# 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

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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 ) $$2a\alpha B$$

Explaination:

"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= $$2a\alpha B$$