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Question

(A) The magnetic field produced by a long straight wire at a distance $X$ is given by $B = \frac{\mu_0 I}{2 \pi X}$.Let $X$ be the distance from point

Vedclass · Accepted Answer

$68 	imes 10^{-6} \,T$

Vedclass · Answer

(A) The magnetic field produced by a long straight wire at a distance $X$ is given by $B = \frac{\mu_0 I}{2 \pi X}$.Let $X$ be the distance from point $P$ to each wire. Since the lines joining $P$ to the wires are perpendicular to each other,the triangle formed by the two wires and point $P$ is a right-angled isosceles triangle with hypotenuse $7 \,cm$.Using the Pythagorean theorem: $X^2 + X^2 = 7^2 \implies 2X^2 = 49 \implies X = \frac{7}{\sqrt{2}} \,cm = \frac{7}{1.4} 	imes 10^{-2} \,m = 5 	imes 10^{-2} \,m$.The magnetic fields due to the two wires are $B_1 = \frac{\mu_0 I_1}{2 \pi X}$ and $B_2 = \frac{\mu_0 I_2}{2 \pi X}$.Since the directions of the currents are opposite,the magnetic field vectors at $P$ are perpendicular to each other. Thus,the net magnetic field is $B_{	ext{net}} = \sqrt{B_1^2 + B_2^2} = \frac{\mu_0}{2 \pi X} \sqrt{I_1^2 + I_2^2}$.Substituting the values: $B_{	ext{net}} = \frac{4 \pi 	imes 10^{-7}}{2 \pi 	imes 5 	imes 10^{-2}} \sqrt{15^2 + 8^2} = \frac{2 	imes 10^{-7}}{5 	imes 10^{-2}} \sqrt{225 + 64} = \frac{2 	imes 10^{-5}}{5} \sqrt{289} = 0.4 	imes 10^{-5} 	imes 17 = 6.8 	imes 10^{-5} \,T = 68 	imes 10^{-6} \,T$.

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