Unit 13 · Topic 13.3

13.3 Induced Currents and Magnetic Forces

When a conductor moves through a magnetic field, an induced current flows, and then the field pushes on that current. This topic covers motional emf, the braking force on sliding bars and moving loops, and how Newton's second law describes their motion.

Key terms

motional emf
induced current
magnetic braking
sliding bar on rails
eddy currents

Motional emf

Picture a metal bar of length $\ell$ sliding at speed v along two rails in a uniform field B that's perpendicular to the rails' plane. The rails and a resistor R close the circuit. As the bar moves, the loop's area grows by $\ell v$ each second, so the flux changes at $\dfrac{d\Phi_B}{dt} = B\ell v$ . The induced emf is $\mathcal{E} = B\ell v$ , and the current is $I = \dfrac{B\ell v}{R}$ .

You can also see it from 12.2: the free charges in the bar move with it, so each feels a force $q\vec{v} \times \vec{B}$ along the bar. That push along the bar is the emf. Both views give the same answer.

The braking force

The induced current flows through the bar, which sits in the field, so the field pushes on it: $F = I\ell B = \dfrac{B^2\ell^2 v}{R}$ . By Lenz's law, this force always opposes the motion. It's called magnetic braking.

The force grows with speed, which is just like air resistance. It's stronger with a stronger field or a longer bar, and weaker with more resistance. With an open circuit (R infinite), there's an emf but no current and no braking force.

Motion and energy

With no other force, Newton's second law gives $m\dfrac{dv}{dt} = -\dfrac{B^2\ell^2}{R}v$ . Separating variables gives $v(t) = v_0 e^{-t/\tau}$ with $\tau = \dfrac{mR}{B^2\ell^2}$ : the bar slows exponentially and never quite stops.

If a constant force pulls the bar, or gravity pulls it down vertical rails, the speed rises until the braking force balances that force. That's terminal velocity: for a bar falling on vertical rails, $mg = \dfrac{B^2\ell^2 v_t}{R}$ , so $v_t = \dfrac{mgR}{B^2\ell^2}$ .

Energy is conserved throughout. At constant speed, the power from the applied force, Fv, equals the power the resistor dissipates, $I^2R$ .

Loops entering a field and eddy currents

A rectangular loop moving into a region of uniform field has a changing flux only while it's crossing the boundary. The field pushes only on the parts of the loop that are inside it. The forces on the two sides that are partly in the field point in opposite directions and cancel, so the net force comes from the leading side, and it opposes the motion. While the loop is completely inside a uniform field, the flux is constant, so there's no current and no force. As it leaves, current flows the other way and the force again opposes the motion.

The same forces can twist a loop instead of pushing it. When a coil spins in a field, as in a generator (13.2), the forces on its sides make a torque that opposes the spin. That's why it takes work to turn a generator, and why it's harder to turn when more current is drawn from it.

