Answer:
a. cosθ b. E.A
Explanation:
a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for 90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function
b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A
Yep. he discovered that coastline from south america and africa fit together like a puzzle, which later became a part of the continential drift theory
It would be C. the color of the pot. its pretty obvious that i would not effect the project.
Explanation:
where is your diagram? lol
Answer:
Current, I = 1000 A
Explanation:
It is given that,
Length of the copper wire, l = 7300 m
Resistance of copper line, R = 10 ohms
Magnetic field, B = 0.1 T
![\mu_o=4\pi \times 10^{-7}\ T-m/A](https://tex.z-dn.net/?f=%5Cmu_o%3D4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5C%20T-m%2FA)
Resistivity, ![\rho=1.72\times 10^{-8}\ \Omega-m](https://tex.z-dn.net/?f=%5Crho%3D1.72%5Ctimes%2010%5E%7B-8%7D%5C%20%5COmega-m)
We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :
![R=\rho \dfrac{l}{A}](https://tex.z-dn.net/?f=R%3D%5Crho%20%5Cdfrac%7Bl%7D%7BA%7D)
![R=\rho \dfrac{l}{\pi r^2}](https://tex.z-dn.net/?f=R%3D%5Crho%20%5Cdfrac%7Bl%7D%7B%5Cpi%20r%5E2%7D)
![r=\sqrt{\dfrac{\rho l}{R\pi}}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%5Cdfrac%7B%5Crho%20l%7D%7BR%5Cpi%7D%7D)
![r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%5Cdfrac%7B1.72%5Ctimes%2010%5E%7B-8%7D%5Ctimes%207300%7D%7B10%5Cpi%7D%7D)
r = 0.00199 m
or
![r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m](https://tex.z-dn.net/?f=r%3D1.99%5Ctimes%2010%5E%7B-3%7D%5C%20m%3D2%5Ctimes%2010%5E%7B-3%7D%5C%20m)
The magnetic field on a current carrying wire is given by :
![B=\dfrac{\mu_o I}{2\pi r}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B%5Cmu_o%20I%7D%7B2%5Cpi%20r%7D)
![I=\dfrac{2\pi rB}{\mu_o}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B2%5Cpi%20rB%7D%7B%5Cmu_o%7D)
![I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B2%5Cpi%20%5Ctimes%200.1%5Ctimes%202%5Ctimes%2010%5E%7B-3%7D%7D%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%7D)
I = 1000 A
So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.