A coil of 50, turns and 4,cm radius carries 2 | vedclass

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Question

$\overrightarrow{\mathrm{B}}=\mathrm{N}\left[\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}ight]=\frac{50 	imes 4 \pi 	imes 10^{-7} 	imes 2}{2 	imes 4

Vedclass · Accepted Answer

$1.57$

Vedclass · Answer

$\overrightarrow{\mathrm{B}}=\mathrm{N}\left[\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}ight]=\frac{50 	imes 4 \pi 	imes 10^{-7} 	imes 2}{2 	imes 4 	imes 10^{-2}}=1.57 	imes 10^{-3} \mathrm{T}$

A coil of $50\, turns$ and $4\,cm$ radius carries $2\,A$ current then magnetic field at its centre is......$mT$

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