Ampère’s observation of magnetic forces between current-carrying wires motivated us to look at the magnetic forces on moving charges. We’re now ready to apply that knowledge to Ampère’s experiment. As a first step, let us find the force exerted by a uniform magnetic field on a long, straight wire carrying current I through the field. As FIGURE 29.42a shows, there’s no force on a current-carrying wire parallel to a magnetic field. This shouldn’t be surprising; it follows from the fact that there is no force on a charged particle moving parallel to $\vec{B}$.

FIGURE 29.42 Magnetic force on a current-carrying wire.

FIGURE 29.42b shows a wire perpendicular to the magnetic field. By the right-hand rule, each charge in the current has a force of magnitude $qvB$ directed to the left. Consequently, the entire length of wire within the magnetic field experiences a force to the left, perpendicular to both the current direction and the field direction.

A current is moving charge, and the magnetic force on a current-carrying wire is simply the net magnetic force on all the charge carriers in the wire. FIGURE 29.43 shows a wire of length l carrying current I and a small segment of length $\text{Δ}x$ in which the charge carriers move with drift velocity $\vec{v}{}_{\text{d}}$. Suppose this small segment of the wire contains charge $\text{Δ}q$. The charge moves through the segment in a time interval $\text{Δ}t=\text{Δ}x/v_{\text{d}}$, during which it experiences a magnetic force—perpendicular to the wire—of magnitude $F=\text{Δ}q{v}_{\text{d}}B$. If we multiply and divide by $\text{Δ}t$, we can write the force on the moving charge in this segment as

In the last step we used the facts that $\text{Δ}q/\text{Δ}t$ is the current I and $v_{\text{d}}\text{Δ}t$ is the segment length $\text{Δ}x$.

FIGURE 29.43 The force on a current is the force on the charge carriers.

If we add up the forces on all the small segments of wire, the force on length l of the wire—if it’s perpendicular to the magnetic field—is $F=IlB$. More generally, the force on length l of a current-carrying wire is

where $\alpha$ is the angle between $\vec{l}$ (the direction of the current) and $\vec{B}$. As an aside, you can see from Equation 29.26 that the magnetic field B must have units of $\text{N}\text{/}\text{A}\text{m}$. This is why we defined $1\text{T}=1\text{N}\text{/}\text{A}\text{m}$ in Section 29.3.

EXAMPLE 29.13 Magnetic levitation

The 0.10 T uniform magnetic field of FIGURE 29.44 is horizontal, parallel to the floor. A straight segment of 1.0-mm-diameter copper wire, also parallel to the floor, is perpendicular to the magnetic field. What current through the wire, and in which direction, will allow the wire to “float” in the magnetic field?

FIGURE 29.44 Magnetic levitation.

FIGURE 29.44 Full Alternative Text

MODEL The wire will float in the magnetic field if the magnetic force on the wire points upward and has magnitude mg, allowing it to balance the downward gravitational force.

SOLVE We can use the right-hand rule to determine which current direction experiences an upward force. With $\vec{B}$ pointing away from us, the direction of the current needs to be from left to right. The forces will balance when

$$F=IlB=mg=\rho (\pi r^{2}l)g$$

where $\rho =8920\text{kg}\text{/}{\text{m}}^3$ is the density of copper. The length of the wire cancels, leading to

$$\begin{array}{ccc}\hfill I& \hfill =\hfill & {\displaystyle \frac{\rho \pi r^{2}g}{B}}={\displaystyle \frac{\left(8920\text{kg}\text{/}{\text{m}}^3\right)\pi {\left(0.00050\text{m}\right)}^2\left(9.80\text{m}\text{/}{\text{s}}^2\right)}{0.10\text{T}}}\hfill \\ \hfill & \hfill =\hfill & 0.69\text{ A}\hfill \end{array}$$

A 0.69 A current from left to right will levitate the wire in the magnetic field.

REVIEW A 0.69 A current is quite reasonable, but this idea is useful only if we can get the current into and out of this segment of wire. In practice, we could do so with wires that come in from below the figure. These input and output wires would be parallel to $\vec{B}$ and not experience a magnetic force. Although this example is very simple, it is the basis for applications such as magnetic levitation trains.