Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current I₁ and COD carries a current I₂. The magnetic field on a point lying at a distance d from O, in a direction perpendicular to the plane of the wires AOB and COD, will be given by (a) (μ₀)/(2π d)(I₁² + I₂²) (b) (μ₀)/(2π)((I₁+I₂)/(d))1/2 (c) (μ₀)/(2π d)(I₁² + I₂²)1/2 (d) (μ₀)/(2π d)(I₁ + I₂)

Answer: (c). The point is at distance d from each long wire, above O. B₁ = (μ₀ I₁)/(2π d) and B₂ = (μ₀ I₂)/(2π d) are perpendicular, since the wires are perpendicular. B = (μ₀)/(2π d)(I₁² + I₂²)1/2 So the answer is (c).

Two identical conducting wires AOB and COD are placed at right angles… | JEE Main 2007 (AIEEE)

Q69JEE Main 2007 · AIEEEMedium

Magnetic Field of Straight Wires

Two identical conducting wires $AOB$ and $COD$ are placed at right angles to each other. The wire $AOB$ carries an electric current $I_1$ and $COD$ carries a current $I_2$ . The magnetic field on a point lying at a distance $d$ from $O$ , in a direction perpendicular to the plane of the wires $AOB$ and $COD$ , will be given by

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Answer: (c)

The point is at distance $d$ from each long wire, above O.
$B_1 = \frac{\mu_0 I_1}{2\pi d}$ and $B_2 = \frac{\mu_0 I_2}{2\pi d}$ are perpendicular, since the wires are perpendicular.
$B = \frac{\mu_0}{2\pi d}\left(I_1^2 + I_2^2\right)^{1/2}$
So the answer is (c).

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