A square of side L meter lies in the x-y plan | vedclass

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(B) The magnetic flux $\phi$ through a surface is given by the dot product of the magnetic field vector $\overrightarrow{B}$ and the area vector $\ove

Vedclass · Accepted Answer

$3 B_0 L^2 	ext{ Wb}$

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(B) The magnetic flux $\phi$ through a surface is given by the dot product of the magnetic field vector $\overrightarrow{B}$ and the area vector $\overrightarrow{A}$.Since the square of side $L$ lies in the $x-y$ plane,its area vector $\overrightarrow{A}$ is directed along the $z$-axis.Therefore,$\overrightarrow{A} = L^2 \hat{k}$.The magnetic field is $\overrightarrow{B} = B_0(2 \hat{i} + 4 \hat{j} + 3 \hat{k})$.The magnetic flux is calculated as:$\phi = \overrightarrow{B} \cdot \overrightarrow{A}$$\phi = [B_0(2 \hat{i} + 4 \hat{j} + 3 \hat{k})] \cdot [L^2 \hat{k}]$$\phi = B_0 L^2 (2 \hat{i} \cdot \hat{k} + 4 \hat{j} \cdot \hat{k} + 3 \hat{k} \cdot \hat{k})$Since $\hat{i} \cdot \hat{k} = 0$,$\hat{j} \cdot \hat{k} = 0$,and $\hat{k} \cdot \hat{k} = 1$,we get:$\phi = B_0 L^2 (0 + 0 + 3(1)) = 3 B_0 L^2 	ext{ Wb}$.

$A$ square of side $L$ meter lies in the $x-y$ plane in a region where the magnetic field is given by $\overrightarrow{B} = B_0(2 \hat{i} + 4 \hat{j} + 3 \hat{k}) \text{ T}$,where $B_0$ is a constant. The magnitude of the magnetic flux passing through the square is . . . . . . .

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