3 protons should be your answer
Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA
