Two long,straight wires carry equal currents | vedclass

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(D) The magnetic field due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 i}{2\pi r}$.Here,the current $i = 10 \ A$ and the di

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$160$

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(D) The magnetic field due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 i}{2\pi r}$.Here,the current $i = 10 \ A$ and the distance of point $P$ from each wire is $r = \frac{5.0 \ cm}{2} = 2.5 \ cm = 2.5 	imes 10^{-2} \ m$.Since the currents are in opposite directions,the magnetic fields produced by both wires at point $P$ point in the same direction (using the right-hand thumb rule).Therefore,the total magnetic field $B_{net} = B_1 + B_2 = 2 	imes \left(\frac{\mu_0 i}{2\pi r}ight) = \frac{\mu_0 i}{\pi r}$.Substituting the values: $B_{net} = \frac{4\pi 	imes 10^{-7} 	imes 10}{\pi 	imes 2.5 	imes 10^{-2}} = \frac{40 	imes 10^{-7}}{2.5 	imes 10^{-2}} = 16 	imes 10^{-5} \ T$.Converting to $\mu T$: $16 	imes 10^{-5} \ T = 160 	imes 10^{-6} \ T = 160 \ \mu T$.

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