Question

Two current-carrying wires are exactly parallel to one another and both carry 2.5A of current. The two wires are separated by a distance of 15cm. The current in wire 1 moves down and the current in wire 2 also moves down. What is the magnitude of the magnetic force per unit length caused by wire 1 on wire 2. What is the direction of the magnetic force caused by wire 2 on wire 1.

Answer 1

1) Magnitude per unit length: $8.3\cdot 10^{-6} N$

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

$\frac{F}{\Delta L}=\frac{\mu_0 I_1 I_2}{2 \pi r}$

where

$\mu_0 = 4\pi \cdot 10^{-7} Tm/A$ is the vacuum permeability

$I_1 =I_2 =2.5 A$ is the current in each wire

$r=15 cm=0.15 m$ is the distance between the two wires

Substituting the numbers into the equation, we find

$\frac{F}{\Delta L}=\frac{(4\pi \cdot 10^{-7} Tm/A)(2.5 A)(2.5 A)}{2 \pi (0.15 m)}=8.3\cdot 10^{-6} N$

2)  direction of the force: attractive

First of all, let's analyze what is the direction of the magnetic field produced by wire 2 at the location of wire 1. Assume that wire 2 is on the left of wire 1. The direction of the current for wire 2 is down, so by using the right-hand rule, we see that the direction of the magnetic field at the location of wire 2 is south.

Now we apply the right-hand rule on wire 2, to find the direction of the force:

- current: down (index finger)

- magnetic field: south (middle finger)

- force: to the left (thumb)

So, the force exerted by wire 2 on wire 1 is towards wire 2 (attractive force)