}{(n \times l)^2}(l)\]
Now when l is tripled, we get
\[ \Rightarrow L \propto {(3l)^3}\]
\[ \Rightarrow L \propto 27{l^3}\]
Self Inductance increases by 27 times. Hence, Option (c) is the right answer.
Note The main difference between self-inductance and Mutual inductance is that in mutual inductance the current changes in one coil induces and emf on the adjacent or neighboring coil which is in the direction that opposes the change in current.
Complete Step By Step Solution
Self inductance is a property of any conducting coil that opposes the change of current flow through the coil. It attains an induction due to the self induced emf produced by coil itself when there is a change in current flow. For our given case, let us assume that the given inductor is shaped rectangular and the coils are wound rectangular in shape.
For a inductor of N turns and length l and area of cross section A, having a magnetic field , the magnetic flux is given as,
\[\phi = \dfrac{{({\mu _0}NIA) \times N}}{l}\], where \[{\mu _0}\]is the magnetic permeability of the material , N is the number of turns and l is the linear dimension of the inductor. Magnetic flux is the product of the magnetic field and the number of turns. Now, Magnetic flux can also be written as a product of the self inductance of the coil and the current through the coil.
\[\phi = LI\]
Equating both, we get
\[ \Rightarrow L = \dfrac{{{\mu _0}{N^2}A}}{l}\]
In our given case, the shape of the inductor is squared having length l as its sides. We know that the total number of turns N is given as the product of the number of turns n and length l. Substituting these in the above equation we get,
\[ \Rightarrow L = \dfrac{{{\mu _0}{{(n \times l)}^2}(l \times l)}}{l}\]
\[ \Rightarrow L = {\mu _0}{(n \times l)^2}(l)\]
Now when l is tripled, we get
\[ \Rightarrow L \propto {(3l)^3}\]
\[ \Rightarrow L \propto 27{l^3}\]
Self Inductance increases by 27 times. Hence, Option (c) is the right answer.
Note The main difference between self-inductance and Mutual inductance is that in mutual inductance the current changes in one coil induces and emf on the adjacent or neighboring coil which is in the direction that opposes the change in current.
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