i think the answer is B police siren.
Answer:
The curl is 
Explanation:
Given the vector function

We can calculate the curl using the definition

Thus for the exercise we will have

So we will get

Working with the partial derivatives we get the curl

opaque glass does not allow light to pass through it.
Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I
To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.
By definition we know that the magnetic field in a sphere can be described as

Where,
a = Radius
z = Distance to the magnetic field
I = Current
Permeability constant in free space
Our values are given as
diameter of the sphere then,

Thus z = a



Re-arrange to find I,



Therefore the current at the pole of this sphere is 