Inductance of a Co-axial cable, Solenoid & Toroid

Inductance of a Co-axial cable:-

Consider a coaxial cable with inner conductor having radius a and outer conductor of radius b and the current is flowing in the cable along z-axis  in which the cable is placed such that the axis of rotation of the cable co-incides with z-axis.

the length of the co-axial cable be d meters.

we know that the \overrightarrow{H_{\phi }} between the region a< \rho < b  is \overrightarrow{H_{\phi }}=\frac{I}{2\pi \rho }\overrightarrow{a_{\phi }}

and \overrightarrow{B_{\phi }}=\frac{\mu _{o}I}{2\pi \rho }\overrightarrow{a_{\phi }}   since \overrightarrow{B_{\phi }}=\mu _{o}\overrightarrow{H_{\phi }}

as L = \frac{\lambda }{I}, here the flux linkage \lambda =1\phi

where \phi-Total flux coming out of the surface

\phi = \oint_{s} \overrightarrow{B}.\overrightarrow{ds}

Since the magnetic flux will be radial plane extending from \rho =a   to   \rho =b and z=0 to z=d.

\overrightarrow{ds_{\phi }} = d\rho dz \overrightarrow{a_{\phi }}

\phi = \oint_{s} \overrightarrow{B}.\overrightarrow{ds}

\phi = \oint_{s}\frac{\mu _{o}I}{2\pi \rho }\overrightarrow{a_{\phi }} .d\rho dz \overrightarrow{a_{\phi }}

\phi = \int_{\rho =a}^{b}\int_{z =0}^{d}\frac{\mu _{o}I}{2\pi \rho } d\rho dz

\phi = \frac{\mu _{o}I}{2\pi } ln d

L=\frac{\phi }{I} =\frac{\mu _{o}d}{2\pi } ln  Henries.

 

Inductance of a Solenoid:-

Consider a Solenoid of N turns and  let the current flowing inside it is  ‘I’ Amperes. The length of the solenoid is ‘l’ meters and ‘A’ is its cross sectional area.

\phi – Total flux coming out of solenoid.

flux linkage \lambda = N\phi      \Rightarrow \lambda = NBA----------EQN(I)        \because \frac{\phi }{A} = B

L = \frac{\lambda }{I}, from the definition

As B is the Magnetic flux density given B= \frac{flux}{unit area} =\frac{\phi }{A}

from EQN(I) ,

\lambda = N\mu _{o}HA-------EQN(II)  because B = \mu _{o}H

The field strength H of a solenoid is H = \frac{NI}{l} A/m

EQN (II) becomes  \lambda = N \mu _{o}\frac{NI}{l}A

\lambda = \frac{N^{2} \mu _{o}IA}{l}

from the inductance definition L = \frac{\lambda }{I}

L= \frac{N^{2} \mu _{o}A}{l}   Henries.

Inductance of a Toroid:-

Consider a toroidal ring with N-turns and carrying current I.

let the radius of the toroid be ‘R’ and the total flux emerging be \phi

then flux linkage  \lambda = N\phi

the magnetic flux density inside a toroid is given by B = \frac{\mu _{o}NI}{2\pi R}

\lambda = NBA

where A is the cross sectional area of the toroid then \lambda = N \frac{\mu_{o} NI}{2\pi R}A

\lambda = \frac{\mu_{o} N^{2}I}{2\pi R}A

L=\frac{\lambda }{I}

L=\frac{\mu_{o} N^{2}A}{2\pi R}  Henries.

if the toroid has a height ‘h’ , inner radius \rightarrow r_{1} and outer radius \rightarrow r_{2} then its Inductance is L=\frac{\mu_{o} N^{2}h}{2\pi }ln   Henries.

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Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.