Question

Suppose that the electric field amplitude of an electromagnetic wave us E_0=120 N// C and that its frequency is 50.0 MHz. (a) Determine B_0,omega,k and lambda, (b) find expressions for E and B.

Answer 1

Given ,

E_○=120N/C

n= 50 MHz

(a) by using equation,

E_○/B_○=c

B_○= E_○/c

B_○= 120/3×10^8

B_○= 4×10^7T

Omega =2pi n

= 2×3.14×5×10^6

=3.14×10^8rad/sec

By using equation

k = omega /c

k=3.14 ×10^8/3×10^8

k=1.05/m

By using equation

Lambda = c/n

= 3×10^8/50×10^6

=6m

(b)In the propagating of wave electric field is along y-axis and magnetic field is along z-axis so both are perpendicular to direction of propagation.

E (vector)= E_○sin (kx-omega t)j^

E (vector)=120sin (1.05x-3.14×10^8t)j^ N/C.......along y-axis

B (vector)=B_○sin (kx-omega t)k^

B (vector)= 4×10^-7 sin(1.05x- 3.14×10^8 t)k^T...along z-axis