Question

A 2.0 m ordinary lamp extension cord carries a 3.0 A current. Such a cord typically consists of two parallel wires carrying equal currents in opposite directions.

a. Find the magnitude of the force that the two segments of this cord exert on each other. (You will need to inspect an actual lamp cord at home and measure or reasonably estimate the quantities needed to do this calculation.)
b. Find the direction (attractive or repulsive) of the force that the two segments of this cord exert on each other. (Opposite or same directional currents, and do the wires attract or repel?)

Answer 1

Our values are given are,

`l = 2m`

`I = 3A`

`r = 3.5mm = 3.5*10^(-3) m`

Electromagnetic force can then be defined as,

`F = (\mu_0 I^2 l)/(2\pi r)`

Here,

`\mu_0` = Permeability free space

I = Current

l = Length

r = Distance between them

Replacing,

`F = ((4\pi *10^(-7))(3)^2(2))/((2\pi)(3.5*10^(-3)))`

`F = 0.001028N`

Therefore the magnitude of the force that two segments of this cord exert on each other is 0.001028N.

Since the magnitude of the force is positive, the direction of both is attractive.