Two long parallel wires are separated by 6.0 mm.The current in one of the wires is twice the other current.If the magnitude of the force on a 3.0-m length of one of the wires is equal to 8.0 $\mu$N,what is the greater of the two currents?
A)0.20 A
B)0.40 A
C)40 mA
D)20 mA
E)0.63 A

Question

Quizplus · Accepted Answer

First, we can use the formula for the force per unit length between two parallel wires with currents I1 and I2:

F = μ₀I₁I₂L/2πd

where μ₀ is the permeability of free space, L is the length of wire being considered, and d is the separation between the wires.

We know that L = 3.0 m and F = 8.0 μN, and we can solve for the product of currents:

8.0 μN = μ₀(2I)(I)(3.0 m) / 2π(6.0 mm)
I² = (8.0 μN)(2π)(6.0 mm) / (μ₀)(3.0 m)(2)
I² = 3.8 x 10^-7 A²
I = 0.62 A (approximately)

Since one current is twice the other, the larger current is 1.24 A and the smaller current is 0.62 A. The answer is the greater of these two currents, which is 0.62 A or choice B.

Two Long Parallel Wires Are Separated by 6 μ\muμ N,what Is the Greater of the Two Currents? A)0

Two Long Parallel Wires Are Separated by 6 $\mu$ N,what Is the Greater of the Two Currents?
A)0