The environment must do work on the wire to pull it. What happens to the energy transferred to the wire by this work? Is energy conserved as the wire moves along the rail? It will be easier to answer this question if we think about power rather than work. Power is the rate at which work is done on the wire. You learned in Chapter 9 that the power exerted by a force pushing or pulling an object with velocity $v$ is $P=Fv$. The power provided to the circuit by pulling on the wire is

This is the rate at which energy is added to the circuit by the pulling force.

But the circuit also dissipates energy by transforming electric energy into the thermal energy of the wires and components, heating them up. The power dissipated by current I as it passes through resistance R is $P=I^{2}R$. Equation 30.4 for the induced current I gives us the power dissipated by the circuit of Figure 30.5:

You can see that Equations 30.6 and 30.7 are identical. The rate at which work is done on the circuit exactly balances the rate at which energy is dissipated. In other words, it’s the work done by the external pulling force that allows the charge separation, creating the emf. The magnetic field moves the charges, but the external force supplies the energy.

If you have to pull on the wire to get it to move to the right, you might think that it would spring back to the left on its own. FIGURE 30.7 shows the same circuit with the wire moving to the left. In this case, you must push the wire to the left to keep it moving. The magnetic force is always opposite to the wire’s direction of motion.

FIGURE 30.7 A pushing force is needed to move the wire to the left.

In both Figure 30.6, where the wire is pulled, and Figure 30.7, where it is pushed, a mechanical force is used to create a current. In other words, we have a conversion of mechanical energy to electric energy. A device that converts mechanical energy to electric energy is called a generator. The slide-wire circuits of Figures 30.6 and 30.7 are simple examples of a generator.

We can summarize our analysis as follows:

1. Pulling or pushing the wire through the magnetic field at speed $v$ creates a motional emf $\epsilon$ in the wire and induces a current $I=\epsilon \text{/}R$ in the circuit.

2. To keep the wire moving at constant speed, a pulling or pushing force must balance the magnetic force on the wire. This force does work on the circuit.

3. The work done by the pulling or pushing force exactly balances the energy dissipated by the current as it passes through the resistance of the circuit.

EXAMPLE 30.3 Lighting a bulb

FIGURE 30.8 shows a circuit consisting of a flashlight bulb, rated $3.0\text{V}\text{/}1.5\text{W}$, and ideal wires with no resistance. The right wire of the circuit, which is 10 cm long, is pulled at constant speed $v$ through a perpendicular magnetic field of strength 0.10 T.

FIGURE 30.8 Circuit of Example 30.3.

FIGURE 30.8 Full Alternative Text

1. What speed must the wire have to light the bulb to full brightness?

2. What force is needed to keep the wire moving?

MODEL Treat the moving wire as a source of motional emf.

VISUALIZE The magnetic force on the charge carriers, $\vec{F}{}_{\text{B}}=q\vec{v}\times \vec{B}$, causes a counterclockwise (ccw) induced current.

SOLVE

1. The bulb’s rating of $3.0\text{V}\text{/}1.5\text{W}$ means that at full brightness it will dissipate 1.5 W at a potential difference of 3.0 V. Because the power is related to the voltage and current by $P=I\text{Δ}V$, the current causing full brightness is

   $$I={\displaystyle \frac{P}{\text{Δ}V}}={\displaystyle \frac{1.5\text{W}}{3.0\text{V}}}=0.50\text{A}$$

   The bulb’s resistance—the total resistance of the circuit—is

   $$R={\displaystyle \frac{\text{Δ}V}{I}}={\displaystyle \frac{3.0\text{V}}{0.50\text{A}}}=6.0\text{Ω}$$

   Equation 30.4 gives the speed needed to induce this current:

   $$v={\displaystyle \frac{IR}{lB}}={\displaystyle \frac{(0.50\text{A})\left(6.0\text{Ω}\right)}{(0.10\text{m})(0.10\text{T})}}=300\text{m}\text{/}\text{s}$$

   You can confirm from Equation 30.6 that the input power at this speed is 1.5 W.

2. From Equation 30.5, the pulling force must be

   $$F_{\text{pull}}={\displaystyle \frac{v{l}^2{B}^2}{R}}=5.0\times {10}^{-3}\text{N}$$

   You can also obtain this result from $F_{\text{pull}}=P\text{/}v$.

REVIEW Example 30.1 showed that high speeds are needed to produce significant potential difference. Thus $300\text{m}\text{/}\text{s}$ is not surprising. The pulling force is not very large, but even a small force can deliver large amounts of power $P=Fv$ when $v$ is large.