SAT Physics Subject Test
Chapter 12 Electromagnetic Induction
In Chapter 11, we learned that electric currents generate magnetic fields. We will now see how magnetism can generate electric currents.
The figure below shows a conducting wire of length ℓ, moving with constant velocity v in the plane of the page through a uniform magnetic field B that’s perpendicular to the page. The magnetic field exerts a force on the moving conduction electrons in the wire. Using the right-hand rule, the direction of the magnetic force, FB, on these electrons (which are negatively charged) is downward.
As a result, electrons will be pushed to the lower end of the wire, which will leave an excess of positive charge at its upper end. This separation of charge creates a uniform electric field, E, within the wire, pointing downward.
A charge q in the wire feels two forces: an electric force, FE = qE, and a magnetic force of magnitude
FB = q v B
If q is negative, FE is upward and FB is downward; if q is positive, FE is downward and FB is upward. So, in both cases, the forces act in opposite directions. Once the magnitude of FE equals the magnitude of FB, the charges in the wire are in electromagnetic equilibrium. This occurs whenqE = qvB—that is, when E = vB.
The presence of the electric field creates a potential difference between the ends of the rod. Since negative charge accumulates at the lower end (which we’ll call point a) and positive charge accumulates at the upper end (point b), point b is at a higher electric potential.
The potential difference Vba is equal to Eℓ and, since E = vB, the potential difference can be written as vB ℓ.
Now imagine that the rod is sliding along a pair of conducting rails connected at the left by a stationary bar. The sliding rod now completes a rectangular circuit, and the potential difference Vba causes current to flow.
The motion of the sliding rod through the magnetic field creates an electromotive force, called motional emf:
ε= vB ℓ
The existence of a current in the sliding rod causes the magnetic field to exert a force on it. Using the formula FB = I ℓ B and the right-hand rule tells us that the direction of FB on the rod is to the left. An external agent must provide this same amount of force to the right to maintain the rod’s constant velocity and keep the current flowing. The power that the external agent must supply is P = Fv = I ℓ Bv, and the electrical power delivered to the circuit is P = IVba = I = IvBℓ. Notice that these two expressions are identical. The energy provided by the external agent is transformed first into electrical energy and then into thermal energy as the conductors making up the circuit dissipate heat.