Two long parallel wires are 0.400 m apart and carry currents of 4.00 A and 6.00 A.What is the magnitude of the force per unit length that each wire exerts on the other wire? (μ₀ = 4π × 10^-7 T ∙ m/A) A)2.00 μN/m B)5.00 μN/m C)12.0 μN/m D)16.0 μN/m E)38.0 μN/m

Question

Quizplus · Accepted Answer

The magnitude of the force per unit length that wire 1 exerts on wire 2 is given by:

$F_{12} = \frac{\mu_0}{2\pi} \frac{I_1 I_2}{d}$

where $\mu_0$ is the permeability constant, $I_1$ and $I_2$ are the currents in wires 1 and 2, respectively, and $d$ is the distance between them.

Substituting the given values, we get:

$F_{12} = \frac{4\pi	imes 10^{-7} 	ext{T}\cdot	ext{m}/	ext{A}}{2\pi} \frac{(4.00	ext{ A})(6.00	ext{ A})}{0.400	ext{ m}} = 12.0	ext{ }\mu	ext{N/m}$

Therefore, the answer is C.

Two Long Parallel Wires Are 0