Two long parallel wires carrying currents I_1 | vedclass

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(D) Let the wires be along the $z$-axis. Since the lines from $P$ to the wires are perpendicular and $P$ is equidistant from both,the distance $r$ fro

Vedclass · Accepted Answer

$2 \sqrt{2} 	imes 10^{-5} \ T$

Vedclass · Answer

(D) Let the wires be along the $z$-axis. Since the lines from $P$ to the wires are perpendicular and $P$ is equidistant from both,the distance $r$ from each wire to $P$ is $r = d / \sqrt{2} = 5 / \sqrt{2} \ cm = 5 / (\sqrt{2} 	imes 100) \ m = 0.05 / \sqrt{2} \ m$.The magnetic field due to a long wire is $B = \mu_0 I / (2 \pi r)$.$B_1 = (4 \pi 	imes 10^{-7} 	imes 4) / (2 \pi 	imes (0.05 / \sqrt{2})) = (8 	imes 10^{-7} 	imes \sqrt{2}) / 0.05 = 160 \sqrt{2} 	imes 10^{-7} \ T = 1.6 \sqrt{2} 	imes 10^{-5} \ T$.$B_2 = (4 \pi 	imes 10^{-7} 	imes 3) / (2 \pi 	imes (0.05 / \sqrt{2})) = (6 	imes 10^{-7} 	imes \sqrt{2}) / 0.05 = 120 \sqrt{2} 	imes 10^{-7} \ T = 1.2 \sqrt{2} 	imes 10^{-5} \ T$.Since the currents are in opposite directions and the lines to $P$ are perpendicular,the magnetic field vectors $B_1$ and $B_2$ are perpendicular to each other.The resultant magnetic field is $B = \sqrt{B_1^2 + B_2^2} = \sqrt{(1.6 \sqrt{2} 	imes 10^{-5})^2 + (1.2 \sqrt{2} 	imes 10^{-5})^2}$.$B = 10^{-5} \sqrt{2} \sqrt{1.6^2 + 1.2^2} = 10^{-5} \sqrt{2} \sqrt{2.56 + 1.44} = 10^{-5} \sqrt{2} \sqrt{4} = 2 \sqrt{2} 	imes 10^{-5} \ T$.

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