A 68 kg hiker walks at 5.0 km/h up a 9% slope. The indicated incline is the ratio of the vertical distance and the horizontal di
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Answer:
I = 69.3 μA
Explanation:
Current through the straight wire, I = 3.45 A
Number of turns, N = 5 turns
Diameter of the coil, D = 1.25 cm
Resistance of the coil,
Distance of the wire from the center of the coil, d = 20 cm = 0.2 m
The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.
Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil
Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345
ΔB = -0.000001725
Induced current,
E = -N (Δ∅)/Δt
Δ∅ = A ΔB
Area, A = πr²
diameter, d = 0.0125 m
Radius, r = 0.00625 m
A = π* 0.00625²
A = 0.0001227 m²
Δ∅ = -0.000001725 * 0.0001227
Δ∅ = -211.6575 * 10⁻¹²
E = -N (Δ∅)/Δt
Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms
I = E/R
I = 0.0000693 A
I = 69 .3 * 10⁻⁶A
I = 69.3 μA