The equation that represents the magnetic fie | vedclass

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(B) The magnetic field of an electromagnetic wave is given by $B = B_0 \sin (\omega t - kx) \hat{k}$.Given $E_0 = 60 \,Vm^{-1}$ and $c = 3 	imes 10^8

Vedclass · Accepted Answer

$\left(2 	imes 10^{-7}ight) \sin [200 \pi(c t-x)] \hat{k} \,T$

Vedclass · Answer

(B) The magnetic field of an electromagnetic wave is given by $B = B_0 \sin (\omega t - kx) \hat{k}$.Given $E_0 = 60 \,Vm^{-1}$ and $c = 3 	imes 10^8 \,ms^{-1}$,the amplitude of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{60}{3 	imes 10^8} = 2 	imes 10^{-7} \,T$.The wave propagates along the $x$-direction and the electric field is in the $y$-direction,so the magnetic field must be in the $z$-direction (unit vector $\hat{k}$).The wave number $k = \frac{2\pi}{\lambda}$. Given $\lambda = 10 \,mm = 10^{-2} \,m$,we have $k = \frac{2\pi}{10^{-2}} = 200\pi \,rad/m$.The general equation is $B = B_0 \sin [k(ct - x)] \hat{k}$.Substituting the values: $B = (2 	imes 10^{-7}) \sin [200\pi(ct - x)] \hat{k} \,T$.

The equation that represents the magnetic field of a plane electromagnetic wave which is propagating along the $x$-direction with a wavelength of $10 \,mm$ and a maximum electric field of $60 \,Vm^{-1}$ in the $y$-direction is (where,$c$ is the speed of light):

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