In the figure, find out the magnetic field at | vedclass

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Question

The magnetic field at $B$ is given by,

$\vec{B}_{ B }=\vec{B}_{	ext {wire }}+\vec{B}_{	ext {ring }}$

$=\left[\frac{\mu_{0} i}{2 r}(-\hat{k})+\

Vedclass · Accepted Answer

$\pi 	imes\left[1+\frac{1}{\pi}ight] 	imes 10^{-5} \;T$

Vedclass · Answer

The magnetic field at $B$ is given by,

$\vec{B}_{ B }=\vec{B}_{	ext {wire }}+\vec{B}_{	ext {ring }}$

$=\left[\frac{\mu_{0} i}{2 r}(-\hat{k})+\frac{\mu_{0} i}{4 \pi r}(-\hat{k})ight]+\left[\frac{\mu_{0} i}{2 \pi r}(-\hat{k})+\frac{\mu_{0} i}{2 r}(-\hat{k})ight]$

$=\frac{\mu_{0} i}{2 \pi r}(-\hat{k})+\left[\frac{\mu_{0} i}{2 \pi r}+\frac{\mu_{0} i}{2 r}ight](-\hat{k})$

$=\frac{\mu_{0} i}{2 r}\left[\frac{1}{\pi}+1ight]$      $....(I)$

Substitute $4 \pi 	imes 10^{-7}$ for $\mu_{0}, 2.5$ for $i$ and $5 	imes 10^{-2}$ for $r$ in equation $(I).$

$\vec{B}_{ B }=\frac{\left(4 \pi 	imes 10^{-7}ight) 2.5}{2\left(5 	imes 10^{-2}ight)}\left[\frac{1}{\pi}+1ight]$

$=10 \pi 	imes 10^{-6}\left[\frac{1}{\pi}+1ight]$

$=\pi 	imes\left[\frac{1}{\pi}+1ight] 	imes 10^{-5} T$

In the figure, find out the magnetic field at $B$ (Given $I =2.5 \;A,r =5\, cm )$

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