A coil has 1000 turns and 500 text{ cm}^2 as | vedclass

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(B) Given: $N = 1000$,$A = 500 	ext{ cm}^2 = 500 	imes 10^{-4} 	ext{ m}^2 = 5 	imes 10^{-2} 	ext{ m}^2$,$B = 2 	imes 10^{-5} 	ext{ Wb/m}^2$,$\D

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$(b)$ $10$

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(B) Given: $N = 1000$,$A = 500 	ext{ cm}^2 = 500 	imes 10^{-4} 	ext{ m}^2 = 5 	imes 10^{-2} 	ext{ m}^2$,$B = 2 	imes 10^{-5} 	ext{ Wb/m}^2$,$\Delta t = 0.2 	ext{ s}$.Since the plane of the coil is perpendicular to the magnetic field,the angle between the area vector and the magnetic field is $	heta_1 = 0^{\circ}$.Initial magnetic flux linked with the coil: $\phi_1 = N B A \cos 0^{\circ} = N B A$.After rotating the coil by $180^{\circ}$,the new angle is $	heta_2 = 180^{\circ}$.Final magnetic flux: $\phi_2 = N B A \cos 180^{\circ} = -N B A$.Change in flux: $\Delta \phi = \phi_2 - \phi_1 = -N B A - N B A = -2 N B A$.The average induced emf is given by: $e = -\frac{\Delta \phi}{\Delta t} = -\frac{-2 N B A}{\Delta t} = \frac{2 N B A}{\Delta t}$.Substituting the values: $e = \frac{2 	imes 1000 	imes 2 	imes 10^{-5} 	imes 5 	imes 10^{-2}}{0.2} = \frac{2 	imes 10^3 	imes 10 	imes 10^{-7}}{0.2} = \frac{2 	imes 10^{-3}}{0.2} = 10 	imes 10^{-3} 	ext{ V} = 10 	ext{ mV}$.

$A$ coil has $1000$ turns and $500 \text{ cm}^2$ as its area. The plane of the coil is placed at right angles to a magnetic induction field of $2 \times 10^{-5} \text{ Wb/m}^2$. The coil is rotated through $180^{\circ}$ in $0.2 \text{ s}$. The average emf induced in the coil,in $\text{mV}$,is

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