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Question

(B) The electric field is given by $\overrightarrow{E} = E_0 \sin(kx - \omega t) \hat{j}$,where $E_0 = 30 \, V/m$,$\omega = 1.5 	imes 10^7 \, rad/s$,

Vedclass · Accepted Answer

$4.8 	imes 10^{-19} \, N$

Vedclass · Answer

(B) The electric field is given by $\overrightarrow{E} = E_0 \sin(kx - \omega t) \hat{j}$,where $E_0 = 30 \, V/m$,$\omega = 1.5 	imes 10^7 \, rad/s$,and $k = 5 	imes 10^{-2} \, rad/m$.The amplitude of the magnetic field $B_0$ is related to the electric field amplitude $E_0$ by $B_0 = \frac{E_0}{c}$.$B_0 = \frac{30}{3 	imes 10^8} = 10^{-7} \, T$.The electron moves along the $y$-axis with velocity $\overrightarrow{v} = 0.1 c \hat{j} = 0.1 	imes 3 	imes 10^8 \hat{j} = 3 	imes 10^7 \hat{j} \, m/s$.The magnetic force is given by $\overrightarrow{F}_{mag} = q(\overrightarrow{v} 	imes \overrightarrow{B})$. Since $\overrightarrow{v}$ is along the $y$-axis and $\overrightarrow{E}$ is along the $y$-axis,the wave propagates along the $x$-axis. The magnetic field $\overrightarrow{B}$ must be perpendicular to both the direction of propagation ($x$-axis) and the electric field ($y$-axis),so $\overrightarrow{B}$ is along the $z$-axis.$F_{max} = q v B_0 = (1.6 	imes 10^{-19} \, C) 	imes (3 	imes 10^7 \, m/s) 	imes (10^{-7} \, T)$.$F_{max} = 1.6 	imes 3 	imes 10^{-19} = 4.8 	imes 10^{-19} \, N$.

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