The magnetic field of a plane electromagnetic wave is given by, $\vec B=3 imes10^{-8}\cos(1.6 imes10^3x+48

Question

The magnetic field of a plane electromagnetic wave is given by, $\vec B=3	imes10^{-8}\cos(1.6	imes10^3x+48	imes10^{10}t)\hat j 	ext{ T}$. The associated electric field will be:

Accepted Answer

1. **Calculate Electric Field Amplitude ($E_0$):** The amplitude of the electric field is related to the magnetic field amplitude ($B_0$) by the speed of light ($c$): $E_0 = c B_0$. 
   Given $B_0 = 3 	imes 10^{-8} 	ext{ T}$ and $c = 3 	imes 10^8 	ext{ m/s}$:
   $$E_0 = (3 	imes 10^8) 	imes (3 	imes 10^{-8}) = 9 	ext{ V/m}$$
2. **Determine Direction:** The direction of propagation of an EM wave is given by $\vec E 	imes \vec B$. The argument $(kx + \omega t)$ indicates the wave is travelling in the negative x-direction ($-\hat i$). The magnetic field is in the y-direction ($\hat j$). We need a direction for $\vec E$ such that $\hat E 	imes \hat j = -\hat i$. Using vector cross product rules, $\hat k 	imes \hat j = -\hat i$. Thus, the electric field is along the z-axis ($\hat k$).
3. **Phase:** In a plane electromagnetic wave in free space, the electric and magnetic fields oscillate in phase. Therefore, the function remains cosine with the same argument.
4. **Conclusion:** Combining the magnitude, direction, and phase, $\vec E = 9 \cos(1.6	imes10^3x+48	imes10^{10}t)\hat k 	ext{ V/m}$.

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