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(A) Given: Frequency $f = 3 	imes 10^{10} \, Hz$,Magnetic field amplitude $B_0 = 10^{-7} \, T$.The amplitude of the electric field is $E_0 = c B_0 =

Vedclass · Accepted Answer

$E_z = 30 \sin 2\pi (100x - 3 	imes 10^{10} t) \, V/m$

Vedclass · Answer

(A) Given: Frequency $f = 3 	imes 10^{10} \, Hz$,Magnetic field amplitude $B_0 = 10^{-7} \, T$.The amplitude of the electric field is $E_0 = c B_0 = (3 	imes 10^8 \, m/s) 	imes (10^{-7} \, T) = 30 \, V/m$.The wave propagates along the $X-$ direction and the magnetic field is along the $Y-$ direction,so the electric field must be along the $Z-$ direction $(E_z)$.The wave number $k = \frac{2\pi}{\lambda} = \frac{2\pi f}{c} = \frac{2\pi (3 	imes 10^{10})}{3 	imes 10^8} = 2\pi (100) \, rad/m$.The angular frequency $\omega = 2\pi f = 2\pi (3 	imes 10^{10}) \, rad/s$.The expression for the electric field is $E_z = E_0 \sin(kx - \omega t) = 30 \sin(2\pi(100x - 3 	imes 10^{10}t)) \, V/m$.

In an $EM$ wave propagating along the $X-$ direction,the magnetic field oscillates at a frequency of $3 \times 10^{10} \, Hz$ along the $Y-$ direction and has an amplitude of $10^{-7} \, T$. The expression for the electric field will be:

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