58 Motional Emf
Learning Objective
By the end of this section, you will be able to:
- Calculate emf, force, magnetic field, and work due to the motion of an object in a magnetic field.
As we have seen, any change in magnetic flux induces an emf opposing that change—a process known as induction. Motion is one of the major causes of induction. For example, a magnet moved toward a coil induces an emf, and a coil moved toward a magnet produces a similar emf. In this section, we concentrate on motion in a magnetic field that is stationary relative to the Earth, producing what is loosely called motional emf. One situation where motional emf occurs is known as the Hall effect and has already been examined. Charges moving in a magnetic field experience the magnetic force F = qvB sin θ, which moves opposite charges in opposite directions and produces an em f = Bℓv. We saw that the Hall effect has applications, including measurements of B and v. We will now see that the Hall effect is one aspect of the broader phenomenon of induction, and we will find that motional emf can be used as a power source. Consider the situation shown in Figure 1. A rod is moved at a speed v along a pair of conducting rails separated by a distance ℓ in a uniform magnetic field B. The rails are stationary relative to B and are connected to a stationary resistor R. The resistor could be anything from a light bulb to a voltmeter. Consider the area enclosed by the moving rod, rails, and resistor. B is perpendicular to this area, and the area is increasing as the rod moves. Thus the magnetic flux enclosed by the rails, rod, and resistor is increasing. When flux changes, an emf is induced according to Faraday’s law of induction.
To find the magnitude of emf induced along the moving rod, we use Faraday’s law of induction without the sign:
[latex]\text{emf} = \text{N}\frac{\Delta\Phi}{\Delta t}\\[/latex].
Here and below, “emf” implies the magnitude of the emf. In this equation, N = 1 and the flux Φ = BA cos θ. We have θ = 0º and cos θ = 1, since B is perpendicular to A . Now ΔΦ = Δ(BA) = BΔA, since B is uniform. Note that the area swept out by the rod is ΔA = ℓΔx. Entering these quantities into the expression for emf yields
[latex]\text{emf}=\frac{B\Delta A}{\Delta t}=B\frac{\ell\Delta x}{\Delta t}\\[/latex].
Finally, note that Δx/Δt = v, the velocity of the rod. Entering this into the last expression shows that
emf = Bℓv (B,ℓ, and v perpendicular)
is the motional emf. This is the same expression given for the Hall effect previously.
Making Connections: Unification of Forces
To find the direction of the induced field, the direction of the current, and the polarity of the induced emf, we apply Lenz’s law as explained in Faraday’s Law of Induction: Lenz’s Law. (See Figure 1(b).) Flux is increasing, since the area enclosed is increasing. Thus the induced field must oppose the existing one and be out of the page. And so the RHR-2 requires that I be counterclockwise, which in turn means the t