A long straight wire carrying current 16 ,A i | vedclass

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(B) The wire is bent at the origin $O$. One segment lies along the positive $x$-axis and the other along the positive $y$-axis.Point $P$ is located at

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(B) The wire is bent at the origin $O$. One segment lies along the positive $x$-axis and the other along the positive $y$-axis.Point $P$ is located at $(-2 \,mm, 0)$, which lies on the negative $x$-axis.$1$. For the segment along the $x$-axis: The point $P$ lies on the line containing this segment. Therefore, the magnetic field produced by this segment at point $P$ is zero.$2$. For the segment along the $y$-axis: This is a semi-infinite wire starting from the origin $O$ and extending along the positive $y$-axis. The magnetic field $B$ at a perpendicular distance $r$ from a semi-infinite wire is given by $B = \frac{\mu_0 I}{4 \pi r}$.Here, $I = 16 \,A$ and $r = 2 \,mm = 2 	imes 10^{-3} \,m$.Substituting the values:$B = 10^{-7} 	imes \frac{16}{2 	imes 10^{-3}}$$B = 10^{-7} 	imes 8 	imes 10^3$$B = 8 	imes 10^{-4} \,T = 0.8 	imes 10^{-3} \,T = 0.8 \,mT$.

Similar Questions

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Two infinite wires carrying opposite electrical currents $I$ and $i$ are placed a distance $x$ apart. $A$ point $P$ at a distance $y$ away from the wire carrying current $i$ is shown in the figure. If the magnetic field is zero at point $P$, then the magnitude of $i$ is

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