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Question

(A) The given magnetic field is $\overrightarrow{B} = B_0 \sin (kx + \omega t) \hat{j}$,where $B_0 = 3 	imes 10^{-8} \; T$.Since the wave propagates

Vedclass · Accepted Answer

$\overrightarrow{E} = 9 \sin (1.6 	imes 10^{3} x + 48 	imes 10^{10} t) \hat{k} \; V/m$

Vedclass · Answer

(A) The given magnetic field is $\overrightarrow{B} = B_0 \sin (kx + \omega t) \hat{j}$,where $B_0 = 3 	imes 10^{-8} \; T$.Since the wave propagates in the negative $x$-direction (as indicated by $+kx$),the direction of propagation is $-\hat{i}$.The relationship between the amplitudes of electric and magnetic fields is $E_0 = c B_0$,where $c = 3 	imes 10^{8} \; m/s$.$E_0 = (3 	imes 10^{8} \; m/s) 	imes (3 	imes 10^{-8} \; T) = 9 \; V/m$.In an electromagnetic wave,the direction of propagation is given by the direction of $\overrightarrow{E} 	imes \overrightarrow{B}$.Here,$(-\hat{i}) = \hat{E} 	imes \hat{j}$.Since $\hat{k} 	imes \hat{j} = -\hat{i}$,the electric field must be in the $\hat{k}$ direction.Thus,$\overrightarrow{E} = 9 \sin (1.6 	imes 10^{3} x + 48 	imes 10^{10} t) \hat{k} \; V/m$.

If the magnetic field in a plane electromagnetic wave is given by $\overrightarrow{B} = 3 \times 10^{-8} \sin (1.6 \times 10^{3} x + 48 \times 10^{10} t) \hat{j} \; T$,then what will be the expression for the electric field?

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