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Question

(D) The force on a stationary charge $Q$ in an electromagnetic wave is due to the electric field $\vec E$ only,as $\vec F = Q\vec E$. For an electroma

Vedclass · Accepted Answer

$0.6 \, N$

Vedclass · Answer

(D) The force on a stationary charge $Q$ in an electromagnetic wave is due to the electric field $\vec E$ only,as $\vec F = Q\vec E$. For an electromagnetic wave,$\vec E = c(\vec B 	imes \hat k)$,where $\hat k$ is the direction of propagation ($z$-axis). Given $\vec B = B_0 \cos(kz - \omega t) \hat i + B_1 \cos(kz - \omega t) \hat j$. Then $\vec E = c [ (B_0 \hat i + B_1 \hat j) 	imes \hat k ] \cos(kz - \omega t) = c [ -B_0 \hat j + B_1 \hat i ] \cos(kz - \omega t)$. The magnitude of the electric field is $E = c \sqrt{B_0^2 + B_1^2} |\cos(kz - \omega t)|$. The maximum force is $F_0 = Q E_{max} = Q c \sqrt{B_0^2 + B_1^2}$. Substituting values: $F_0 = 10^{-4} 	imes 3 	imes 10^8 	imes \sqrt{(3 	imes 10^{-5})^2 + (2 	imes 10^{-6})^2} \approx 10^{-4} 	imes 3 	imes 10^8 	imes 3 	imes 10^{-5} = 0.9 \, N$. The rms force is $F_{rms} = \frac{F_0}{\sqrt{2}} = \frac{0.9}{1.414} \approx 0.636 \, N$. Thus,the closest value is $0.6 \, N$.

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