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A loop, made of straight edges has six corners at $\mathrm{A}(0,0,0)$, $\mathrm{B}(\mathrm{L}, 0,0), \mathrm{C}(\mathrm{L}, \mathrm{L}, 0), \mathrm{D}(0, \mathrm{~L}, 0) \mathrm{E}(0, \mathrm{~L}, \mathrm{~L})$ and $\mathrm{F}(0,0, \mathrm{~L}) . \mathrm{A}$ magnetic field $\mathrm{B}=\mathrm{B}_0(\hat{\mathrm{i}}+\hat{\mathrm{k}}) \mathrm{T}$ is present in the region. The flux passing through the loop $\mathrm{ABCDEFA}$ (in that order) is
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The correct answer is:
$2 \mathrm{~B}_0 \mathrm{~L}^2 \mathrm{~Wb}$
$2 \mathrm{~B}_0 \mathrm{~L}^2 \mathrm{~Wb}$
The loop can be considered in two planes, Plane of ABCDAlies $x$-y planewhoseareavedor $A_1=|A| \hat{k}, A_1=L^2 \hat{k}$ whereas plane of ADEFA lies in y-z plane whose area vector $\mathrm{A}_2=|\mathrm{A}| \hat{\mathrm{i}}, \mathrm{A}_2=\mathrm{L}^2 \hat{\mathrm{i}}$.
Then the magnetic flux linked with uniform surface of area
A in uniform magnetic field is
$\begin{aligned} \phi=& B \cdot A \\ A &=A_1+A_2 \\ &=\left(L^2 \hat{k}+L^2 \hat{i}\right) \\ \text { and } B &=B_0(\hat{i}+\hat{k}) \end{aligned}$
$\begin{aligned} \text { Now, } \phi &=B \cdot A=B_0(\hat{i}+\hat{k}) \cdot\left(L^2 \hat{k}+L^2 \hat{i}\right) \\ &=2 B_0 L^2 W b \end{aligned}$
Then the magnetic flux linked with uniform surface of area
A in uniform magnetic field is
$\begin{aligned} \phi=& B \cdot A \\ A &=A_1+A_2 \\ &=\left(L^2 \hat{k}+L^2 \hat{i}\right) \\ \text { and } B &=B_0(\hat{i}+\hat{k}) \end{aligned}$
$\begin{aligned} \text { Now, } \phi &=B \cdot A=B_0(\hat{i}+\hat{k}) \cdot\left(L^2 \hat{k}+L^2 \hat{i}\right) \\ &=2 B_0 L^2 W b \end{aligned}$
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