Question

Two parallel wires are separated by 6.10 cm, each carrying 2.85 A of current in the same direction. (a) What is the magnitude of the force per unit length between the wires

Answer 1

Answer:

The force per unit length is $2.66 \times 10^{-5} \ N/m$

Explanation:

The current carrying by each wires = 2.85 A

The current in both wires flows in same direction.

The gap between the wires = 6.10 cm

Now we will use the below expression for the force per unit length. Moreover, before using the below formula we have to change the unit centimetre into meter. So, we just divide the centimetre with 100.

$F/l = \frac{\mu _0i_1 i_2}{2\pi d} \\i_1 = 2.85 \\i_ 2 = 2.85 \\\mu _0 = 4\pi \times 10^{-7} \\d = 0.061 \\F/l = \frac{4\pi \times 10^{-7} \times 2.85 \times 2.85}{2 \pi \times 0.061} \\= 2.66 \times 10^{-5} \ N/m$