A square loop of side 1,m and resistance 1,Om | vedclass

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(A) Given:Side length of the square loop,$l = 1\,m$.Area of the loop,$A = l^2 = (1\,m)^2 = 1\,m^2$.Magnetic field,$B = 0.5\,T$.The plane of the loop i

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$0.5$

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(A) Given:Side length of the square loop,$l = 1\,m$.Area of the loop,$A = l^2 = (1\,m)^2 = 1\,m^2$.Magnetic field,$B = 0.5\,T$.The plane of the loop is perpendicular to the magnetic field,which means the area vector $\overrightarrow{A}$ is parallel to the magnetic field vector $\overrightarrow{B}$.Therefore,the angle $	heta$ between $\overrightarrow{B}$ and $\overrightarrow{A}$ is $0^\circ$.The magnetic flux $\phi$ is given by the formula:$\phi = B A \cos	heta$Substituting the values:$\phi = 0.5 	imes 1 	imes \cos(0^\circ)$Since $\cos(0^\circ) = 1$,$\phi = 0.5 	imes 1 	imes 1 = 0.5\,Wb$.

$A$ square loop of side $1\,m$ and resistance $1\,\Omega$ is placed in a magnetic field of $0.5\,T$. If the plane of the loop is perpendicular to the direction of the magnetic field,the magnetic flux through the loop is $\dots\dots$ weber.

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