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(A) The magnetic field due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.For wire $X$ $(I_X = 5 	ext{ A})$,the d

Vedclass · Accepted Answer

$3$

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(A) The magnetic field due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.For wire $X$ $(I_X = 5 	ext{ A})$,the distance to point $P$ is $r_X = 6 	ext{ cm} + 4 	ext{ cm} = 10 	ext{ cm} = 0.1 	ext{ m}$. Using the right-hand rule,the magnetic field $B_X$ at $P$ is directed into the page.$B_X = \frac{\mu_0 	imes 5}{2\pi 	imes 0.1} = \frac{2 	imes 10^{-7} 	imes 5}{0.1} = 100 	imes 10^{-7} 	ext{ T} = 10^{-5} 	ext{ T}$.For wire $Y$ $(I_Y = 4 	ext{ A})$,the distance to point $P$ is $r_Y = 4 	ext{ cm} = 0.04 	ext{ m}$. Using the right-hand rule,the magnetic field $B_Y$ at $P$ is directed into the page.$B_Y = \frac{\mu_0 	imes 4}{2\pi 	imes 0.04} = \frac{2 	imes 10^{-7} 	imes 4}{0.04} = 200 	imes 10^{-7} 	ext{ T} = 2 	imes 10^{-5} 	ext{ T}$.Since both fields are in the same direction (into the page),the resultant magnetic field is $B_{net} = B_X + B_Y = 1 	imes 10^{-5} + 2 	imes 10^{-5} = 3 	imes 10^{-5} 	ext{ T}$.Comparing this with $x 	imes 10^{-5} 	ext{ T}$,we get $x = 3$.

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