The magnitude and direction of the electric field in the wire are mathematically given as
![L &=[(v / L) v / m] \hat{i}](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D)
<h3>What is the magnitude and direction of the electric field in the wire?</h3>
Generally, the equation for is mathematically given as
A cylindrical wire that is straight and parallel to the x-axis has the following dimensions: length L, diameter d, resistivity p, diameter d, potential v, and z length. combining elements from both sides
E d 
![\begin{aligned}&-E \int_0^L d x=\int_v^0 d v \\\therefore E \cdot L &=v \\L &=[(v / L) v / m] \hat{i}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26-E%20%5Cint_0%5EL%20d%20x%3D%5Cint_v%5E0%20d%20v%20%5C%5C%5Ctherefore%20E%20%5Ccdot%20L%20%26%3Dv%20%5C%5CL%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D%5Cend%7Baligned%7D)
In conclusion, the magnitude and direction of the electric field in the wire are given as
![L &=[(v / L) v / m]](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D)
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Answer:
Utilization, effects
Explanation:
The conductors that carry the current to electrical devices and utilization equipment are the heart of all electrical systems. There are associated effects whenever current flows through a conductor.
The strength of the gravitational field is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
And now we can use the previous equation to calculate the field strength at that altitude:
And we can see this value is a bit less than the gravitational strength at the surface, which is
.
It spins one bc it takes 365 days to make a year and to make a year the earth has to mov some
