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 $$\vec{B} = (3\hat{i} + 4\hat{k})$$ T. The quantity of flux through the loop ABCDEFA (in Wb) is

We are asked to find the magnetic flux through the wire loop that follows the edges A → B → C → D → E → F → A, where the corner coordinates are

$$$A(0,0,0),\, B(5,0,0),\, C(5,5,0),\, D(0,5,0)$$$, $$E(0,5,5),\, F(0,0,5).$$

The magnetic field in the region is uniform and given by

$$\vec B = 3\hat i + 4\hat k\;\text{ T}.$$

The loop is not planar, so instead of splitting the surface into several rectangles, it is quicker to use a vector formula that gives the net area vector directly from the boundary curve. For any closed contour $$\mathcal C$$, an oriented surface bounded by that contour has area vector

$$\vec A \;=\;\frac12\oint_{\mathcal C}\vec r\times d\vec r,$$

where $$\vec r$$ is the position vector of a point on the contour and $$d\vec r$$ is the differential displacement along the contour. This follows from Stokes’ theorem and is valid no matter how the surface is bent; the result depends only on the boundary.

We therefore evaluate the line integral $$\displaystyle\oint\vec r\times d\vec r$$ one edge at a time, always moving in the given order A → B → C → D → E → F → A.

Edge AB: $$$\vec r=\langle x,0,0\rangle,\;0\le x\le5,\;d\vec r=\langle dx,0,0\rangle.$$$ So $$\vec r\times d\vec r=\vec0,$$ and the integral along AB is $$\vec0.$$

Edge BC: $$$\vec r=\langle5,y,0\rangle,\;0\le y\le5,\;d\vec r=\langle0,dy,0\rangle.$$$ Then $$$\vec r\times d\vec r=\begin{vmatrix}\hat i&\hat j&\hat k\\5&y&0\\0&dy&0\end{vmatrix}=5\,dy\,\hat k.$$$ Integrating, $$\int_{0}^{5}5\,dy\,\hat k=25\hat k.$$

Edge CD: $$$\vec r=\langle x,5,0\rangle,\;5\ge x\ge0,\;d\vec r=\langle dx,0,0\rangle.$$$ Here $$$\vec r\times d\vec r=\begin{vmatrix}\hat i&\hat j&\hat k\\x&5&0\\dx&0&0\end{vmatrix}=-5\,dx\,\hat k.$$$ Integrating from 5 down to 0, $$\int_{5}^{0}(-5)\,dx\,\hat k=25\hat k.$$

Edge DE: $$$\vec r=\langle0,5,z\rangle,\;0\le z\le5,\;d\vec r=\langle0,0,dz\rangle.$$$ Thus $$$\vec r\times d\vec r=\begin{vmatrix}\hat i&\hat j&\hat k\\0&5&z\\0&0&dz\end{vmatrix}=5\,dz\,\hat i.$$$ Integrating, $$\int_{0}^{5}5\,dz\,\hat i=25\hat i.$$

Edge EF: $$$\vec r=\langle0,y,5\rangle,\;5\ge y\ge0,\;d\vec r=\langle0,dy,0\rangle.$$$ Then $$$\vec r\times d\vec r=\begin{vmatrix}\hat i&\hat j&\hat k\\0&y&5\\0&dy&0\end{vmatrix}=-5\,dy\,\hat i.$$$ Integrating from 5 down to 0, $$\int_{5}^{0}(-5)\,dy\,\hat i=25\hat i.$$

Edge FA: $$$\vec r=\langle0,0,z\rangle,\;5\ge z\ge0,\;d\vec r=\langle0,0,dz\rangle.$$$ The cross‐product is zero, so the contribution is $$\vec0.$$

Adding all six contributions we get

$$$\oint\vec r\times d\vec r=(25+25)\hat i+(25+25)\hat k= \langle50,\,0,\,50\rangle.$$$

Dividing by 2, the area vector bounded by the loop is

$$$\vec A=\frac12\langle50,\,0,\,50\rangle=\langle25,\,0,\,25\rangle\text{ m}^2.$$$

Magnetic flux is defined as

$$\Phi=\vec B\cdot\vec A.$$

Substituting $$\vec B=\langle3,\,0,\,4\rangle\text{ T}$$ and $$\vec A=\langle25,\,0,\,25\rangle\text{ m}^2,$$ we have

$$$\Phi=(3)(25)+(0)(0)+(4)(25)=75+100=175\;\text{Wb}.$$$

So, the answer is $$175$$.