In a solid sheet of metal moving through a field, induced currents swirl in loops called eddy currents. They create braking forces too. Eddy current brakes slow some trains and roller coasters, and a magnet dropped through a copper pipe falls slowly for the same reason.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
A bar pulled at constant speed
A 0.50 m metal bar slides at a steady 4.0 m/s along rails connected by a 2.0 Ω resistor. A uniform 0.80 T field points perpendicular to the plane of the rails. Find the induced emf, the current, the magnetic force on the bar, and the power needed to keep it moving.
Show the solutionHide the solution
1. Step 1: $\mathcal{E} = B\ell v = (0.80)(0.50)(4.0) = 1.6\text{ V}$ .
2. Step 2: $I = \dfrac{\mathcal{E}}{R} = \dfrac{1.6}{2.0} = 0.80\text{ A}$ .
3. Step 3: $F = I\ell B = (0.80)(0.50)(0.80) = 0.32\text{ N}$ , opposing the motion.
4. Step 4: To keep the speed constant, the pulling force must also be 0.32 N, so the power is $Fv = (0.32)(4.0) = 1.28\text{ W}$ . Check: $I^2R = (0.80)^2(2.0) = 1.28\text{ W}$ . The energy you supply becomes thermal energy in the resistor.
Answer: 1.6 V, 0.80 A, 0.32 N opposing the motion, 1.28 W
Example 2Calculator allowed
Terminal speed on vertical rails
A 0.020 kg bar of length 0.25 m slides without friction down vertical rails connected at the top by a 0.50 Ω resistor. A 0.40 T horizontal field is perpendicular to the plane of the rails. Find the bar's terminal speed.
Show the solutionHide the solution
1. Step 1: As the bar falls, the braking force $\dfrac{B^2\ell^2v}{R}$ points up and grows with speed.
2. Step 2: At terminal speed, it balances gravity: $mg = \dfrac{B^2\ell^2v_t}{R}$ .
3. Step 3: $v_t = \dfrac{mgR}{B^2\ell^2} = \dfrac{(0.020)(9.8)(0.50)}{(0.40)^2(0.25)^2} = \dfrac{0.098}{0.010} = 9.8\text{ m/s}$ .
Answer: 9.8 m/s
Example 3Calculator allowed
A bar coasting to a stop (classic trap)
A bar of mass m and length $\ell$ on frictionless horizontal rails is given speed $v_0$ and released. The rails are connected by a resistor R, and a uniform field B is perpendicular to their plane. Find v(t).
Show the solutionHide the solution
1. Step 1: The only horizontal force is magnetic braking: $m\dfrac{dv}{dt} = -\dfrac{B^2\ell^2}{R}v$ .
2. Step 2: Separate and integrate: $\displaystyle\int_{v_0}^{v} \frac{dv'}{v'} = -\frac{B^2\ell^2}{mR}\int_0^t dt'$ , so $\ln\dfrac{v}{v_0} = -\dfrac{B^2\ell^2t}{mR}$ .
3. Step 3: $v(t) = v_0 e^{-B^2\ell^2t/(mR)}$ .
4. Step 4: The trap is treating the braking force as constant and using a constant-acceleration equation. The force shrinks as the bar slows, so the speed decays exponentially.
Answer: $v(t) = v_0 e^{-B^2\ell^2t/(mR)}$

Common mistakes

Using kinematics equations for constant acceleration. The magnetic force depends on speed, so the acceleration keeps changing.
Getting the direction of the magnetic force wrong. By Lenz's law, it always opposes the motion that causes the induced current.
Putting a force on the parts of a loop outside the field, or assuming a loop fully inside a uniform field still has a current.
Forgetting that a bigger resistance means a smaller current and a weaker braking force.

On the exam

Sliding bars and loops crossing field boundaries are favorite free-response setups. Expect to find the emf, current, force, a differential equation for v, and a graph of v against t.
Energy questions often ask where the work goes. Answer: into thermal energy in the resistor, with Fv = I²R at constant speed.

Connected topics

Videos

Topic 13.3 - Induced Currents and Magnetic Forces
Lessons With LondotWatch on YouTube (opens in a new tab)
Induced Forces on Current Carrying Loops
Flipping PhysicsWatch on YouTube (opens in a new tab)
Emf induced in rod traveling through magnetic field | Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Motional emf via Newton's Second Law
Flipping PhysicsWatch on YouTube (opens in a new tab)
Electromagnetic Induction: Square Loop Across a Magnetic Field
Physics NinjaWatch on YouTube (opens in a new tab)
Induced EMF In Moving Conductor, Sliding Bar Generator - Faraday's Law of Electromagnetic Induction
The Organic Chemistry TutorWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 13.3 Induced Currents and Magnetic Forces. Pick an answer to see if you got it, and why.

Two long horizontal conducting rails are 0.50 m apart and are connected at their left ends by a 2.0 Ω resistor. A conducting bar lies across the rails. A uniform 0.40 T magnetic field points into the page, perpendicular to the plane of the rails. Take +x to the right and +y toward the top of the page.

An external force pulls the bar to the right at a constant 3.0 m/s. The rails and bar have negligible resistance, and friction is negligible.

Described setup

Question 1 of 4Calculator allowed

What is the current in the resistor?

Question 2 of 4Calculator allowed

In which direction does the current flow in the bar?

Question 3 of 4Calculator allowed

What external force is needed to keep the bar moving at constant speed?

Question 4 of 4Calculator allowed

At what rate does the external force do work on the bar?

0 of 4 answered