Answer:
2.54 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the radial width of 13.1 μm, dr = 13.1 μm. r = 1.50 mm. We have a differential current dI. We remove the first integral by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ ⇒ dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr evaluate at r = 1.50 mm = 1.50 × 10⁻³ m and dr = 13.1 μm = 0.013 mm = 0.013 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.34 × 10 A/m³ × (1.50 × 10⁻³ m)² × 0.013 × 10⁻³ m = 2544.69 × 10⁻⁹ A = 2.54 × 10⁻⁶ A = 2.54 μA
Answer:
what are the options for me
The correct answer is picture 1 because the grey arrow represents the real object size and the orange - yellowish arrow represents the image size. So since the image size is bigger than the object size. So this means picture 1 has a larger image.
Answer:
Explanation:
Displacement is the shortest distance or path between two points.
1) Displacement = √(36² + 45²) = 57.63 miles
2) Displacement = √(100² + 500²) = 509.9 meters
3) Displacement = √(60² + 40²) = 72.11 miles
4) Displacement = √(700² + 500²) = 860.23 miles
5) Displacement = 300 - 300 = 0 miles
6) Displacement = 200 + 100 = 300 miles
7) Displacement = √(650² + 650²) = 919.24 miles
8) Yes, since a distance is moved in a direction