A square loop of side 2 m lies in the Y-Z pl | vedclass

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(A) The area vector $\vec{A}$ of a square loop of side $2 \ m$ lying in the $Y-Z$ plane is directed along the $X$-axis.$\vec{A} = (2 	imes 2) \hat{i}

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The area vector $\vec{A}$ of a square loop of side $2 \ m$ lying in the $Y-Z$ plane is directed along the $X$-axis.$\vec{A} = (2 	imes 2) \hat{i} = 4 \hat{i} \ m^2$.The magnetic field is given by $\vec{B} = (5 \hat{i} + 3 \hat{j} - 4 \hat{k}) \ T$.The magnetic flux $\phi$ is given by the dot product of the magnetic field and the area vector:$\phi = \vec{B} \cdot \vec{A}$$\phi = (5 \hat{i} + 3 \hat{j} - 4 \hat{k}) \cdot (4 \hat{i})$$\phi = (5 	imes 4) + (3 	imes 0) + (-4 	imes 0)$$\phi = 20 \ Wb$.Thus,the magnitude of the magnetic flux is $20 \ Wb$.

$A$ square loop of side $2 \ m$ lies in the $Y-Z$ plane in a region having a magnetic field $\vec{B}=(5 \hat{i}+3 \hat{j}-4 \hat{k}) \ T$. The magnitude of magnetic flux through the square loop is (in $Wb$)

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