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Asked in GSEB Board March 2023 · Flux through a surface
Given: a square of side L in the xy-plane, so its area vector is ⃗A = L²̂k; ⃗B = B₀(2̂i + 4̂j + 3̂k) T.
Idea: flux is Φ = ⃗B · ⃗A; only the component of ⃗B along the normal to the surface counts.
Φ = B₀(2̂i + 4̂j + 3̂k) · L²̂k.
Since ̂i · ̂k = ̂j · ̂k = 0 and ̂k · ̂k = 1: Φ = 3B₀L².
So the flux through the square is 3B₀L² Wb.
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