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The magnetic flux through the triangular loop shown in the figure below:
where a uniform magnetic field of strength $2 \mathrm{~T}$ points perpendicularly into the plane of the triangle.
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where a uniform magnetic field of strength $2 \mathrm{~T}$ points perpendicularly into the plane of the triangle.
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Verified Answer
The correct answer is:
$2 \times 10^{-4} \mathrm{~Wb}$
$\begin{aligned} & \phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}} \\ & =\mathrm{BA} \cos 0^{\circ} \\ & =2 \times \frac{1}{2} \times 2 \times 1 \times 10^{-4} \\ & =2 \times 10^{-4} \mathrm{~Wb}\end{aligned}$
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