Answer:
i = 0.477 10⁴ B
the current flows in the counterclockwise
Explanation:
For this exercise let's use the Ampere law
∫ B . ds = μ₀ I
Where the path is closed
Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.
From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.
We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field
Segment on the y axis
L₀ = (y2-y1)
L₀ = 3-0 = 3 cm
Segment on the point x = 2 cm
L₁ = 3-0
L₁ = 3cm
B L = μ₀ I
B 2L = μ₀ I
i = 2 L B /μ₀
i= 2 0.03 / 4π 10⁻⁷ B
i = 4.77 10⁴ B
The current is perpendicular to the magnetic field whereby the current flows in the counterclockwise
Well let’s put it this way. To find the neutrons you subtract the atomic atomic Nuremberg from the atomic mass. So
Mass=81-Number=28
81-28=53
Final answer is 53.
Answer:
The tube should be held vertically, perpendicular to the ground.
Explanation:
As the power lines of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its potential.
And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.
