A transverse electromagnetic plane wave with a frequency of 2.18 MHz travels dir

Question

A transverse electromagnetic plane wave with a frequency of 2.18 MHz travels directly away from you (- z direction). At time t = 0, your magnetic field detector measures a magnetic field of 5.29 × 108 Tesla pointing down (- y direction) which further measurement reveals to be the maximum magentic field strength for the wave.

a) Also at time t = 0 and at the location of your detector, what is the electric field associated with this wave? (Enter + if rigthward and - if leftward.)

b) At time t = 0, what is the electric field 32.9 meters directly behind (further away from you) your detector?

c) At time t = 4.29 × 107 seconds, what is the electric field at your detector?

d) At time t = 4.29 × 107 seconds, what is the electric field 32.9 centimeters directly to the right of your detector?

Explanation / Answer

given

f = 2.18 Mhz

w = 2*pi*f

= 2*pi*2.18*10^6

= 1.37*10^7 rad/s

Bmax = 5.29*10^-8 T

c = lamda*f

==> lamda = c/f

= 3*10^8/(2.18*10^6)

= 137.6 m

k = 2*pi/137.6

= 0.0456 rad/m

a) Emax = Bmax*c

= 5.29*10^-8*3*10^8

= 15.87 N/c (towards +x axis)

so, E = +15.87 N/c

b) E = Emax*cos(kz + w*t)

at t = 0, z = -32.9 m

E = 15.87*cos(0.0456*32.9 + 1.37*10^7*0)

= 1.1 N/c

c)

at t = 4.29*10^-7, z = 0 m

E = 15.87*cos(0.0456*0 + 1.37*10^7*4.29*10^-7)

= 14.6 N/c

d) at t = 4.29*10^-7, z = 32.9 m

E = 15.87*cos(0.0456*32.9 + 1.37*10^7*4.29*10^-7)

= 7.28 N/c

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