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Question

Magnetic field at centre of currentcarrying coil,

$B_{c}=\frac{\mu_{0} I}{2 r}.........(i)$

Magnetic field at axial point due to a current-carrying

Vedclass · Accepted Answer

$2 \sqrt 2 :1 $

Vedclass · Answer

Magnetic field at centre of currentcarrying coil,

$B_{c}=\frac{\mu_{0} I}{2 r}.........(i)$

Magnetic field at axial point due to a current-carrying coil at distance of $r$

$\mathrm{d}=r$

$\mathrm{B}=\frac{\mu_{0} \operatorname{Ir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{d}^{2}ight)^{3 / 2}}$

$\Rightarrow \mathrm{B}_{\mathrm{a}}=\frac{\mathrm{\mu}_{0} \mathrm{Ir}^{2}}{2\left(2 \mathrm{r}^{2}ight)^{3 / 2}}.........(ii)$

Now, $\frac{\mathrm{B}_{\mathrm{c}}}{\mathrm{B}_{\mathrm{a}}}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}} 	imes \frac{2\left(2 \mathrm{r}^{2}ight)^{3 / 2}}{\mu_{0} \mathrm{Ir}^{2}}=2 \sqrt{2}$

$\mathrm{B}_{\mathrm{c}}: \mathrm{B}_{\mathrm{a}}=2 \sqrt{2}: 1$

The magnetic field at the centre of a circular current carrying-conductor of radius $r$ is $B_c$. The magnetic field on its axis at a distance $r$ from the centre is $B_a$. The value of $B_c : B_a$ will be :-

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