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A sinusoidal electromagnetic wave is propagating in a vacuum in the z-direction. If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 4.50 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time

Respuesta :

Answer:

Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction

Explanation:

From maxwell's equations;

B = E/v

Where;

B is maximum magnitude of magnetic field

E is maximum electric field

v is speed of light which has a constant value of 3 x 10^(8) m/s

We are given, E = 4.5 V/m

Thus; B = 4.5/(3 x 10^(8))

B = 1.5 x 10^(-8) T

Now, for Electric field, vector E to be in the positive x-direction, the product of vector E and vector B will have to be in the positive z-direction when vector B is in the positive y-direction

Thus,

Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction

Magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex]  in the positive y-direction.

From Maxwell's equations,

[tex]\bold {B = \dfrac Ev}[/tex]

Where;

B - maximum magnitude of magnetic field = ?

E- maximum electric field  = 4.5 V/m

v- speed of light =  3 x 10^(8) m/s

Put the values in the formula,

[tex]\bold {B = \dfrac {4.5}{3 x 10^8}}\\\\\bold {B = 1.5 x 10^{-8}\ T}[/tex]

When Electric field, is in the positive x-direction, vector B is in the positive y-direction and  the product of vector E and vector B will have to be in the positive z-direction.

Therefore, magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex]  in the positive y-direction.

To know more about Maxwell's equation,

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