Inductance of a Co-axial cable, Solenoid & Toroid

Advertisements

Advertisements3

Advertisements4

Advertisements6

Advertisements66

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.