Question

The intensity of an electromagnetic wave is 8 x 10^7 W/m^2. What is the amplitude of the magnetic field of this wave?

Answer 1

An electromagnetic wave is composed by E and H, where E is the electric field and H is the magnetic field. The amplitude is the peak value.

Answer 2

Which is found by using this equation: I = (c*Bmax²)/(2μ0), where c is the speed of light, μ0 is the permeability, I is the intensity of the electromagnetic wave and Bmax is the peak value. Just plug in the values and solve for Bmax!

Answer 3

Similarly, we can found the max value of the electric field by doing c x Bmax (or Hmax), where x denotes the cross product. Since they are perpendicular to each other (the angle is 90°), sin(x) is just 1.

Answer 4

find*

Answer 5

wow! and for permeability is a constant or i have to use another formula?

Answer 6

It is a constant. µ0 = 4π x 10^-7 T•m/A

Answer 7

mmmm... so is 8x10^7= (299792458)(Bmax)^2/2(4pi)10^-7

Answer 8

mmmm... so is 8x10^7= (299792458)(Bmax)^2/2(4pi)10^-7

Answer 9

intensity = E^2/uoC

E^2 = 8*10^7 *2* 4pi*10^-7 * 3*10^8

E = 24.55*10^4 N/C

Answer 10

Here!
intensity = E^2/uoC

E^2 = 8*10^7 * 4pi*10^-7 * 3*10^8

E = 17.36*10^4 N/C

now since E/B = c

B = E/c = 17.36*10^4/3*10^8 = 5.78 *10^-4 T

is this right?

Answer 11

\[I=8\times 10^{7}=\frac{1}{2 \mu _{0}}E _{m}B _{m}...........(1)\]
\[E _{m}=cB _{m}..........(2)\]
Substituting for Em in equation (1) gives
\[8\times 10^{7}=\frac{1}{2 \mu _{o}}cB _{m}...........(3)\]
Rearranging equation (3) we get
\[B _{m}=\sqrt{\frac{8\times 10^{7}\times 2\times 4 \pi \times 10^{-7}}{3\times 10^{8}}}\ T\]

Answer 12

so is 8.2 x 10^-4 ???

Answer 13

Sorry, my bad. Equation (3) should be written
\[8\times 10^{7}=\frac{1}{2 \mu _{0}}cB _{m}^{2}\]
My calculation resulted in 8.185 * 10^-4 T for the amplitude of the magnetic wave.

Answer 14

nice! :) thanks!

Answer 15

You're welcome :)