We’ll start our investigation of electromagnetic induction by looking at situations in which the magnetic field is fixed while the circuit moves or changes. Consider a conductor of length $l$ that moves with velocity $\vec{v}$ through a perpendicular uniform magnetic field $\vec{B}$, as shown in FIGURE 30.2. The charge carriers inside the wire—assumed to be positive—also move with velocity $\vec{v}$, so they each experience a magnetic force $\vec{F}{}_{\text{B}}=q\vec{v}\times \vec{B}$ of strength $F_{\text{B}}=qvB$. This force causes the charge carriers to move, separating the positive and negative charges. The separated charges create an electric field inside the conductor.

FIGURE 30.2 The magnetic force on the charge carriers in a moving conductor creates an electric field inside the conductor.

The charge carriers continue to separate until the electric force $F_{\text{E}}=qE$ exactly balances the magnetic force $F_{\text{B}}=qvB$, creating an equilibrium situation. This balance happens when the electric field strength is

In other words, the magnetic force on the charge carriers in a moving conductor creates an electric field $\mathit{\text{E}}\mathbf{=}\mathit{v}\mathit{B}$ inside the conductor.

The charge separation also creates an electric potential difference between the two ends of the moving conductor. FIGURE 30.3a defines a coordinate system in which $\vec{E}=-vBĵ$. Using the connection between the electric field and the electric potential,

FIGURE 30.3 Generating an emf.

Full Alternative Text

Thus the motion of the wire through a magnetic field induces a potential difference $\mathit{\text{vlB}}$ between the ends of the conductor.

There’s an important analogy between this potential difference and the potential difference of a battery. FIGURE 30.3b reminds you that a battery uses a nonelectric force—the charge escalator—to separate positive and negative charges. The emf $\epsilon$ of the battery was defined as the work performed per charge $\left(W\text{/}q\right)$ to separate the charges. An isolated battery, with no current, has a potential difference $\text{Δ}{V}_{\text{bat}}=\epsilon$. We could refer to a battery, where the charges are separated by chemical reactions, as a source of chemical emf.

The moving conductor also develops a potential difference because of the work done to separate the charges—although what’s doing the work is a subtle issue because you learned in Chapter 29 that magnetic fields do no work. We’ll return to this later in the section. Regardless of what does the work, you can think of the moving conductor as a “battery” that stays charged only as long as it keeps moving but “runs down” if it stops. The emf of the conductor is due to its motion, rather than to chemical reactions inside, so we can define the motional emf of a conductor moving with velocity $\vec{v}$ perpendicular to a magnetic field $\vec{B}$ to be

EXAMPLE 30.2 Potential difference along a rotating bar

A metal bar of length $l$ rotates with angular velocity $\omega$ about a pivot at one end of the bar. A uniform magnetic field $\vec{B}$ is perpendicular to the plane of rotation. What is the potential difference between the ends of the bar?

VISUALIZE FIGURE 30.4 is a pictorial representation of the bar. The magnetic forces on the charge carriers will cause the outer end to be positive with respect to the pivot.

FIGURE 30.4 Pictorial representation of a metal bar rotating in a magnetic field.

FIGURE 30.4 Full Alternative Text

SOLVE Even though the bar is rotating, rather than moving in a straight line, the velocity of each charge carrier is perpendicular to $\vec{B}$. Consequently, the electric field created inside the bar is exactly that given in Equation 30.1, $E=vB$. But $v$, the speed of the charge carrier, now depends on its distance from the pivot. Recall that in rotational motion the tangential speed at radius r from the center of rotation is $v=\omega r$. Thus the electric field at distance r from the pivot is $E=\omega rB$. The electric field increases in strength as you move outward along the bar.

The electric field $\vec{E}$ points toward the pivot, so its radial component is $E_r=-\omega rB$. If we integrate outward from the center, the potential difference between the ends of the bar is

$$\begin{array}{ccc}\hfill \text{Δ}V& \hfill =\hfill & V_{\text{tip}}-V_{\text{pivot}}=-{\displaystyle {\int }_0^l}{E}_{r}dr\hfill \\ \hfill & \hfill =\hfill & -{\displaystyle {\int }_0^l}(-\omega rB)dr=\omega B{\displaystyle {\int }_0^l}rdr=\frac{1}{2}\omega l^{2}B\hfill \end{array}$$

REVIEW $\frac{1}{2}\omega l$ is the speed at the midpoint of the bar. Thus $\text{Δ}V$ is $v_{\text{mid}}lB$, which seems reasonable.