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Scorpion4ik [409]
3 years ago
15

A sinusoidal electromagnetic wave is propagating in a vacuum in the z-direction. Part A If at a particular instant and at a cert

ain point in space the electric field is in the x-direction and has a magnitude of 4.70 V/mV/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
Physics
1 answer:
kicyunya [14]3 years ago
4 0

Answer:

the magnitude of the magnetic field is 1.23 × 10⁻⁸ T  { in the +y direction }

Explanation:

Given the data in the question;

the relation between the electric field and the magnetic field in an electromagnetic wave can be written as follows;

c = E₀/B₀

where c is speed of light, E₀ is the amplitude of the electric field and B₀ is the amplitude of the magnetic field.

given that;  E₀ = 4.70 V/m, we know that; speed of light c = 3 × 10⁸ m/s

so from our relation; c = E₀/B₀

B₀ = E₀/c

we substitute

B₀ = 4.70 / 3 × 10⁸

B₀ = 1.23 × 10⁻⁸ T  { in the +y direction }

Therefore, the magnitude of the magnetic field is 1.23 × 10⁻⁸ T  { in the +y direction }

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The <u>centripetal acceleration</u> a_{c} of an object moving in a uniform circular path is given by the following equation:

a_{c}=\frac{V^{2}}{r}  (1)

Where:

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On the other hand, the tangential velocity  is expressed as:

V=\omega r (2)

Where \omega is the angular velocity, which can be found knowing the child makes 5 revolutions in 13.4s:

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Substituting (5) in (1):

a_{c}=\frac{(0.39 m/s)^{2}}{1.05 m}  (6)

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