Question

A 7.0 m ordinary lamp extension cord carries a current, I = 7.0 A. Such a cord typically consists of two parallel wires carrying equal currents in opposite directions. The two wires are separated by about 4 mm. Find the magnitude and direction of the force that the two segments of this cord exert on each other. A) 2 N , The wires repel B) 2 N, The wires attract C) 0.02 N , The wires repel D) 0.02 N, The wires attract

Answer 1

(C)
`0.02 N` , The wires repel

Step-by-step explanation:

We are given that

Length of wire=L=7 m

`I_1=I_2=I=7 A`

`r=4 mm=4* 10^(-3) m`

`1 mm=10^(-3) m`

We have to find the magnitude and direction of the force that the two segments of this cord exert on each other.

`F=(2\mu_0I_1I_2L)/(4\pi r)`

Where
`(\mu_0)/(4\pi)=10^(-7)`

Using the formula

`F=(2* 10^(-7)* 7* 7* 7)/((4* 10^(-3))`

`F=0.017 N\approx 0.02 N`

When the wires carrying current in equal current in opposite direction then, the wire repel to each other.

Hence, the wires repel to each other.