# Physics Magnetism?

## 1.) In this question, you will calculate the magnetic force between two current carrying wires.Two long parallel wires, each carrying a current of 25mA, are held 130mm apart in vacuum, The currents in the wires flow in opposite directions. What is the force per unit length on each wire due to the other? Take the magnetic permeability of free space to be μ0= 4πx10-7. 2.) Is this force between the wires attractive or repulsive ?

Mar 3, 2018

This is a complicated answer. The force will be repulsive since the currents are moving in opposite directions.

#### Explanation:

First, we want to determine the magnitude of the magnetic field, B.

B={\mu_0*I}/{2\pir}->{4\pi*10^-7*2.5x10^-2A}/ [2*\pi*1.3x10^-1m}->{5x10^-9}/{1.3*10^-1}T

$B = 3.84 x {10}^{-} 8 T$

Then, the force will be ${F}_{M} = I \cdot L x B$ or ${F}_{M} = I \cdot L \sin \left(\setminus \Theta\right) B$

Assuming $\setminus \Theta$ = 90 degrees, that simplifies the equation to
${F}_{M} = I \cdot L \cdot B$

So, ${F}_{M} = 2.5 x {10}^{-} 2 A \cdot L \cdot 3.84 x {10}^{-} 8 T$

${F}_{M} = 9.61 x {10}^{-} 10 \cdot L$

These two forces are going to be equal, so you can have ${F}_{12} \mathmr{and} {F}_{21}$

Using the right hand rule, these forces will point outward from each other, so therefore the repulsion.

I hope that helps!