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

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(A) The magnetic flux $\Phi$ through a surface is given by the dot product of the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$.Given,

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(A) The magnetic flux $\Phi$ through a surface is given by the dot product of the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$.Given,$\vec{B} = B_0(2 \hat{i} + 3 \hat{j} + 4 \hat{k})$.The square lies in the $x-y$ plane,so its area vector $\vec{A}$ is perpendicular to the $x-y$ plane,which is along the $z$-axis.Thus,$\vec{A} = L^2 \hat{k}$.The magnetic flux $\Phi$ is calculated as:$\Phi = \vec{B} \cdot \vec{A}$$\Phi = [B_0(2 \hat{i} + 3 \hat{j} + 4 \hat{k})] \cdot [L^2 \hat{k}]$$\Phi = B_0 L^2 (2 \hat{i} \cdot \hat{k} + 3 \hat{j} \cdot \hat{k} + 4 \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:$\Phi = B_0 L^2 (0 + 0 + 4(1)) = 4 B_0 L^2$.Therefore,the magnitude of the flux is $4 B_0 L^2$.

$A$ square of side $L$ metre lies in the $x-y$ plane in a region where the magnetic field is $\vec{B} = B_0(2 \hat{i} + 3 \hat{j} + 4 \hat{k})$,where $B_0$ is a constant. The magnitude of the magnetic flux passing through the square (in weber) is: (in $B_0 L^2$)

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