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United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How do induced currents create forces, and how does energy conservation govern them?

Topic 13.3 Induced Currents and Magnetic Forces: analyze the forces on induced currents, the energy and power in induction, and eddy-current effects.

A calculus-based answer to AP Physics C E&M Topic 13.3, covering the force on an induced current, the energy balance of a sliding rod, the power dissipated, eddy currents and magnetic braking.

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  1. What this topic is asking
  2. The force on an induced current
  3. The sliding rod and energy balance
  4. Eddy currents and braking
  5. Try this

What this topic is asking

The College Board (Topic 13.3) wants you to analyze the forces that act on induced currents, track the energy and power in an induction setup, and understand eddy currents and magnetic braking. The unifying idea is that induction always opposes its cause, so external work is needed and it becomes electrical (then thermal) energy.

The force on an induced current

Once a changing flux induces a current, that current flows through a magnetic field and feels a force F⃗=IL⃗×B⃗\vec{F} = I\vec{L}\times\vec{B} (magnitude F=BILF = BIL). Lenz's law guarantees this force opposes the change: a rod pushed to widen a loop feels a force resisting its motion, and a coil entering a field feels a force pushing it out. The induced effects always fight back.

The sliding rod and energy balance

The canonical example is a conducting rod sliding on rails in a perpendicular field:

  • EMF: the growing enclosed area gives ε=BLv\varepsilon = BLv.
  • Current: I=εR=BLvRI = \dfrac{\varepsilon}{R} = \dfrac{BLv}{R}.
  • Retarding force: F=BIL=B2L2vRF = BIL = \dfrac{B^2 L^2 v}{R}, opposing the motion.

To move the rod at constant speed, an applied force must balance this. The mechanical power it supplies is

Pmech=Fv=B2L2v2RP_{mech} = Fv = \frac{B^2 L^2 v^2}{R}

and the electrical power dissipated in the resistance is

Pelec=I2R=(BLvR)2R=B2L2v2RP_{elec} = I^2 R = \left(\frac{BLv}{R}\right)^2 R = \frac{B^2 L^2 v^2}{R}

These are equal: the mechanical work done against the magnetic braking force becomes electrical energy, then heat. This is exactly where induced electrical energy comes from.

Eddy currents and braking

In a solid conductor moving through a changing field (rather than a thin wire loop), the induced currents swirl in closed loops called eddy currents. They obey Lenz's law too, opposing the relative motion, so a metal plate swinging through a magnetic field is rapidly damped (magnetic braking, used in trains and exercise machines). The eddy currents dissipate the kinetic energy as heat, the same principle that warms a pan on an induction cooktop.

Try this

Q1. A rod carries an induced current of 2.02.0 A, has length 0.250.25 m, and sits in a 0.500.50 T field. Find the force on it. [2 points]

  • Cue. F=BIL=(0.50)(2.0)(0.25)=0.25F = BIL = (0.50)(2.0)(0.25) = 0.25 N, opposing the motion.

Q2. State where the mechanical work done on a rod sliding at constant speed ends up. [1 point]

  • Cue. Dissipated as heat in the circuit's resistance (the induced current's I2RI^2 R loss).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I (multiple choice). A conducting rod slides along frictionless rails at constant velocity in a magnetic field, driven by an applied force. The applied force does work that becomes (A) kinetic energy of the rod (B) magnetic potential energy (C) electrical energy dissipated in the circuit's resistance (D) zero. Justify your reasoning.
Show worked answer →

A 1-point MCQ on the energy balance of a sliding rod. The answer is (C).

At constant velocity the kinetic energy does not change, so the applied force's work cannot go to (A). There is no magnetic potential energy store, ruling out (B). The induced current dissipates energy in the resistance as heat, P=I2RP = I^2 R, and the applied work exactly supplies it. The trap is (A): kinetic energy is constant here.

AP 2024 (style)6 marksSection II (FRQ, quantitative). A rod of length L=0.40L = 0.40 m and the circuit's resistance R=0.50 ΩR = 0.50\,\Omega slide on rails at constant v=3.0v = 3.0 m/s in a 0.250.25 T field perpendicular to the plane. (a) Calculate the motional EMF. (b) Calculate the induced current and the force needed to keep the rod moving. (c) Show the mechanical power input equals the electrical power dissipated.
Show worked answer →

A 6-point FRQ on the force and energy of a sliding rod.

(a) EMF (1 point): ε=BLv=(0.25)(0.40)(3.0)=0.30\varepsilon = BLv = (0.25)(0.40)(3.0) = 0.30 V.
(b) Current and force (3 points): I=εR=0.300.50=0.60I = \dfrac{\varepsilon}{R} = \dfrac{0.30}{0.50} = 0.60 A. The field exerts a retarding force F=BIL=(0.25)(0.60)(0.40)=0.060F = BIL = (0.25)(0.60)(0.40) = 0.060 N; to keep constant velocity the applied force equals this, 0.0600.060 N.
(c) Power balance (2 points): mechanical P=Fv=(0.060)(3.0)=0.18P = Fv = (0.060)(3.0) = 0.18 W; electrical P=I2R=(0.60)2(0.50)=0.18P = I^2 R = (0.60)^2(0.50) = 0.18 W. They match, as energy conservation requires.

Markers reward the EMF, the current and balancing force, and the equal mechanical and electrical powers.

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