$\overrightarrow{\text{dB}}= \frac{\mu_{0}}{4\pi}\frac{\text{I}\delta\text{l}\sin\alpha}{\text{r}^{2}}$ - - - - - - (1)
The direction of $\overrightarrow{\text{dB}}$ is perpendicular to the plane containing $\overrightarrow{\text{dl}}$ and $\overrightarrow{\text{r}}$ and is given by right hand screw rule. As the angle between I $\overrightarrow{\text{dl}}$ and $\overrightarrow{\text{r}}$ is 90°, the magnitude of the magnetic induction $\overrightarrow{\text{dB}}$ is given by,
$\overrightarrow{\text{dB}} = \frac{\mu_{0}\text{I}}{4\pi} \frac{\text{dl}\sin90^{o}}{\text{r}^{2}} = \frac{\mu_{0}\text{I}\text{dl}}{4\pi\text{r}^{2}}$ - - - - (2)
If we consider the magnetic induction produced by the whole of the circular coil, then by symmetry the components of magnetic induction perpendicular to the axis will be cancelled out, while those parallel to the axis will be added up. Thus the resultant magnetic induction $\overrightarrow{\text{B}}$ at axial point P is along the axis and may be evaluated as follows:
The component of $\overrightarrow{\text{dB}}$ along the axis,
$\overrightarrow{\text{dB}}_{x} = \frac{\mu_{0}\text{I dl}}{4\pi\text{r}^{2}}\sin\alpha$ - - - - - - (3)
But sin $\alpha =\frac{\text{R}}{\text{r}}\text{ and }\text{r} = (\text{R}^{2} + \text{x}^{2})^{1/2}$
$ \therefore\overrightarrow{\text{dB}}_{x} =\frac{\mu_{0}\text{I dl}}{4\pi\text{r}^{2}}. \frac{\text{R}}{\text{r}} =\frac{\mu_{0}\text{IR}}{4\pi\text{r}^{3}}\text{dl} = \frac{\mu_{0}\text{IR}}{4\pi(\text{R}^{2} +\text{x}^{2})^{3/2}}\text{dl}$ - - - - (4)
Therefore the magnitude of resultant magnetic induction at axial point P due to the whole circular coil is given by
$ \overrightarrow{\text{B}} = \oint\frac{\mu_{0}\text{IR}}{4\pi(\text{R}^{2} + \text{x}^{2})^{3/2}}\text{dl} = \frac{\mu_{0}\text{IR}}{4\pi(\text{R}^{2} + \text{x}^{2})^{3/2}}\oint\text{dl}$
But $\oint\text{dl} = \text{ length of the loop } = 2\pi\text{R}$ - - - - - - - (5)
Therefore, $\text{B} =\frac{\mu_{0}\text{IR}}{4\pi(\text{R}^{2} + \text{x}^{2})^{3/2}}(2\pi\text{R})$
$\overrightarrow{\text{B}} = \text{B}_{x}\hat{i} =\frac{\mu_{0}\text{IR}^{2}}{2(\text{R}^{2} + \text{x}^{2})^{3/2}}\hat{\text{i}}.$
Figure shows a toroidal solenoid of average radius ‘r’ and of N turns.
For points inside the core of toroid Current I, flowing through it, set up a magnetic field within the core.
According to Ampere’s circuital law
$\oint\overrightarrow{\text{B}}.\overrightarrow{\text{d}l} = \mu_{0}\text{I}$
where ‘I’ is the current in the toroid.
Net current =NI
$\therefore\oint\overrightarrow{\text{B}}.\overrightarrow{\text{d}l} = \mu_{0}\text{NI}$
$\Rightarrow|\text{B}|2\pi\text{r} = \mu_{0}\text{NI}$
$\Rightarrow|\text{B}| = \frac{\mu_{0}\text{NI}}{2\pi\text{r}}$
$ = \mu_{0}\text{n} \text{I}$
$\bigg[\because\text{n} = \frac{\text{N}}{2\pi\text{r}}\bigg]$
$\oint\overrightarrow{\text{B}}. \overrightarrow{\text{d}l} = \mu_{0|}\text{I} = 0 $
$\Rightarrow|\text{B}|_{inside} = 0 $
$\oint\overrightarrow{\text{B}}.\overrightarrow{\text{d}l} = 0$
$\Rightarrow|\text{B}|_{exterior}= 0.$
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