for an electromagnetic wave moving through vacuum, what is the ratio of the magnitude of the electric field to the magnetic field equal to?

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08-12-2022
Physics

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for an electromagnetic wave moving through vacuum, what is the ratio of the magnitude of the electric field to the magnetic field equal to?

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jaiswaljyoti jaiswaljyoti · Answer 1 · 2022-12-18T03:40:27+01:00

The ratio of the magnitude of the electric field to the magnetic field of an electromagnetic wave moving through a vacuum is equal to the speed of light, c = 3.00 × 10^8 m/s.

The ratio of the magnitude of the electric field to the magnetic field for an electromagnetic wave moving through vacuum is equal to the speed of light, c.

The electric field and the magnetic field are two different components of an electromagnetic wave. The electric field is a vector field that describes the force exerted on charged particles by the wave and the magnetic field is a vector field that describes the force exerted on magnetic objects by the wave.

The speed of light, c, is the speed at which the wave propagates through a vacuum. This is equal to the ratio of the propagation speed of the electric field to the propagation speed of the magnetic field. As these two fields propagate together as electromagnetic waves, they must move at the same speed, which is the speed of light.

Therefore, the ratio of the magnitude of the electric field to the magnetic field for an electromagnetic wave moving through vacuum is equal to the speed of light, c.

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