91Ó°ÊÓ

The force per unit length between two current carrying wires is given by

$\frac{\mathrm{F}}{\mathrm{L}}=\frac{{\mathrm{μ}}_0{âˆ\yen }_1}{2\mathrm{Ï€}d}$

Where,${\mathrm{μ}}_0$ is the permeability of vaccum,$l_1$ and$l_2$ is the current through the two wires and d is the distance between wire.

Attraction and repulsion between two parallel current carrying wires

When the current is flowing in same direction in both parallel wires then attraction occurs between two wires.

When the current is flowing in opposite direction in both parallel wires then repulsion occurs between two wires.

Relation between operating power, applied voltage and current in circuit

The relationship between power voltage and current is given by

Power = voltage x current

P = VI

Where, P is the operating power , V is the voltage across the bulb and l is the current in the circuit.

Here, the current is flowing in opposite directions in both wires.

Therefore, the force between two wires must be repulsive in nature.

Given : Potential difference across bulb is V = 120W .

Operating power of bulb is P = 100W.

The distance between two current carrying wires is

Using

P = VI

Now, putting the values of constants in above equation

$\begin{array}{r}100\mathrm{W}=120\mathrm{V}×\mathrm{I}\\ \mathrm{I}=\frac{100}{120}\\ \mathrm{I}=0.83\mathrm{A}\end{array}$

Thus, the current through the both wires is l = 0.83A

Using

$\begin{array}{r}\frac{\mathrm{F}}{\mathrm{L}}=\frac{{\mathrm{μ}}_0{d}_2}{2\mathrm{π}d}\\ \frac{\mathrm{F}}{\mathrm{L}}=\frac{{\mathrm{μ}}_{0}d}{2\mathrm{π}d}\end{array}$

Now, putting the values of constants in above equation

$\begin{array}{r}\frac{F}{L}=\frac{4\mathrm{π}×{10}^{−7}\mathrm{T}.\mathrm{m}/\mathrm{A}×0.83\mathrm{A}×0.83\mathrm{A}}{2\mathrm{π}×3×{10}^{−3}\mathrm{m}}\\ \frac{F}{L}=4.6×{10}^{−5}\mathrm{N}/\mathrm{m}\end{array}$

Thus, The force per meter does each wire of the cord exert on the other is$4.6×{10}^{-5}N/m$ .