Q.Consider a closed loop $C$ in a magnetic field. The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula $\phi = \mathbf{B}_1 \cdot d\mathbf{A}_1 + \mathbf{B}_2 \cdot d\mathbf{A}_2 + \ldots$. Now if we choose two different surfaces $S_1$ and $S_2$ having $C$ as their edge, would we get the same answer for flux? Justify your answer.

Concept: Magnetic Flux and Solenoidal Nature of $\mathbf{B}$

The magnetic field is solenoidal: $\nabla \cdot \mathbf{B} = 0$. This means magnetic field lines have no sources or sinks — they always form closed loops.

Reasoning:

1. For two surfaces $S_1$ and $S_2$ sharing the same boundary $C$, their union forms a closed surface $S = S_1 \cup S_2$ (with opposite orientation on $C$).
2. By the divergence theorem, the net flux through any closed surface is zero: $\oint_{S} \mathbf{B} \cdot d\mathbf{A} = \int_V (\nabla \cdot \mathbf{B})\, dV = 0$ …

The question touches on a subtle but beautiful point about magnetic fields: whether the flux through a loop depends on which surface you stretch across it. Intuitively, you might worry that different surfaces could give different answers — after all, the magnetic field varies from point to point, and two surfaces might cut through different field configurations. But the answer is no: the flux is the same for any surface with the same boundary. The reason lies in a fundamental property of magnetism: magnetic field lines have no sources or sinks — they always form closed loops. Mathematically, this is expressed as $\nabla \cdot \mathbf{B} = 0$, Gauss's law for magnetism.

Let’s see why this guarantees a unique flux.

1. Set up the two surfaces.

   Take the closed loop $C$ and two different surfaces $S_1$ and $S_2$, both having $C$ as their boundary. The flux through $S_1$ is $\Phi_1 = \int_{S_1} \mathbf{B} \cdot d\mathbf{A}$, and through $S_2$ is $\Phi_2 = \int_{S_2} \mathbf{B} \cdot d\mathbf{A}$. We want to check if $\Phi_1 = \Phi_2$.

2. Combine them into a closed surface.

   Consider the surface formed by joining $S_1$ and $S_2$ along their common boundary $C$. This creates a single closed surface $S = S_1 \cup S_2$ (with the orientation of $S_2$ reversed so that the outward normals are consistent). The total flux through this closed surface is:

The minus sign appears because the outward normal on $S_2$ points opposite to the orientation we originally used for flux through $S_2$ alone.

3. Apply Gauss’s law for magnetism. For any closed surface, the net magnetic flux is zero:

since $\nabla \cdot \mathbf{B} = 0$ everywhere. Therefore:

Method: Proving a Flux Value Is Independent of the Chosen Surface

Use this method for "does the flux depend on the surface" style conceptual questions involving $\vec{B}$ (never for $\vec{E}$, where enclosed charge WOULD make it surface-dependent unless no charge lies between the two surfaces).

Steps

Step 1: Recall the defining property of $\vec{B}$

Magnetic field lines never begin or end — there is no magnetic monopole. This is expressed as

Step 2: Build a closed surface from the two open surfaces

If $S_1$ and $S_2$ both share the same boundary loop $C$, then $S = S_1 \cup S_2$ (with $S_2$'s normal reversed so the whole thing has one consistent outward orientation) is a single CLOSED surface.

Step 3: Apply Gauss's law of magnetism to that closed surface

$\oint_S \vec{B}\cdot d\vec{A} = \Phi_1 - \Phi_2 = 0$ …