Q.A loop, made of straight edges, has six corners at $A(0,0,0)$, $B(L,0,0)$, $C(L,L,0)$, $D(0,L,0)$, $E(0,L,L)$ and $F(0,0,L)$. A magnetic field $\mathbf{B} = B_0(\hat{i} + \hat{k})\ \text{T}$ is present in the region. The flux passing through the loop $ABCDEFA$ (in that order) is

Concept: Magnetic Flux — the surface integral of $\mathbf{B}$ over the loop’s area. Since $\mathbf{B}$ is uniform, flux is $\Phi = \mathbf{B} \cdot \mathbf{A}$, where $\mathbf{A}$ is the area vector of the loop.

The loop $ABCDEFA$ is a 3D shape: a square in the $xy$-plane ($ABCD$) plus a square in the $yz$-plane ($DEFA$). The total area vector is the sum of the area vectors of these two planar faces, each oriented by the right-hand rule along the given path order.

Step 1: For face $ABCD$ (in $z=0$ plane), the path goes $A \to B \to C \to D$. Using the right-hand rule, the area vector points along $+\hat{k}$. Area $= L^2$, so $\mathbf{A}_1 = L^2 \hat{k}$.

Step 2: For face $DEFA$ (in $x=0$ plane), the path goes $D \to E \to F \to A$. The right-hand rule gives area vector along $+\hat{i}$. Area $= L^2$, so $\mathbf{A}_2 = L^2 \hat{i}$.

Step 3: Total area vector $\mathbf{A} = \mathbf{A}_1 + \mathbf{A}_2 = L^2(\hat{i} + \hat{k})$.

Flux $\Phi = \mathbf{B} \cdot \mathbf{A} = B_0(\hat{i} + \hat{k}) \cdot L^2(\hat{i} + \hat{k}) = B_0 L^2 (1 + 1) = 2 B_0 L^2$.

The path $A\to B\to C\to D\to E\to F\to A$ is non-planar, so choose a convenient open surface bounded by it: two adjoining faces of the cube of side $L$.

Face 1 — $ABCD$ in the plane $z=0$. Traversed $A\to B\to C\to D$ (counterclockwise seen from $+z$), its area vector is

Face 2 — $ADEF$ in the plane $x=0$. Along the loop this face is traversed $D\to E\to F\to A$. Using two consecutive edges $\overrightarrow{DE}=L\hat k$ and $\overrightarrow{EF}=-L\hat j$:

The shared edge $AD$ is interior to this surface, so it is not part of the boundary.

Total area vector and flux. For a uniform field the flux is $\vec B\cdot\vec A$ summed over the planar pieces:

The $\hat k$-component of $\vec B$ threads face 1 and the $\hat i$-component threads face 2, each contributing $B_0L^2$.

Method: Flux Through a Non-Planar (3D) Loop — Split Into Flat Faces

When a closed loop's corners don't all lie in one plane, you can't write down a single area vector for it directly. This method shows how to still compute the flux linked by such a loop, using the fact that flux depends only on the loop's boundary, not on which surface you choose to span it with.

Steps

Step 1: Recognise that any surface bounded by the loop gives the same flux

For a uniform field, the flux linked by a closed loop equals $\vec B \cdot \vec A_{\text{total}}$ for any surface bounded by that loop — so choose the most convenient one. The easiest choice is almost always the set of flat, coordinate-plane faces that the loop's straight edges naturally trace out (here, two adjoining faces of a cube).

Step 2: Break the loop into its flat, planar segments

Trace the loop corner by corner and group consecutive edges into flat faces — e.g. four corners that all share the same coordinate value (all $z=0$, or all $x=0$) form one flat rectangular/square face each.

Step 3: Find each face's area vector, respecting the loop's own traversal direction

For each flat face, use the right-hand rule on the loop's stated direction of travel around that face (e.g. $A\to B\to C\to D$) to fix which way its area vector points — this consistency matters because a face traversed the "wrong way" flips the sign of its contribution.

using two consecutive edge vectors of that face (or simply "area $\times$ outward normal" if the orientation is obvious by inspection).

Step 4: Add the face area vectors, then dot with $\vec B$

Any edge shared between two chosen faces (interior to your surface, not part of the outer loop boundary) doesn't need separate treatment — it's already accounted for once each face is added.

This "span the loop with convenient flat pieces, then add area vectors" trick works for any polygonal 3D loop, not just cube edges — the loop's own corner coordinates always tell you how to choose the flat faces.