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(A) Magnetic flux $\Phi$ through a surface area $A$ in a uniform magnetic field $B$ is given by $\Phi = B \cdot A \cdot \cos(	heta)$. Since the magne

Vedclass · Accepted Answer

Case $I: \Phi = (L^2 + l^2)B; 	ext{ Case } II: \Phi = (L^2 - l^2)B$

Vedclass · Answer

(A) Magnetic flux $\Phi$ through a surface area $A$ in a uniform magnetic field $B$ is given by $\Phi = B \cdot A \cdot \cos(	heta)$. Since the magnetic field is perpendicular to the plane of the loops,$	heta = 0^\circ$ and $\cos(0^\circ) = 1$,so $\Phi = B \cdot A$.In Case $I$,the total area enclosed by the wire frame is the sum of the areas of the two squares: $A_I = L^2 + l^2$. Thus,$\Phi_I = (L^2 + l^2)B$.In Case $II$,the smaller square is inside the larger square. The effective area enclosed by the wire frame is the area of the larger square minus the area of the smaller square: $A_{II} = L^2 - l^2$. Thus,$\Phi_{II} = (L^2 - l^2)B$.Therefore,the correct option is $A$.

The adjoining figure shows two different arrangements in which two square wire frames are placed in a uniform constantly decreasing magnetic field $B$. The value of magnetic flux in each case is given by:

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