Question

Two parallel rods are each 0.38 m in length. They are attached at their centers to a spring that is initially neither stretched nor compressed. The spring has a spring constant of 210 N/m. When 1400 A of current is in each rod in the same direction, the spring is observed to be compressed by 3.0 cm. Treat the rods as long, straight wires and find the separation between them when the current is present.

Answer 1

1.88cm

Step-by-step explanation:

F = 210 N/m

L = 0.38m

I₁ = I₂ = I = 1400A

r = ?

μ₀ = 4π * 10⁻⁷ N/A²

From Bio-Savart law,

F/L = μ₀I₁I₂ / 4πr²

but I₁ = I₂ = I

(210 / 0.38) = (4π * 10⁻⁷ * 1400²) / 4πr²

552.63 = (1*10⁻⁷ * 1960000) / r²

552.63r² = 0.196

r² = 0.196 / 552.63

r² = 3.546*10⁻⁴

r = √(3.546*10⁻⁴)

r = 0.0188m = 1.88cm

Answer 2

The separation between the rods is 0.024 m

Step-by-step explanation:

Given:

L = length of the rods = 0.38 m

k = spring constant = 210 N/m

I = current = 1400 A

Δx = 3 cm = 0.03 m

The separation between the rods can be get from:

`IL((\mu I)/(2\pi r) )=-k*delta-x\\r=(\mu I^(2) L)/(2\pi k*delta-x) =(4\pi x10^(-7)*1400^(2)*0.38 )/(2\pi *210*0.03) =0.024m`