The magnetic field in a plane electromagnetic | vedclass

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Question

(D) The relationship between the amplitude of the electric field $E_0$ and the magnetic field $B_0$ is given by $E_0 = B_0 c$,where $c$ is the speed o

Vedclass · Accepted Answer

$E_z = 105 \sin (1.5 	imes 10^3 x + 0.5 	imes 10^{11} t) \ Vm^{-1}$

Vedclass · Answer

(D) The relationship between the amplitude of the electric field $E_0$ and the magnetic field $B_0$ is given by $E_0 = B_0 c$,where $c$ is the speed of light in vacuum $(c \approx 3 	imes 10^8 \ ms^{-1})$.Given $B_0 = 3.5 	imes 10^{-7} \ T$,we calculate $E_0 = (3.5 	imes 10^{-7}) 	imes (3 	imes 10^8) = 105 \ Vm^{-1}$.The wave propagates in the negative $x$-direction (indicated by the $+kx$ term). The magnetic field is in the $y$-direction $(B_y)$. Since the electric field,magnetic field,and direction of propagation are mutually perpendicular,the electric field must be in the $z$-direction $(E_z)$.Thus,the electric field is $E_z = 105 \sin (1.5 	imes 10^3 x + 0.5 	imes 10^{11} t) \ Vm^{-1}$.

The magnetic field in a plane electromagnetic wave is given by $B_y = (3.5 \times 10^{-7}) \sin (1.5 \times 10^3 x + 0.5 \times 10^{11} t) \ T$. The corresponding electric field will be:

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