Answer:
a) B = 10⁻¹ r
, b) B = 4 10⁻⁹ / r
, c) B=0
Explanation:
For this exercise let's use Ampere's law
∫ B. ds = μ₀ I
Where I is the current locked in the path. Let's take a closed path as a circle
ds = 2π dr
B 2π r = μ₀ I
B = μ₀ I / 2μ₀ r
Let's analyze several cases
a) r <Rw
Since the radius of the circumference is less than that of the wire, the current is less, let's use the concept of current density
j = I / A
For this case
j = I /π Rw² = I’/π r²
I’= I r² / Rw²
The magnetic field is
B = (μ₀/ 2π) r²/Rw² 1 / r
B = (μ₀ / 2π) r / Rw²
calculate
B = 4π 10⁻⁷ /2π r / 0.002²
B = 10⁻¹ r
b) in field between Rw <r <Rs
In this case the current enclosed in the total current
I = 0.02 A
B = μ₀/ 2π I / r
B = 4π 10⁻⁷ / 2π 0.02 / r
B = 4 10⁻⁹ / r
c) the field outside the coaxial Rs <r
In this case the waxed current is zero, so
B = 0
There is no correct description on that list of choices.
Answer:
The angular velocity is slowing down.
Explanation:
- By convention, if a rigid body is rotating clockwise, the angular velocity is negative.
- If the angular acceleration has a positive sign, since the angular acceleration and the angular velocity have opposite signs, this means that the angular velocity is slowing down.
In order to solve this problem, we must first find out the value of each line on the number line. However, we can make this problem more simple by ignoring every interval except for the ones between 0 and 6. There are three total intervals in between 0 and 6 (including 6 and excluding 0). Therefore, we can do 6/2, and get an interval value of 2. This means that each line adds a value of 2. Since the car is only one line past zero, we only have to add one value of 2. Since 0 + 2 = 2, our final answer is C. 2.
Hope this helps!
I would say 150 joules, i don't know if its right though check
