Answer: B
Explanation: the teacher just told us the answer
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step by step explanation:
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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
Explanation:
d = Diameter of wheel = 27 cm
r = Radius = 
m = Mass of wheel = 800 g
= Initial angular velocity = 
Equation of rotational motion

Moment of inertia is given by

Torque is given by

The torque the friction exerts is -0.0037406448 Nm
For more information on torque and moment of inertia refer
brainly.com/question/13936874
brainly.com/question/3406242