Topic 13.3: Induced Currents and Magnetic Forces

What Topic 13.3 requires

Topic 13.3 has one learning objective and four essential-knowledge statements. It prints no boundary statement.

13.3.A, describe the force exerted on a conductor due to the interaction between an external magnetic field and an induced current within that conductor.

• 13.3.A.1 states that when an induced current is created in a conductive loop, the already-present magnetic field will exert a magnetic force on the moving charge carriers within the loop, with the relevant equation $\vec{F}_B = \int I \left(d\vec{\ell} \times \vec{B}\right)$.
• 13.3.A.2 states that when current is induced in a conducting loop, magnetic forces are only exerted on the segments of the loop that are within the external magnetic field, and that these magnetic forces may cause translational or rotational acceleration.
• 13.3.A.3 states that the force on a conducting loop is proportional to the induced current in the loop, which depends on the rate of change of magnetic flux, the resistance of the loop, and the velocity of the loop.
• 13.3.A.4 states that Newton's second law can be applied to a conducting loop moving in a magnetic field as it experiences an induced emf.

The suggested skills are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.B, calculate or estimate an unknown quantity with units from known quantities by selecting and following a logical computational pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Topic 13.3 is the only topic in Unit 13 that lists skill 2.D, functional dependence, and reading 13.3.A.3 next to that listing tells you the exam's angle here. The statement names four things the force depends on, and 2.D asks for factors of change. Expect to be asked what happens to the force when the speed doubles, or the resistance halves, or the field is tripled, more often than to be asked for a number.

This is also the topic where the electromagnetic half of the course hands back to mechanics. Statement 13.3.A.4 says so outright, and the C: Mechanics table reprinted on the E&M sheet supplies the tools: $\vec{F}_{\text{net}} = d\vec{p}/dt$, $\vec{a}_{\text{sys}} = \vec{F}_{\text{net}}/m_{\text{sys}}$, $P_{\text{avg}} = W/\Delta t = \Delta E/\Delta t$, and $K = \frac{1}{2}mv^2$.

The chain, in the order the CED builds it

Every problem in this topic is the same five links. Say them in order and the physics writes itself.

1. Something changes the flux. Usually a loop moving into, out of, or across the boundary of a field region.
2. Faraday's law gives an emf. $\lvert \mathcal{E} \rvert = \lvert d\Phi_B/dt \rvert$, from Topic 13.2.
3. The circuit gives a current. $I = \mathcal{E}/R$, using the loop's total resistance.
4. The field pushes on that current. $\vec{F}_B = \int I \left(d\vec{\ell} \times \vec{B}\right)$, statement 13.3.A.1, which the sheet prints.
5. Lenz's law fixes the direction, and it always opposes the motion. No exceptions, because the alternative would create energy.

Run the chain symbolically for the standard case and one expression comes out that is worth knowing on sight. A loop whose straddling side has length $\ell$ moves at speed $v$ perpendicular to a uniform field $B$, with loop resistance $R$:

That last expression is statement 13.3.A.3 made explicit. Read the dependences off it:

The two squared factors are the functional-dependence question this topic exists to ask. The field enters twice because it both drives the current and pushes on it, and the length enters twice for the same reason.

One caution. That single expression is a result, not a starting point. It assumes a uniform field, a straight straddling segment perpendicular to both the field and the velocity, and a constant resistance. When any of those fails, go back to steps 2 through 4 and redo them. The CED prints the integral, not the shortcut.

Only the segments inside the field feel a force

Statement 13.3.A.2 is the one that decides most answers: magnetic forces are only exerted on the segments of the loop that are within the external magnetic field. It follows directly from the integral at 13.3.A.1, since $\vec{B} = 0$ on any segment outside the field region and those contributions vanish.

Three situations, and they behave completely differently.

A loop entirely inside a uniform field. Move it however you like: the flux through it does not change, so there is no emf, no induced current, and no force. Even if you drive a current around it with a battery, the four forces on the four sides cancel in a uniform field, so the net force is zero. The forces are real, they just add to nothing, and they can still produce a torque.

A loop entirely outside the field. Nothing at all.

A loop straddling the boundary. Now the flux changes as the loop moves, a current flows, and only the one side inside the field is pushed. That single unbalanced force is the whole answer, and it always points so as to oppose the motion. Entering a field region: the force pulls backwards. Leaving it: the force pulls backwards. Both, because Lenz's law does not care which direction you are going, only that you are changing the flux.

The crossing therefore has three phases, and a graph of the force against position has three regions: a constant retarding force while the leading side is inside and the trailing side is outside, zero while the whole loop is inside, and a constant retarding force again while the trailing side is still inside and the leading side has left. Sketching that three-region graph is a standard skill 1.A task.

Statement 13.3.A.2 also adds that these magnetic forces may cause translational or rotational acceleration. The rotational case is what happens when the unbalanced forces do not act through the centre of mass, or when opposite sides of a loop are pushed in opposite directions. A rectangular loop in a uniform field with a current in it feels equal and opposite forces on its two opposite sides, and if the loop's plane is not perpendicular to the field those two forces are offset, producing a torque that turns the loop. That is a motor, which is the first essential question the CED prints for this unit.

Newton's second law on a moving loop

Statement 13.3.A.4 says Newton's second law can be applied to a conducting loop moving in a magnetic field as it experiences an induced emf. That sentence is short and it is the hardest thing in the topic, because the force depends on the velocity and the velocity is what you are solving for.

Write it out for a loop of mass $m$ falling out of a field region under gravity, with $g = 9.8 \ \mathrm{m/s^2}$, taking downward as positive:

That is a first-order differential equation, and it behaves the same way a resistive-force problem in mechanics does. Three things can be asked about it, and they need different amounts of work.

The initial acceleration. At the instant of release, $v = 0$, so the magnetic force is zero and $a = g$. No calculus.

The terminal velocity. Set $dv/dt = 0$, so the two forces balance:

Algebra only. This is the most-asked version, and the second worked example does it numerically.

The full motion. Solving the equation gives an exponential approach to $v_t$. Nothing in Unit 13 prints that solution, and no boundary statement in Topic 13.3 says anything either way. What calibrates the expectation is the CED's own phrasing in its sample free-response set, where Question 4 asks students to "Derive, but do not solve, a differential equation" for a rate of change, and then, in a later part, to justify a prediction "by referring to the differential equation you wrote". Setting the equation up and reasoning from it is the assessed skill; integrating it is not the point of the question.

Reasoning from the equation without solving it is a real technique, and it is worth practising. From $m\,dv/dt = mg - B^2\ell^2 v/R$ you can read off, with no integration at all: that the acceleration starts at $g$ and decreases; that it never reverses, because the magnetic term cannot exceed $mg$ once $v$ has stopped growing; that the speed approaches $v_t$ from below and never exceeds it; and that a larger resistance gives a larger terminal speed, because a poorer conductor carries less induced current. Each of those is a defensible claim supported by the equation, which is exactly what skill 3.B asks for.

The energy audit, and why it always balances

The cleanest justification available in this topic is the energy one, and it is worth doing at least once with numbers so you trust it.

Hold a loop moving at constant speed across a field boundary. