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Two square wire frames of side L and side l (with l < L) and of equal resistance are placed in a uniform magnetic field B that is directed perpendicular to their planes. In Case I the magnetic fluxes through the two squares add, while in Case II they oppose each other. The total magnetic flux through the combination in each case is:

1. Case I: Phi = (L² + l²)B; Case II: Phi = (L² - l²)B
2. Case I: Phi = pi*(L² + l²)B; Case II: Phi = pi*(L² - l²)B
3. Case I: Phi = pi*(L² - l²)B; Case II: Phi = pi*(L² + l²)B
4. Case I: Phi = (L² + l²)B; Case II: Phi = pi*(L - l)² B

Correct answer: Case I: Phi = (L² + l²)B; Case II: Phi = (L² - l²)B

Solution

For a planar loop in a uniform field perpendicular to it, flux equals B times the enclosed area. The squares have areas L² and l². The factor pi appears only for circular areas, so any option containing pi is wrong. Adding (Case I) gives B(L² + l²); opposing (Case II) gives B(L² - l²).

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