State Biot – Savart law. Deduce the expression for the magnetic field at a point on the axis of a current carrying circular loop of radius ‘R’, distant ‘x’ from the centre. Hence write the magnetic field at the centre of a loop.
CBSE OUTSIDE DELHI - SET 3 PANCHKULA 2015
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Biot Savart’s law $\overrightarrow{\text{dB}}\propto\text{I}\frac{\overrightarrow{\text{dl}}\times\overrightarrow{\text{r}}}{{{\text{r}^{3}}}}$
$\overrightarrow{\text{dB}} = \frac{\mu_{\circ}}{4\pi}\text{I}\frac{\overrightarrow{\text{dl}}\times\hat{\text{r}}}{{{\text{r}^{2}}}}$
[?? ∝ ?, ?? ∝ ?? ??? ?? ∝ $\frac{1}{\text{r}^{2}} ]$
Derivation:
The resultant magnetic field will be along the axis as the perpendicular (to the axis) components cancel out in pairs.
$\text{B} = \int_{\circ}^{e\pi\text{R}}\text{dB}\cos\theta$
$\int_{\circ}^{2\pi\text{R}}\frac{\mu_{\circ}}{4\pi}\frac{\text{Idl}}{(\text{R}^{2} + \text{x}^{2})}\frac{\text{R}}{(\text{R}^{2} + \text{x}^{2})^{1/2}}$
$=\frac{\mu_o{\text{I}}}{4\pi} \frac{2\pi{\text{R}}^{2}}{{(R}^{2}+{\text{x}}^{2})^{3/2}}=\frac{\mu_o{\text{I}{\text{R}}}^{2}}{{2(R}^{2}+{\text{x}}^{2})^{3/2}}$
At centre, x= 0
$\therefore \text{B}_{\circ} = \frac{\mu_{\circ}\text{I}}{2\text{R}}$.
$\overrightarrow{\text{dB}} = \frac{\mu_{\circ}}{4\pi}\text{I}\frac{\overrightarrow{\text{dl}}\times\hat{\text{r}}}{{{\text{r}^{2}}}}$
[?? ∝ ?, ?? ∝ ?? ??? ?? ∝ $\frac{1}{\text{r}^{2}} ]$
Derivation:
The resultant magnetic field will be along the axis as the perpendicular (to the axis) components cancel out in pairs.
$\text{B} = \int_{\circ}^{e\pi\text{R}}\text{dB}\cos\theta$
$\int_{\circ}^{2\pi\text{R}}\frac{\mu_{\circ}}{4\pi}\frac{\text{Idl}}{(\text{R}^{2} + \text{x}^{2})}\frac{\text{R}}{(\text{R}^{2} + \text{x}^{2})^{1/2}}$
$=\frac{\mu_o{\text{I}}}{4\pi} \frac{2\pi{\text{R}}^{2}}{{(R}^{2}+{\text{x}}^{2})^{3/2}}=\frac{\mu_o{\text{I}{\text{R}}}^{2}}{{2(R}^{2}+{\text{x}}^{2})^{3/2}}$
At centre, x= 0
$\therefore \text{B}_{\circ} = \frac{\mu_{\circ}\text{I}}{2\text{R}}$.
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