Question

Two long, parallel wires are separated by a distance of 0.400 m shown in fig.The currentsand have the directions shown. Each current is doubled, so thatbecomes 10.0 A andbecomes 4.00 A. Now what is the magnitude of the force that each wire exerts on a 1.20-m length of the other?

Short answer

The magnitude of the force that each wire exerts on a 1.20-m length of the other is $24.0\times {10}^{-6}N$ .

Step-by-step solution

Step 1: The force per unit length between two parallel current carrying wires

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

$\frac{\mathrm{F}}{\mathrm{L}}=\frac{{\mu }_0{\mathrm{d}}_2}{2\pi d}$

Where, ${\mu }_0$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.

Step 2:  The calculation of magnitude of the force that each wire exerts on a 1.20m  length of the other

Given : Distance between parallel wires is d = 0.400m

The currents in wires are $l_1=10.0A$and $l_2=4.00A$

Using formula

$\frac{\mathrm{F}}{\mathrm{L}}=\frac{{\mu }_0{\parallel }_2}{2\pi d}$

Now, putting the values of constants in above equation

$\begin{array}{r}\frac{F}{1.20m}=\frac{4\pi \times {10}^{-7}\times 10.0\times 4.00}{2\pi \left(0.400\right)}\\ F=2.00\times {10}^{-8}\times 1.20\\ F=24.0\times {10}^{-8}\mathrm{N}\end{array}$

Thus, the magnitude of the force that each wire exerts on a 1.20-m length of the other is $24.0\times {10}^{-6}\mathrm{N}$.