C. Forces have mass and take up space
Answer:
It's the duration between successive new moons. Also called a lunation or synodic month, it has a mean period of 29.53059 days (29 days 12 hours and 44 minutes).
The observable universe consists of galaxies and other matter that can, principally, be seen from Earth because the light signals have had time to reach us. Not everything in the sky is the way it is when we see it, because of the distance the light travels to reach us.
Hope this helps :)
To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:
![\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}](https://tex.z-dn.net/?f=%5Coint%20%5Cvec%7BB%7D%5Cvec%7Bdl%7D%20%3D%20%5Cmu_0%20%5Cepsilon_0%20%5Cfrac%7Bd%5CPhi_E%7D%7Bdt%7D)
Where,
B= Magnetic Field
l = length
= Vacuum permeability
= Vacuum permittivity
Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to
![B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}](https://tex.z-dn.net/?f=B%282%5Cpi%20r%29%20%3D%20%5Cmu_0%20%5Cepsilon_0%20%5Cfrac%7Bd%28EA%29%7D%7Bdt%7D)
Recall that the speed of light is equivalent to
![c^2 = \frac{1}{\mu_0 \epsilon_0}](https://tex.z-dn.net/?f=c%5E2%20%3D%20%5Cfrac%7B1%7D%7B%5Cmu_0%20%5Cepsilon_0%7D)
Then replacing,
![B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}](https://tex.z-dn.net/?f=B%282%5Cpi%20r%29%20%3D%20%5Cfrac%7B1%7D%7BC%5E2%7D%20%28%5Cpi%20r%5E2%29%20%5Cfrac%7Bd%28E%29%7D%7Bdt%7D)
![B = \frac{r}{2C^2} \frac{dE}{dt}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7Br%7D%7B2C%5E2%7D%20%5Cfrac%7BdE%7D%7Bdt%7D)
Our values are given as
![dE = 2150N/C](https://tex.z-dn.net/?f=dE%20%3D%202150N%2FC)
![dt = 5s](https://tex.z-dn.net/?f=dt%20%3D%205s)
![C = 3*10^8m/s](https://tex.z-dn.net/?f=C%20%3D%203%2A10%5E8m%2Fs)
![D = 0.440m \rightarrow r = 0.220m](https://tex.z-dn.net/?f=D%20%3D%200.440m%20%5Crightarrow%20r%20%3D%200.220m)
Replacing we have,
![B = \frac{r}{2C^2} \frac{dE}{dt}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7Br%7D%7B2C%5E2%7D%20%5Cfrac%7BdE%7D%7Bdt%7D)
![B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B0.220%7D%7B2%283%2A10%5E8%29%5E2%7D%20%5Cfrac%7B2150%7D%7B5%7D)
![B =5.25*10^{-16}T](https://tex.z-dn.net/?f=B%20%3D5.25%2A10%5E%7B-16%7DT)
Therefore the magnetic field around this circular area is ![B =5.25*10^{-16}T](https://tex.z-dn.net/?f=B%20%3D5.25%2A10%5E%7B-16%7DT)