A long thin walled pipe of radius R carries a | vedclass

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Question

$d I=\frac{I}{2 \pi} d t h \eta$

Field at center due to very long wire carrying current $dI$

$d B=\frac{\mu_{0}}{2 \pi}=\frac{I}{R}=\frac{\mu_{0

Vedclass · Accepted Answer

$\frac{{{\mu _0}I\sqrt 2 }}{{4{\pi ^2}R}}$

Vedclass · Answer

$d I=\frac{I}{2 \pi} d t h \eta$

Field at center due to very long wire carrying current $dI$

$d B=\frac{\mu_{0}}{2 \pi}=\frac{I}{R}=\frac{\mu_{0}}{4 \pi^{2}} \frac{I}{R} d t h \eta$

$\vec{B}=\int d \vec{B}=\left[\int_{0}^{\pi / 2} d B \cos 	heta \hat{i}+\int_{0}^{\pi / 20 i} d B \sin 	heta \hat{i}ight]$

$=\frac{\mu_{0}}{4 \pi^{2} R} \frac{I}{R}\left[\int_{0}^{\pi / 2} \cos 0 d t h \eta \hat{i}+\int_{0}^{\pi / 20 i} \sin 0 d 	heta \hat{j}ight]$

$\vec{B}=\frac{\mu_{0}}{4 \pi^{2} R} \frac{I}{R}[\hat{i}+\hat{j}] \Rightarrow B=\frac{\mu_{0}}{4 \pi^{2} R} \frac{I}{R} \sqrt{2}$

A long thin walled pipe of radius $R$ carries a current $I$ along its length. The current density is uniform over the circumference of the pipe. The magnetic field at the center of the pipe due to quarter portion of the pipe shown, is

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