If the electric field of a plane electromagne | vedclass

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Question

(A) The given electric field is $E_z = 60 \sin(0.5 	imes 10^3 x + 1.5 	imes 10^{11} t) \ Vm^{-1}$.Comparing this with the standard equation $E_z = E

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$B_y = 2 	imes 10^{-7} \sin(0.5 	imes 10^3 x + 1.5 	imes 10^{11} t) \ T$

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(A) The given electric field is $E_z = 60 \sin(0.5 	imes 10^3 x + 1.5 	imes 10^{11} t) \ Vm^{-1}$.Comparing this with the standard equation $E_z = E_0 \sin(kx + \omega t)$,we get $E_0 = 60 \ Vm^{-1}$.The amplitude of the magnetic field $B_0$ is given by $B_0 = \frac{E_0}{c}$,where $c = 3 	imes 10^8 \ ms^{-1}$ is the speed of light.$B_0 = \frac{60}{3 	imes 10^8} = 20 	imes 10^{-8} = 2 	imes 10^{-7} \ T$.Since the wave propagates in the negative $x$-direction (indicated by the $+kx$ term) and the electric field is in the $z$-direction,the magnetic field must be in the $y$-direction to satisfy the direction of propagation $\vec{E} 	imes \vec{B} \propto \vec{v}$.Specifically,$\hat{k} 	imes \hat{i} = \hat{j}$ (for propagation in $-x$,$\hat{E} 	imes \hat{B} = -\hat{i}$,so $\hat{k} 	imes \hat{j} = -\hat{i}$).Thus,the magnetic field is $B_y = B_0 \sin(kx + \omega t) = 2 	imes 10^{-7} \sin(0.5 	imes 10^3 x + 1.5 	imes 10^{11} t) \ T$.

If the electric field of a plane electromagnetic wave is $E_z = 60 \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t) \ Vm^{-1}$,then the magnetic field of the wave is

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