Question

A 0.474 m long wire carrying 1.39 A ofcurrent is 0.0535 m from a secondparallel wire carrying 0.884 A ofcurrent in the same direction. What isthe (+) magnitude of the magneticforce between the wires?[?] × 10?] N

Answer 1

Given:

Length of wire, L = 0.474 m

Current carried by wire 1, I1 = 1.39 A

DIstance between wires, r = 0.0535 m

Current carried by second wire, I2 = 0.884 A

Let's find the magnitude of the magnetic force between the wires.

Apply the formula:

`F=(\mu_oI_1I_2L)/(2\pi d)`

Where:

F is the force between the wires.

μo = 4π x 10⁻⁷

d is the distance between wires = 0.0535

We have:

`\begin{gathered} F=(4\pi*10^(-7)*1.39*0.884*0.474)/(2\pi *0.0535) \\ \\ F=(2*10^(-7)*1.39*0.884*0.474)/(0.0535) \\ \\ F=2.18*10^(-6)\text{ N} \end{gathered}`

Therefore, the magnitude of magnetic force between the wires is 2.18 x 10⁻⁶ N.

ANSWER:

2.18 x 10⁻⁶ N