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Suppose a 6.00 nC electric charge is in the beam. What is the maximum electric force (in N) it experiences? If the static charge moves at 300 m/s, what maximum magnetic force (in N) can it feel?

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Complete Question

Electromagnetic radiation from a 3.00 mW laser is concentrated on a 2.00 mm2 area. Suppose a 6.00 nC electric charge is in the beam. What is the maximum electric force (in N) it experiences? If the static charge moves at 300 m/s, what maximum magnetic force (in N) can it feel?

Answer:

The value are


F = 6.378 *10^(-6) \ N

and


F_k = 6.381 *10^(-12) \ N

Step-by-step explanation:

From the question we are told that

The power of the laser is
P = 3.00 mW = 3.00*10^(-3) \ W

The area is
A = 2.00 \ mm^2 = 2.0 *10^(-6) \ m^2]

The speed is
v = 300 \ m/s

The magnitude of the electric charge is
Q = 6.00 nC = 6.00 *10^(-9) \ C

Generally the intensity of the beam is mathematically represented as


I = (P)/(A)

=>
I = (3.00 *10^(-3))/( 2.0 *10^(-6))

=>
I = 1500 \ W/m^2

Generally the magnitude of the electric field is mathematically represented as


E_m = \sqrt{(2I)/(c \epsilon_o) }

Here c is the speed of light with value
c = 3.0*10^(8) \ m/s


\epsilon_o is the permittivity of free space with value
\epsilon_o = 8.85*10^(-12) C/(V \cdot m)


E_m = \sqrt{(2 * 1500 )/(3.0*10^(8) *8.85*10^(-12) ) }

=>
E_m = 1063 \ V/m

Generally the maximum electric force is mathematically represented as


F = Q * E_m

=>
F = 6*10^(-9 ) * 1063

=>
F = 6.378 *10^(-6) \ N

Generally the intensity of the beam can also be mathematically represented as


I = ( c * B^2 )/( 2 * \mu_o)

Here B is the magnetic field


\mu_o is the permeability of free space with value
\mu_o = 4\pi * 10^(-7) N/A^2

So


B = \sqrt{(2 * \mu_o * I )/(c) }

=>
B = \sqrt{(2 * 4 \pi *10^(-7) * 1500 )/(3.0*10^(8)) }

=>
B = 3.545*10^(-6) \ T

Generally the maximum magnetic force is mathematically represented as


F_k = Q * v * B

=>
F_k = 6.00 *10^(-9) * 300 * 3.545*10^(-6)

=>
F_k = 6.381 *10^(-12) \ N

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