Two infinitely long wires are placed at (1 te | vedclass

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(A) The wires are located at $A(1, 1)$ and $B(1, -1)$. The distance of each wire from the origin $O(0, 0)$ is $r = \sqrt{1^2 + 1^2} = \sqrt{2} 	ext{

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$\sqrt{2}$

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(A) The wires are located at $A(1, 1)$ and $B(1, -1)$. The distance of each wire from the origin $O(0, 0)$ is $r = \sqrt{1^2 + 1^2} = \sqrt{2} 	ext{ cm}$.Since both wires carry current in the same direction (perpendicular to the $xy$-plane),the magnetic field vectors at the origin are perpendicular to the position vectors $OA$ and $OB$.The angle of $OA$ with the $x$-axis is $45^\circ$ and $OB$ is $-45^\circ$. The magnetic field $B_A$ due to wire $A$ is perpendicular to $OA$ at $135^\circ$ to the $x$-axis,and $B_B$ due to wire $B$ is perpendicular to $OB$ at $45^\circ$ to the $x$-axis.The angle between $B_A$ and $B_B$ is $90^\circ$.The magnitude of the field due to one wire is $B_0 = \frac{\mu_0 I}{2 \pi r}$.The net magnetic field magnitude is $|B| = \sqrt{B_0^2 + B_0^2 + 2 B_0 B_0 \cos(90^\circ)} = \sqrt{2 B_0^2} = \sqrt{2} B_0$.Therefore,$\frac{|B|}{B_0} = \sqrt{2}$.

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