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A square of side L metres lies in the xy-plane in a region, where the magnetic field is given by $B=B_0(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \mathrm{T}$, where $\mathrm{B}_0$ is constant. The magnitude of flux passing through the square is
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Verified Answer
The correct answer is:
$4 \mathrm{~B}_0 \mathrm{~L}^2 \mathrm{~Wb}$
$4 \mathrm{~B}_0 \mathrm{~L}^2 \mathrm{~Wb}$
As we know that, the magnetic flux linked with uniform surface of area $A$ in uniform magnetic field is $\phi=B$ B.A The direction of $A$ is perpendicular to the plane of square and square line in $x$-y plane in a region.
$$
\begin{array}{ll}
\mathrm{A}=\mathrm{L}^2 \mathrm{k} \\
\text { As given that, } & \mathrm{B}=\mathrm{B}_0(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \\
\text { So, } & \phi=\mathrm{B}^2 \cdot \mathrm{A}=\mathrm{B}_0(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot \mathrm{L}^2 \hat{\mathrm{k}} \\
& =4 \mathrm{~B}_0 \mathrm{~L}^2 \mathrm{~Wb}
\end{array}
$$
$$
\begin{array}{ll}
\mathrm{A}=\mathrm{L}^2 \mathrm{k} \\
\text { As given that, } & \mathrm{B}=\mathrm{B}_0(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \\
\text { So, } & \phi=\mathrm{B}^2 \cdot \mathrm{A}=\mathrm{B}_0(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot \mathrm{L}^2 \hat{\mathrm{k}} \\
& =4 \mathrm{~B}_0 \mathrm{~L}^2 \mathrm{~Wb}
\end{array}
$$
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