Question

A loop $ABCDEFA$ of straight edges has six corner points$A(0,0,0),B(5,0,0),C(5,5,0),D(0,5,0),E(0,5,5),F(0,0,5)$. The magnetic field in this region is$B=(3\hat{i}+4\hat{k})T$. The quantity of flux through the loop$ABCDEFA$(in $Wb$) is

Answer 1

Step 1. Given data

Value of magnetic field, $B=(3\hat{i}+4\hat{k})T$

Step 2. Find the area vector, $\overrightarrow{A}$

${\overrightarrow{A}}_{ABCD}=\overrightarrow{AB}\times \overrightarrow{BC}$

${\overrightarrow{A}}_{ABCD}=5\hat{i}\times 5\hat{j}$ [$\hat{i}\times \hat{j}=\hat{k}$]

${\overrightarrow{A}}_{ABCD}=25\hat{k}$

${\overrightarrow{A}}_{ADEF}=\overrightarrow{AD}\times \overrightarrow{DE}$

${\overrightarrow{A}}_{ADEF}=5\hat{j}\times 5\hat{k}$ [$\hat{j}\times \hat{k}=\hat{i}$]

${\overrightarrow{A}}_{ADEF}=25\hat{i}$

${\overrightarrow{A}}_{net}=(25\hat{k}+25\hat{i})m^2$

Step 3: Calculating flux

Now, flux, $\varphi$

$\varphi =B.A$

$$\begin{gathered} =(3\hat{i}+4\hat{k})T.\left(25\hat{k}+25\hat{i}\right)m^2 \\ =25\times 3+25\times 4 \\ =175Wb \end{gathered}$$

Hence, the quantity of flux through the loop$ABCDEFA$ is $175Wb$.