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A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate \(\alpha \) The induced emf in the loop at an instant when its side is 'a' is
A) \(2a\alpha B\)
B) \({{a}^{2}}\alpha B\)
C)\(2{{a}^{2}}\alpha B\)
D)\(a\alpha B\)

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Correct solution is option A ) \(2a\alpha B\) 


"a" at any time "t" , the side of the square a= \((a_0-\alpha_t)\) , where \(\alpha_0\) side at t=0 . 

at this instant , flux through the square 

\(\phi = BA\cos0^o = B(a_o-\alpha_t)^2 \)

\(\therefore emf \ induced \ E= -\frac{d\phi}{dt}\) 

\(E=-B\times 2 (a_0-\alpha_t)\) 

⇒E= \(2a\alpha B\) 



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