A long, straight wire is turned into a loop o | vedclass

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Question

$\mathrm{B}$ at the centre of a coil carrying a current, $i$ is

$\mathrm{B}_{	ext {coil }}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}} \quad 	ext { (

Vedclass · Accepted Answer

$3.4 	imes {10^{ - 5}}$ $newton/(amp-meter)$, outwards

Vedclass · Answer

$\mathrm{B}$ at the centre of a coil carrying a current, $i$ is

$\mathrm{B}_{	ext {coil }}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}} \quad 	ext { (outwards) }$

$B$ due to wire,

$\mathrm{B}_{	ext {wire }}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$

${	ext { Given }: \mathrm{i}=8\, \mathrm{A}, \mathrm{r}=10 	imes 10^{-2} \,\mathrm{m}}$

$\frac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}}$

Magnetic field at centre $\mathrm{C}$

$\mathrm{B}_{\mathrm{C}}=\mathrm{B}_{	ext {coil }}+\mathrm{B}_{	ext {wire }}$

${=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}(\mathrm{upward})+\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}}$ (inwards)

${=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}-\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}\left(1-\frac{1}{\pi}ight)}$ (outwards)

$=\frac{4 \pi 	imes 10^{-7} 	imes 8}{2 	imes 10 	imes 10^{-2}}\left(1-\frac{1}{3.14}ight) \quad$ (outwards)

$=\frac{4 	imes 3.14 	imes 10^{-7} 	imes 8 	imes 2.14}{2 	imes 10 	imes 10^{-2} 	imes 3.14} \quad$ (outwards)

$=3.424 	imes 10^{-5} \quad$ (outwards)

A long, straight wire is turned into a loop of radius $10\,cm$ (see figure). If a current of $8\, A$ is passed through the loop, then the value of the magnetic field and its direction at the centre $C$ of the loop shall be close to

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