Answer
The rate at which the magnetic field is changing is
Explanation
From the question we are told that
The electric field strength is 
The radius is 
The rate of change of the magnetic field is mathematically represented as

Where
is change of a unit length

Where A is the area which is mathematically represented as

So
where L is the circumference of the circle which is mathematically represented as

So
![E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]](https://tex.z-dn.net/?f=E%20%282%20%5Cpi%20r%20%29%20%3D%20%20%28%5Cpi%20r%5E2%20%29%20%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D)
![E = \frac{r}{2} [\frac{dB}{dt} ]](https://tex.z-dn.net/?f=E%20%20%3D%20%20%20%5Cfrac%7Br%7D%7B2%7D%20%20%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D)
![[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }](https://tex.z-dn.net/?f=%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D%20%3D%20%5Cfrac%7BE%7D%7B%20%5Cfrac%7Br%7D%7B2%7D%20%7D)
substituting values
![[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }](https://tex.z-dn.net/?f=%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D%20%3D%20%5Cfrac%7B3.5%20%2A10%5E%7B-3%7D%7D%7B%20%5Cfrac%7B15%7D%7B2%7D%20%7D)
Answer:
Explanation:
The electron has a negative charge. Proton is positive and neutron is neutral.
Answer:
a) v = 2,9992 10⁸ m / s
, b) Eo = 375 V / m
, B = 1.25 10⁻⁶ T,
c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz
, T = 1.05 10⁻¹⁵ s
, UV
Explanation:
In this problem they give us the equation of the traveling wave
E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]
a) what the wave velocity
all waves must meet
v = λ f
In this case, because of an electromagnetic wave, the speed must be the speed of light.
k = 2π / λ
λ = 2π / k
λ = 2π / 1.99 10⁷
λ = 3,157 10⁻⁷ m
w = 2π f
f = w / 2 π
f = 5.97 10¹⁵ / 2π
f = 9.50 10¹⁴ Hz
the wave speed is
v = 3,157 10⁻⁷ 9.50 10¹⁴
v = 2,9992 10⁸ m / s
b) The electric field is
Eo = 375 V / m
to find the magnetic field we use
E / B = c
B = E / c
B = 375 / 2,9992 10⁸
B = 1.25 10⁻⁶ T
c) The period is
T = 1 / f
T = 1 / 9.50 10¹⁴
T = 1.05 10⁻¹⁵ s
the wavelength value is
λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm
this wavelength corresponds to the ultraviolet
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