A loop ABCDEFA of straight edges has six corner points A(0, 0, 0), B(5, 0, 0), C(5, 5, 0), D (0, 5, 0), E(0, 5, 5) and F(0, 0, 5). The magnetic field in this region is B⃗ = ( 3î + 4k̂ )T . The quantity of flux through the loop ABCDEFA (in Wb) is _______.

Solution

Physics sir · 4 steps

This loop is bent along AD, so split it into two flat squares and add the flux through each.

1. Split the loop. Add an imaginary edge AD. The loop becomes square ABCDA in the plane z = 0 and square ADEFA in the plane x = 0. Edge AD is crossed once as D to A and once as A to D, so it cancels and the two squares together are the original loop.
2. Area vectors. ABCDA runs counterclockwise seen from +z, so its area vector is 25 k̂ m². For ADEFA, going A to D is along +ĵ and D to E is along +k̂. Since ĵ × k̂ = î, its area vector is 25 î m².

   φ = B⃗·(A⃗₁ + A⃗₂) = (3î + 4k̂)·(25î + 25k̂)

3. Put in numbers. The î part gives 3 × 25 = 75 Wb through the vertical square. The k̂ part gives 4 × 25 = 100 Wb through the horizontal square.
4. So. Total flux = 75 + 100 = 175 Wb.

The common slip: 100. Counting only the square in the xy-plane gives 100. The vertical square ADEFA also carries flux from the 3î part of the field.

The idea to keep. For a loop that is not flat, break it into flat pieces that share an edge. Keep the same sense of travel in every piece, then add B⃗·A⃗ for each one.

Answer from the NTA final answer key. Solution by PrepWiser.