# A square of side L meters lies in the x-y plane in a region, where the magnetic field is given by $B = B_o (2i + 3j + 4k)T$ where Bo is constant. The magnitude of flux passing through the square is$(a) 2 B_o L^2 Wb.$ $(b) 3 B_o L^2 Wb.$ $(c) 4 B_o L^2 Wb.$ $(d) \sqrt{}2 9B_o L^2 Wb.$

Answers (2)

Answer: The answer is the option (c)

For elementary area dA of a surface, flux linked $d\phi=BdA\cos \theta\:or\: d\phi=\vec{B}.\vec{dA}$

So-net flux through the surface

$\phi = \oint \vec{B}\times \vec{dA}=BA \cos \theta$

In this problem
$A= L^2\hat{k}\:and\:B=B_0\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right )T\\ \phi = \vec{B}.\vec{A}=B_0\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right ).L^2\hat{k}= 4 B_o L^2 Wb.$

Answer: The answer is the option (c)

For elementary area dA of a surface, flux linked $d\phi=BdA\cos \theta\:or\: d\phi=\vec{B}.\vec{dA}$

So-net flux through the surface

$\phi = \oint \vec{B}\times \vec{dA}=BA \cos \theta$

In this problem

$A= L^2\hat{k}\:and\:B=B_0\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right )T\\ \phi = \vec{B}.\vec{A}=B_0\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right ).L^2\hat{k}= 4 B_o L^2 Wb.$

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