The magnetic field of an electromagnetic wave | vedclass

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(B) Given: Magnetic field $B = 10^{-12} \sin(5 	imes 10^6 t) 	ext{ T}$,number of turns $N = 300$,and area $A = 20 	ext{ cm}^2 = 20 	imes 10^{-4} \

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$-3 	imes 10^{-6} \cos(5 	imes 10^6 t) 	ext{ V}$

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(B) Given: Magnetic field $B = 10^{-12} \sin(5 	imes 10^6 t) 	ext{ T}$,number of turns $N = 300$,and area $A = 20 	ext{ cm}^2 = 20 	imes 10^{-4} 	ext{ m}^2$.According to Faraday's law of induction,the induced emf $e$ is given by $e = -N \frac{d\phi}{dt}$,where $\phi = BA$.Since the coil is perpendicular to the field,$\phi = BA \cos(0^\circ) = BA$.Substituting the values:$e = -N \frac{d}{dt}(BA) = -NA \frac{dB}{dt}$$e = -300 	imes (20 	imes 10^{-4}) 	imes \frac{d}{dt} [10^{-12} \sin(5 	imes 10^6 t)]$$e = -300 	imes 20 	imes 10^{-4} 	imes 10^{-12} 	imes \cos(5 	imes 10^6 t) 	imes (5 	imes 10^6)$$e = -6000 	imes 10^{-16} 	imes 5 	imes 10^6 	imes \cos(5 	imes 10^6 t)$$e = -30000 	imes 10^{-10} 	imes \cos(5 	imes 10^6 t)$$e = -3 	imes 10^4 	imes 10^{-10} 	imes \cos(5 	imes 10^6 t)$$e = -3 	imes 10^{-6} \cos(5 	imes 10^6 t) 	ext{ V}$.

The magnetic field of an electromagnetic wave in a certain region obeys the relation $B = 10^{-12} \sin(5 \times 10^6 t) \text{ T}$,where $t$ is the time. Then,the induced emf in a coil of $300$ turns and area $20 \text{ cm}^2$,oriented perpendicular to the field,is:

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