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Question:

Two identical long conducting wires AOB and COD are placed at right angles to each other, with one above the other such that 'O' is their common point. The wires carry I₁ and I₂ currents, respectively. Point 'P' is lying at distance 'd' from 'O' along a direction perpendicular to the plane containing the wires. The magnetic field at the point 'P' will be:

μ₀/(2πd)(I₁+I₂)

μ₀/(2πd)(I₂² - I₂²)

μ₀/(2πd)(I₂²+I₂²)¹/²

μ₀/(2πd)(I₁I₁)

Solution:

→B due to wire (1) →B₁ = (μ₀I₂/(2πd))^j
→B due to wire (2) →B₂ = (μ₀I₁/(2πd))(-^i)
|Bnet| = μ₀/(2πd)√(I₁² + I₂²)