A long solenoid is carrying a current I = I_0 | vedclass

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(C) The magnetic field inside a long solenoid is given by $B = \mu_0 N I$. Substituting $I = I_0 \sin(\omega t)$,we get $B = \mu_0 N I_0 \sin(\omega t

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$2 \mu_0 N I_0 R^2 \omega \cos(\omega t)$

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(C) The magnetic field inside a long solenoid is given by $B = \mu_0 N I$. Substituting $I = I_0 \sin(\omega t)$,we get $B = \mu_0 N I_0 \sin(\omega t)$.The square loop has its corners touching the solenoid,meaning the diagonal of the square is equal to the diameter of the solenoid. Let $l$ be the side of the square. Then,the diagonal $d = l\sqrt{2} = 2R$.Thus,$l = \frac{2R}{\sqrt{2}} = R\sqrt{2}$.The area of the square loop is $A = l^2 = (R\sqrt{2})^2 = 2R^2$.The magnetic flux linked with the loop is $\phi = B \cdot A = (\mu_0 N I_0 \sin(\omega t)) \cdot (2R^2) = 2 \mu_0 N I_0 R^2 \sin(\omega t)$.The induced emf is $e = -\frac{d\phi}{dt} = -\frac{d}{dt} [2 \mu_0 N I_0 R^2 \sin(\omega t)]$.Taking the magnitude,$|e| = 2 \mu_0 N I_0 R^2 \omega \cos(\omega t)$.

$A$ long solenoid is carrying a current $I = I_0 \sin(\omega t)$,having $N$ turns per unit length and radius $R$. $A$ square loop is placed inside the solenoid with its plane perpendicular to the solenoid axis,and its corners touching the solenoid. Find the emf induced in the square coil.

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