12.3 Magnetism and Current-Carrying Wires

Magnetic Field Produced by a Current-Carrying Wire

Current-carrying wires generate magnetic fields around them, creating one of the most fundamental relationships in electromagnetism. This relationship demonstrates how moving electric charges produce magnetic effects. 🧲

• Magnetic field vectors around a long, straight, current-carrying wire form concentric circles perpendicular to the wire
• Field vectors are tangent to these circles at every point
• No field component points toward, away from, or parallel to the wire itself

The magnetic field strength at any point near a long, straight, current-carrying wire depends on two key factors: the current magnitude and the distance from the wire.

• Field magnitude is directly proportional to current (double the current, double the field strength)

• Field magnitude is inversely proportional to perpendicular distance from wire's central axis (double the distance, half the field strength)

• Calculate using the equation: $B=\frac{\mu_{0}}{2 \pi} \frac{I}{r}$

  Where:

  • $B$ = magnetic field strength (in teslas, T)
  • $\mu_0$ = permeability of free space constant ($4π × 10^{-7}$ T·m/A)
  • $I$ = current in the wire (in amperes, A)
  • $r$ = perpendicular distance from wire (in meters, m)

To determine the direction of a current-carrying wire's magnetic field, we use the right-hand rule:

1. Point your thumb in the direction of conventional current (positive to negative)
2. Your fingers naturally curl in the direction of the magnetic field
3. This creates a circular pattern around the wire

For a current loop (like a circular wire), the magnetic field at its center points along the loop's axis. Again, the right-hand rule helps determine the specific direction - curl your fingers in the direction of the current flow, and your thumb points in the field direction at the center.

When multiple current-carrying wires are present in a region, the net magnetic field at any location is found by vector addition of the individual fields from each wire. This means accounting for both magnitude and direction of each contributing field.

Force on a Current-Carrying Wire in a Magnetic Field

When a current-carrying wire is placed in an external magnetic field, it experiences a force. This force is the basis for electric motors and many other electromagnetic devices. ⚡

The force magnitude depends on four critical factors:

1. Current in the wire (more current means more force)
2. Length of wire within the magnetic field (longer wire means more force)
3. Magnetic field strength (stronger field means more force)
4. Angle between the current direction and magnetic field direction (maximum force when perpendicular)

This relationship is quantified by the equation:

$F_{B}=I \ell B \sin \theta$

Where:

• $F_B$ = magnetic force on the wire (in newtons, N)
• $I$ = current in the wire (in amperes, A)
• $\ell$ = length of wire in the magnetic field (in meters, m)
• $B$ = external magnetic field strength (in teslas, T)
• $\theta$ = angle between current and field directions (in degrees or radians)

The force is maximum when the current and magnetic field are perpendicular ($\theta = 90°$, $\sin 90° = 1$), and zero when they're parallel ($\theta = 0°$, $\sin 0° = 0$).

To determine the direction of the magnetic force on the current-carrying wire, we use another version of the right-hand rule:

1. Point your fingers in the direction of the conventional current
2. Orient your palm facing the direction of the magnetic field
3. Your extended thumb now points in the direction of the force

This rule helps visualize the three-dimensional relationship between current, field, and force, which are all perpendicular to each other in the maximum force scenario.

Practice Problem 1: Magnetic Field from a Wire

Solution

To find the magnetic field strength at a point near a long, straight current-carrying wire, we use the equation:

$B=\frac{\mu_{0}}{2 \pi} \frac{I}{r}$

Given:

• Current, $I = 5.0$ A
• Distance from wire, $r = 10$ cm = 0.10 m
• Permeability of free space, $\mu_0 = 4\pi \times 10^{-7}$ T·m/A

Substituting these values:

$B=\frac{4\pi \times 10^{-7}}{2 \pi} \frac{5.0}{0.10}$

$B=\frac{4\pi \times 10^{-7}}{2 \pi} \times 50$

$B= 2 \times 10^{-7} \times 50$

$B= 1.0 \times 10^{-5}$ T

The magnetic field at a distance of 10 cm from the wire is $1.0 \times 10^{-5}$ T or 10 μT.

Practice Problem 2: Force on a Current-Carrying Wire

Solution

To find the magnetic force on a current-carrying wire in a magnetic field, we use the equation:

$F_{B}=I \ell B \sin \theta$

Given:

• Current, $I = 3.0$ A
• Wire length, $\ell = 25$ cm = 0.25 m
• Magnetic field strength, $B = 0.50$ T
• Angle between current and field, $\theta = 30°$

Substituting these values:

$F_{B}= 3.0 \times 0.25 \times 0.50 \times \sin 30°$

$F_{B}= 3.0 \times 0.25 \times 0.50 \times 0.5$

$F_{B}= 0.1875$ N

The magnetic force on the wire is 0.1875 N or approximately 0.19 N.