Question

Two long parallel wires carry currents of 3.05 A 3.05 A and 6.69 A 6.69 A . The magnitude of the force per unit length acting on each wire is 7.33 × 10 − 5 N / m 7.33×10−5 N/m . Find the separation distance d d of the wires expressed in millimeters.

Answer 1

Answer:

The separation between wires is 55.6 mm.

Explanation:

Given that,

Current in wire 1, $I_1=3.05\ A$

Current in wire 2, $I_2=6.69\ A$

The magnitude of the force per unit length acting on each wire is, $\dfrac{F}{l}=7.33\times 10^{-5}\ N/m$

We know that the formula of magnitude of the force per unit length is given by :

$\dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi d}$

Here, d is the separation between wires.

$d=\dfrac{\mu_o I_1I_2}{2\pi (F/l)}\\\\d=\dfrac{4\pi \times 10^{-7}\times 3.05\times 6.69}{2\pi \times 7.33\times 10^{-5}}\\\\d=0.0556\ m\\\\d=55.6\ mm$

So, the separation between wires is 55.6 mm.