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(D) The electric field amplitude is given by $E_0 = 800 	ext{ V/m}$.The relationship between the electric field amplitude $E_0$ and magnetic field am

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$12.8 	imes 10^{-18} 	ext{ N}$

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(D) The electric field amplitude is given by $E_0 = 800 	ext{ V/m}$.The relationship between the electric field amplitude $E_0$ and magnetic field amplitude $B_0$ is $B_0 = E_0 / c$,where $c = 3 	imes 10^8 	ext{ m/s}$ is the speed of light.$B_0 = \frac{800}{3 	imes 10^8} 	ext{ T}$.The magnetic force on a moving charge is $F = qvB \sin 	heta$. For maximum force,the electron moves normal to the magnetic field,so $	heta = 90^\circ$ and $\sin 90^\circ = 1$.$F_{\max} = e v B_0 = (1.6 	imes 10^{-19} 	ext{ C}) 	imes (3 	imes 10^7 	ext{ m/s}) 	imes \left( \frac{800}{3 	imes 10^8} 	ext{ T} ight)$.$F_{\max} = 1.6 	imes 10^{-19} 	imes 800 	imes 10^{-1} = 1.6 	imes 8 	imes 10^{-18} = 12.8 	imes 10^{-18} 	ext{ N}$.

$A$ light beam is described by $E = 800 \sin \omega (t - x/c)$. An electron is allowed to move normal to the propagation of the light beam with a speed of $3 \times 10^{7} \text{ m/s}$. What is the maximum magnetic force exerted on the electron?

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