Question

A stationary particle of charge q = 2.1 × 10-8 c is placed in a laser beam (an electromagnetic wave) whose intensity is 2.9 × 103 w/m2. determine the maximum magnitude of the (a) electric and (b) magnetic forces exerted on the charge. if the charge is moving at a speed of 3.7 × 104 m/s perpendicular to the magnetic field of the electromagnetic wave, find the maximum magnitudes of the (c) electric and (d) magnetic forces exerted on the particle.

Answer 1

(a) The intensity of the electromagnetic wave is related to the amplitude of the electric field by
$I= \frac{1}{2} c \epsilon_0 E^2$
where
I is the intensity
c is the speed of light
$\epsilon_0$ 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:
$E= \frac{2 I}{c \epsilon_0} = \frac{2 ( 2.9 \cdot 10^3 Wm^{-2})}{(3 \cdot 10^8 m/s)(8.85 \cdot 10^{-12} Fm^{-1})} =2.2 \cdot 10^6 N/C$

Now that we have the intensity of the electric field, we can calculate the electric force on the charge:
$F=qE=(2.1 \cdot 10^{-8} C)(2.2 \cdot 10^6 N/C)=0.046 N$

(b) We can calculate the amplitude of the magnetic field starting from the amplitude of the electric field:
$B= \frac{E}{c}= \frac{2.2 \cdot 10^6 N/C}{3 \cdot 10^8 m/s}=7.3 \cdot 10^{-3} T$

The magnetic force is given by
$F=qvB \sin \theta$
where v is the particle's speed, B the magnetic field intensity and $\theta$ 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 $v=3.7 \cdot 10^4 m/s$, in a direction perpendicular to the magnetic field (so, the angle $\theta$ is $90^{\circ}$), and so by using the intensity of the magnetic field we found in point (b), we can calculate the magnetic force on the particle:
$F=qvB \sin \theta = (2.1 \cdot 10^{-8}C)(3.7 \cdot 10^4 m/s)(7.3 \cdot 10^{-3} T)(\sin 90^{\circ} )=$
$=5.7 \cdot 10^{-6} N$