Complete Question:
A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?
Answer:
The magnitude of the magnetic field = 7.24 μT
Explanation:
Inner radius, a = 4.0 mm = 0.004 m
Outer radius, b = 30 mm = 0.03 m
Radius, r = 12 mm = 0.012 m
let h² = b² - a²
h² = 0.03² - 0.004²
h² = 0.000884
Let d² = r² - a²
d² = 0.012² - 0.004²
d² = 0.000128
Current I = 3A
μ = 4π * 10⁻⁷
The magnitude of the magnetic field is given by:

B = 7.24 * 10⁻⁶T
B = 7.24 μT
The y-component of the acceleration is 
Explanation:
The y-component of the ice skater acceleration can be calculated with the equation

where
is the y-component of the final velocity
is the y-component of the initial velocity
t is the time elapsed
Here we have:
- Initial velocity is
at
, so its y-component is 
- Final velocity is
at
, so its y-component is 
The time elapsed is
t = 8.33 s
Therefore, the y-component of the acceleration is

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Answer:
volume
Explanation:
it depends on the volume of the items
hope this helps
Answer: 168.75 N
Explanation:
first, let's convert microcoulombs to coulombs
q1 = 1e-4 C
q2 = 3e-5 C
r = 0.4 m
then use the equation Fe = 
plug in values --> F = (9e9*1e-4*3e-5)/(0.4)^2
F = 168.75 N