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

Answer:

(C) $0.02 N$ , The wires repel

Explanation:

We are given that

Length of wire=L=7 m

$I_1=I_2=I=7 A$

$r=4 mm=4\times 10^{-3} m$

$1 mm=10^{-3} m$

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

$F=\frac{2\mu_0I_1I_2L}{4\pi r}$

Where $\frac{\mu_0}{4\pi}=10^{-7}$

Using the formula

$F=\frac{2\times 10^{-7}\times 7\times 7\times 7}{(4\times 10^{-3}}$

$F=0.017 N\approx 0.02 N$

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.