Animated Solution for Physics - Electromagnetic Induction: 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 $\mathbf{B}=(3\hat{i}+4\hat{k})\text{ T}$. The quantity of flux through the loop ABCDEFA (in Wb) is ………… .

The Power of Superposition

When you first look at a 3D loop like ABCDEFA, it can feel a bit overwhelming. Calculating magnetic flux through a surface that bends through multiple planes isn't as straightforward as a simple flat circle. But here is where the beauty of physics comes in: we can use the principle of superposition to break this complex problem into incredibly simple, bite-sized pieces.

Let's analyze the coordinates given: $A(0,0,0)$, $B(5,0,0)$, $C(5,5,0)$, $D(0,5,0)$, $E(0,5,5)$, and $F(0,0,5)$. If you plot these points, you'll notice something magical. The loop can be perfectly split into two distinct 2D squares that share a common edge, the line segment AD.

The first part is the square ABCDA, which lies entirely flat in the xy-plane. The second part is the square ADEFA, which stands vertically in the yz-plane. By imagining a line from A to D, we've transformed one confusing 3D surface into two simple 2D surfaces!

To calculate flux, we need the area vector $\mathbf{A}$ for our surfaces. Let's tackle them one by one.

For the square ABCDA, the side length is 5 units, making its area $5\times 5=25$. To find the direction, we use the right-hand rule. Following the path A $\to$ B $\to$ C $\to$ D, your fingers curl from the x-axis to the y-axis, making your thumb point straight up along the positive z-axis.

Now for the square ADEFA. Its area is also 25. Following the path A $\to$ D $\to$ E $\to$ F, your fingers curl from the y-axis to the z-axis, so your thumb points along the positive x-axis.

The total area vector for our entire 3D loop is simply the vector sum of these two individual area vectors:

With our total area vector ready, we can bring in the master equation for magnetic flux. The flux $\Phi$ is the dot product of the uniform magnetic field vector $\mathbf{B}$ and the area vector $\mathbf{A}$.

We are given the magnetic field $\mathbf{B}=(3\hat{i}+4\hat{k})\text{ T}$. Let's substitute our vectors into the equation:

Evaluating the dot product is straightforward. We multiply the $\hat{i}$ components together and the $\hat{k}$ components together:

And there we have it! By breaking down a 3D geometry into 2D planes, a seemingly complex flux problem turns into a simple vector addition and a dot product. 175 Wb is our final answer.