Physics /Electromagnetism /Module 1.2

Magnetic Force on Current-Carrying Conductors

Explore how microscopic electron drift rates sum up to drive macroscopic conductor deflection and levitation under the Laplace force.

When a single charged particle moves through a magnetic field, it experiences the Lorentz force. But what happens when trillions of these charges are locked inside a wire? Instead of escaping, their collective microscopic deflections transfer momentum directly to the metal lattice, causing the entire macroscopic conductor to bend, jump, or levitate.

Oersted's Discovery and RHR-2

Before calculating the magnetic force acting on a current, we must examine the connection Hans Oersted discovered in 1819: electric currents produce their own magnetic fields. A current-carrying conductor generates circular loops of magnetic field lines concentric to the wire.

To determine the direction of this current-generated field, we use the Second Right-Hand Rule (RHR-2). Point your right thumb in the direction of the electric current $I$ , and your fingers will naturally wrap around the wire in the direction of the circular magnetic field lines $\vec{B}$ .

FIGURE 2.1: CONCENTRIC MAGNETIC FIELD LINES (RHR-2) Hover to Animate

By convention, field vectors pointing directly out of the page are represented by a dot ( $\cdot$ ) representing the tip of an incoming arrow, while vectors pointing into the page are represented by a cross ( $\times$ ) representing the tail feathers of an arrow flying away.

Summing Microscopic Forces

Electric current $I$ is not a single entity; it is the ordered drift of individual charge carriers (free electrons in metals) moving with drift velocity $\vec{v}_d$ . If a wire of length $L$ contains $n$ free charge carriers per unit volume, the total number of moving charges in that wire segment is:

N = n A L

where $A$ is the cross-sectional area of the wire. Each individual charge carrier experiences a microscopic Lorentz force $\vec{F}_{micro} = q(\vec{v}_d \times \vec{B})$ . Summing these microscopic forces across all charge carriers gives the net macroscopic force acting on the wire segment:

\vec{F} = N \vec{F}_{micro} = (n A L) q (\vec{v}_d \times \vec{B})

By grouping the variables, we notice that the term $I = n q A v_d$ is precisely the definition of electric current! Substituting this current into the force sum yields the classical macroscopic equation.

FIGURE 2.2: MICROSCOPIC DRIFT & MOMENTUM TRANSFER Hover to Animate

As visualized in Figure 2.2, conventional current flows to the right, which means the actual charge carriers (electrons) are drifting to the left. When the magnetic field points into the page, the Right-Hand Rule dictates that these leftward-drifting negative electrons experience an upward Lorentz force. As the electrons are deflected, they collide with the positive copper ions of the solid lattice. This continuous, microscopic momentum transfer acts as a vertical drag, pulling the entire physical wire upward—manifesting as the macroscopic Laplace force.

Deflecting the Conductor

To determine the direction of the magnetic force on a wire, we once again employ the Right-Hand Rule (RHR-1). Point your flat fingers in the direction of the current flow $I$ (positive charge drift), curl them in the direction of the magnetic field lines $\vec{B}$ , and your thumb will point in the direction of the resulting macroscopic force $\vec{F}$ .

This physical principle has major real-world applications. By suspending a wire horizontally in a perpendicular magnetic field, we can run a current through it to generate an upward magnetic force. If this magnetic force perfectly matches the downward force of gravity, the wire will float in place, reducing the mechanical tension in its supporting leads to absolute zero.

Closed Loops & Uniform Fields

An important theoretical baseline occurs when current flows in a closed wire loop inside a uniform magnetic field. If you integrate the force vector around any closed shape:

\vec{F}_{net} = \oint I (d\vec{l} \times \vec{B}) = I \left( \oint d\vec{l} \right) \times \vec{B} = 0

Because the closed path vector sum $\oint d\vec{l}$ is always exactly zero (you end where you started), the net translation force on any closed loop in a uniform field is zero.

However, do not let this deceive you. While the loop cannot fly off in a straight line, different segments of the loop experience forces in opposite directions. This does not push the loop, but it creates a net torque , causing the loop to spin. This torque forms the bedrock mechanics behind DC electric motors.

Play with the loop below. Notice how the red force arrows ALWAYS point in opposite directions, canceling out translational movement, but their leverage creates rotational torque.

The Geometry of Torque

Why does the loop spin? In the simulation above, look at the flat state ( $90^\circ$ ). There are two forces pulling on opposite sides of the wire. From basic mechanics, a twisting force (Torque, $\tau$ ) is calculated by multiplying the Force by the distance between them (the width of the loop, $w$ ).

We know the magnetic force on one wire segment is $F = I L B$ . If we multiply that force by the width of the loop to get the torque, we get:

\tau = (I L B) \times w = I (L \cdot w) B

Notice the geometry hidden in the math: Length ( $L$ ) multiplied by width ( $w$ ) is simply the Area ( $A$ ) of the loop! The equation simplifies beautifully to $\tau = I A B$ .

The Magnetic Dipole Moment ( $\mu$ )

Because the combination of "Current $\times$ Area" universally determines how much twisting leverage a loop has, physicists gave it a special name: the Magnetic Dipole Moment ( $\vec{\mu}$ ). If the coil has multiple turns of wire ( $N$ ), we multiply by that as well:

\mu = N I A

The direction of $\vec{\mu}$ points perfectly perpendicular to the surface area of the loop (the blue arrow in the simulator). With this vector defined, the total torque acting on any loop at any angle is simply a cross product:

\vec{\tau} = \vec{\mu} \times \vec{B} \quad \implies \quad |\tau| = \mu B \sin\theta

Concept Checks

Balancing Gravity

A copper wire segment of length

L = 0.50\text{ m}

and mass

m = 10\text{ g}

is suspended horizontally in a uniform magnetic field

B = 0.50\text{ T}

pointing perfectly into the page. What current is required to perfectly balance the gravitational pull? (Use

g = 9.8\text{ m/s}^2

)

Parallel Force Dropoff

A conductor wire carrying a powerful 10.0 A current is laid perfectly parallel to a very strong 3.0 Tesla magnetic field. What is the magnetic force per unit length acting on the wire?

Flexible Wire Deflection

A straight, flexible copper wire is suspended horizontally inside a uniform magnetic field pointing vertically into the page. If a strong electric current is run through the wire from left to right (

+x

-direction), which way will the wire bend? What if the current is reversed?

Calculating Dipole Moment

A circular coil of wire with 50 turns has a radius of

0.10\text{ m}

. It carries a current of

2.0\text{ A}

. What is the magnitude of its magnetic dipole moment?

Torque Orientations

A rectangular current loop is placed in a uniform magnetic field. At what angle between the loop's surface normal (

\vec{\mu}

) and the magnetic field (

\vec{B}

) does the loop experience maximum torque? At what angle does it experience zero torque?