Question

Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine,B0,w, k, and l. (b) Find expressions for E and B.

Answer 1

Given: The electric field amplitude of an electromagnetic wave is 120 N/C and the frequency is 50.0 MHz.

a)

The magnetic field amplitude is given as,

B 0 = E 0 c

Where, the electric field strength is E 0 and the speed of the light is c.

By substituting the given values in the above expression, we get

B 0 = 120 3× 10 8 =4× 10 −7  T =400 nT

The angular frequency is given as,

ω=2πν

Where, the angular frequency is ω and the frequency is ν.

By substituting the given values in the above expression, we get

ω=2×3.14×50× 10 6 =3.14× 10 8   rad/s

The wave vector is given as,

k= ω c

Where, the wave vector is k.

By substituting the given values in the above expression, we get

k= 3.14× 10 8 2× 10 8 =1.05  m −1

The wavelength of the wave is given as,

λ= c ν

Where, the frequency is ν.

By substituting the given values in the above expression, we get

λ= 3× 10 8 50× 10 6 =6.0 m

Thus, the value of the B 0 , ω, k and λ is 400 nTesla, 3.14× 10 8   rad/s , 1.05  m −1 and 6.0 mrespectively.

b)

The expression for the electric field vector along the y-axis is given as,

E → = E 0 sin( kx−ωt ) j ^

By substituting the given values in the above expression, we get

E → =120sin( 1.05x−3.14× 10 8 t ) j ^  N/C

The expression for the magnetic field vector along the z-axis is given as,

B → = B 0 sin( kx−ωt ) k ^

By substituting the given values in the above expression, we get

B → =( 4× 10 −7 )sin( 1.05x−3.14× 10 8 t ) k ^ =400sin( 1.05x−3.14× 10 8 t ) k ^  nT

Thus, the expression for the electric field vector along the y-axis and the magnetic field vector along the z-axis is 120sin( 1.05x−3.14× 10 8 t ) j ^  N/C and 400sin( 1.05x−3.14× 10 8 t ) k ^  nT respectively.