(a) The intensity of the electromagnetic wave is related to the amplitude of the electric field by
[tex]I= \frac{1}{2} c \epsilon_0 E^2 [/tex]
where
I is the intensity
c is the speed of light
[tex]\epsilon_0[/tex] is the electric permittivity
E is the amplitude of the electric field
By substituting the numbers of the problem and re-arranging the equation, we can find E:
[tex]E= \frac{2 I}{c \epsilon_0} = \frac{2 ( 2.9 \cdot 10^3 Wm^{-2})}{(3 \cdot 10^8 m/s)(8.85 \cdot 10^{-12} Fm^{-1})} =2.2 \cdot 10^6 N/C [/tex]
Now that we have the intensity of the electric field, we can calculate the electric force on the charge:
[tex]F=qE=(2.1 \cdot 10^{-8} C)(2.2 \cdot 10^6 N/C)=0.046 N[/tex]
(b) We can calculate the amplitude of the magnetic field starting from the amplitude of the electric field:
[tex]B= \frac{E}{c}= \frac{2.2 \cdot 10^6 N/C}{3 \cdot 10^8 m/s}=7.3 \cdot 10^{-3} T [/tex]
The magnetic force is given by
[tex]F=qvB \sin \theta[/tex]
where v is the particle's speed, B the magnetic field intensity and [tex]\theta [/tex] the angle between B and v.
In this case the charge is stationary, so v=0, and so the magnetic force is zero: F=0.
(c) The electric force has not changed compared to point (a), because it does not depend on the speed of the particle, so we have again F=0.046 N.
(d) This time, the particle is moving with speed [tex]v=3.7 \cdot 10^4 m/s[/tex], in a direction perpendicular to the magnetic field (so, the angle [tex]\theta[/tex] is [tex]90^{\circ}[/tex]), and so by using the intensity of the magnetic field we found in point (b), we can calculate the magnetic force on the particle:
[tex]F=qvB \sin \theta = (2.1 \cdot 10^{-8}C)(3.7 \cdot 10^4 m/s)(7.3 \cdot 10^{-3} T)(\sin 90^{\circ} )=[/tex]
[tex]=5.7 \cdot 10^{-6} N[/tex]
