
Self-inductance of a coaxial cable: A coaxial cable consists of a long cylinder of radius which is surrounded by a hollow coaxial cylinder of the radius . Find the self-induction per unit length of such a cable.
Answer
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Hint: An emf induced in a circuit due to the changes in the circuit's magnetic field is known as self-inductance. Here we have a coaxial cable consisting of two long cylinders of radius and . We have to use ampere’s circuital law to find the self-inductance of the coaxial cable.
Formula used:
where stands for the field,
stands for the permeability of free space,
stands for the current flowing through the circuit,
stands for the radius of the cable.
where stands for the induced emf,
stands for the field, and
stands for the small area considered.
Complete step by step answer:
The radius of the coil inside is given as , and the coil surrounding has a radius .
The field in the circuit is given by,
The emf induced in a small area is given by,
Substituting the value of in the above equation, we get
(Considering a small rectangular area, )
To find the total induction, we have to integrate the above equation between the radii of the two cables, and .
Taking the constants out,
Integrating, we get
Applying the limits, we get
This can be written as,
When a current passes through a coil, flux is associated with the coil. The flux is proportional to the current through the coil.
That is,
This can be written as,
where is called the coefficient of self-induction or self-inductance.
Substituting this value for ,
Common terms on both sides are cancelled, we get
The self-inductance of a coaxial coil is thus,
Note: The phenomenon by which a coil opposes the growth or decay of current through it by producing an emf. When current passes through the coil, the coefficient of self-induction of a coil is numerically equal to the flux linked with the coil.
Formula used:
where
where
Complete step by step answer:
The radius of the coil inside is given as
The field in the circuit is given by,
The emf induced in a small area is given by,
Substituting the value of
To find the total induction, we have to integrate the above equation between the radii of the two cables,
Taking the constants out,
Integrating, we get
Applying the limits, we get
This can be written as,
When a current passes through a coil, flux is associated with the coil. The flux
That is,
This can be written as,
where
Substituting this value for
Common terms on both sides are cancelled, we get
The self-inductance of a coaxial coil is thus,
Note: The phenomenon by which a coil opposes the growth or decay of current through it by producing an emf. When
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