## **Problem**

Consider two current-carrying wires, one with current $ I_1 $, and one with current $ I_2 $. Using the right-hand rule, one can determine the direction of the magnetic fields across each wire. The right-hand rule is also used to determine the direction of the magnetic forces exerted on each current-carrying wire. Try it!

**Part 1:**If the currents through each wire are travelling in the same direction, will the wires attract or repel?

**Part 2:** Determine the magnitude of the magnetic force density (magnetic force per unit length) exerted on each wire.

## **Solution**

**Part 1**

Refer to the diagram on the left. Using the right-hand rule, if the two currents are travelling in the same direction, the magnetic fields will be curling in the same direction (counter-clockwise in the diagram because both currents point up). Thus the generated forces are pointing towards each other, which means the wires will attract.

**Part 2**

First determine the magnitude of the magnetic field for each wire. One can use Ampere's Law.

- $ \oint_C \mathbf{B} \cdot d\boldsymbol{l} = \mu_0 I $

- $ |\mathbf{B_1}| 2\pi r = \mu_0 I_1 $

- $ |\mathbf{B_1}| = \frac{\mu_0 I_1}{2\pi r} $

Similarly,

- $ |\mathbf{B_2}| = \frac{\mu_0 I_2}{2\pi r} $.

The magnetic force of a current-carrying wire is

- $ \mathbf{F} = I\boldsymbol{l} \times \mathbf{B} $.

The magnitude of the magnetic force is thus

- $ |\mathbf{F}| = |Il||B|\sin \theta $.

Therefore,

- $ |\mathbf{F}| = |I_1 l|\left|\frac{\mu_0 I_2}{2\pi r} \right|\sin {90}^{\circ} $

- $ \frac{|\mathbf{F}|}{l} = \frac{\mu_0 I_1 I_2}{2\pi r} $.