# 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 $\overrightarrow{H_{\phi&space;}}$ between the region $a<&space;\rho&space;<&space;b$  is $\overrightarrow{H_{\phi&space;}}=\frac{I}{2\pi&space;\rho&space;}\overrightarrow{a_{\phi&space;}}$

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

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

where $\phi$-Total flux coming out of the surface

$\phi&space;=&space;\oint_{s}&space;\overrightarrow{B}.\overrightarrow{ds}$

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

$\overrightarrow{ds_{\phi&space;}}&space;=&space;d\rho&space;dz&space;\overrightarrow{a_{\phi&space;}}$

$\phi&space;=&space;\oint_{s}&space;\overrightarrow{B}.\overrightarrow{ds}$

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

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

$\phi&space;=&space;\frac{\mu&space;_{o}I}{2\pi&space;}&space;ln&space;d$

$L=\frac{\phi&space;}{I}&space;=\frac{\mu&space;_{o}d}{2\pi&space;}&space;ln$  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 $\lambda&space;=&space;N\phi$      $\Rightarrow&space;\lambda&space;=&space;NBA----------EQN(I)$        $\because&space;\frac{\phi&space;}{A}&space;=&space;B$

$L&space;=&space;\frac{\lambda&space;}{I}$, from the definition

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

from EQN(I) ,

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

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

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

$\lambda&space;=&space;\frac{N^{2}&space;\mu&space;_{o}IA}{l}$

from the inductance definition $L&space;=&space;\frac{\lambda&space;}{I}$

$L=&space;\frac{N^{2}&space;\mu&space;_{o}A}{l}$   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  $\lambda&space;=&space;N\phi$

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

$\lambda&space;=&space;NBA$

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

$\lambda&space;=&space;\frac{\mu_{o}&space;N^{2}I}{2\pi&space;R}A$

$L=\frac{\lambda&space;}{I}$

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

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

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