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
Answer:
The angle between the electric field lines and the equipotential surface is 90 degree.
Explanation:
The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.
The electric field lines are always perpendicular to the equipotential surface.
As
For equipotential surface, dV = 0 so
The dot product of two non zero vectors is zero, if they are perpendicular to each other.
It is B: ionosphere. This has the ability to bounce radio signals
It's C, with the 2/7/4/6 in front of each reactant and product.
