Question

The wires in a household lamp cord are typically 3.5 mm apart center to center and carry equal currents in opposite directions. If the cord carries direct current to a 100-W light bulb connected across a 120-V potential difference, what force per meter does each wire of the cord exert on the other? Is the force attractive or repulsive? Is this force large enough so it should be considered in the design of the lamp cord? (Model the lamp cord as a very long straight wire.)

Answer 1

Explanation:

It is given that,

Distance between wires, d = 3.5 mm = 0.0035 m

Power of light bulb, P = 100 W

Potential difference, V = 120 V

(a) We need to find the force per unit length each wire of the cord exert on the other. It is given by :

$\dfrac{F}{l}=\dfrac{\mu_o I^2}{2\pi r}$

Power, P = V × I

$I=\dfrac{P}{V}=\dfrac{100}{120}=0.83\ A$

This gives, $\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times (0.83)^2}{2\pi \times 0.0035}$

$\dfrac{F}{l}=0.0000393\ N/m$

$\dfrac{F}{l}=3.93\times 10^{-5}\ N/m$

(b) Since, the two wires carry equal currents in opposite directions. So, teh force is repulsive.

(c) This force is negligible.

Hence, this is the required solution.