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A circular coil of radius ' $r$ ' and number of turns ' $n$ ' carries a current ' $I$ '. The magnetic fields at a small distance ' $h$ ' along the axis of the coil $\left(B_a\right)$ and at the centre of the coil $\left(\mathrm{B}_{\mathrm{c}}\right)$ are measured. The ralation between $B_c$ and $B_a$ is
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Verified Answer
The correct answer is:
$\quad \mathrm{B}_{\mathrm{c}}=\mathrm{B}_{\mathrm{a}}\left(1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right)^{\frac{3}{2}}$
Magnetic field along the axis of the coil is:
$$
\mathrm{B}_{\mathrm{a}}=\frac{\mu_0}{4 \pi}\left[\frac{2 \pi \mathrm{nIr} \mathrm{r}^2}{\left(\mathrm{r}^2+\mathrm{h}^2\right)^{\frac{3}{2}}}\right]
$$
Magnetic field at the centre of the coil is:
$$
\begin{aligned}
\mathrm{B}_{\mathrm{C}} & =\frac{\mu_0}{4 \pi}\left(\frac{2 \pi \mathrm{nI}}{\mathrm{r}}\right) \\
\frac{\mathrm{B}_{\mathrm{C}}}{\mathrm{B}_{\mathrm{a}}} & =\frac{\left(\mathrm{r}^2+\mathrm{h}^2\right)^{\frac{3}{2}}}{\mathrm{r}^3} \\
\therefore \quad \mathrm{B}_{\mathrm{C}} & =\mathrm{B}_{\mathrm{a}}\left[1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right]^{\frac{3}{2}}
\end{aligned}
$$
$$
\mathrm{B}_{\mathrm{a}}=\frac{\mu_0}{4 \pi}\left[\frac{2 \pi \mathrm{nIr} \mathrm{r}^2}{\left(\mathrm{r}^2+\mathrm{h}^2\right)^{\frac{3}{2}}}\right]
$$
Magnetic field at the centre of the coil is:
$$
\begin{aligned}
\mathrm{B}_{\mathrm{C}} & =\frac{\mu_0}{4 \pi}\left(\frac{2 \pi \mathrm{nI}}{\mathrm{r}}\right) \\
\frac{\mathrm{B}_{\mathrm{C}}}{\mathrm{B}_{\mathrm{a}}} & =\frac{\left(\mathrm{r}^2+\mathrm{h}^2\right)^{\frac{3}{2}}}{\mathrm{r}^3} \\
\therefore \quad \mathrm{B}_{\mathrm{C}} & =\mathrm{B}_{\mathrm{a}}\left[1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right]^{\frac{3}{2}}
\end{aligned}
$$
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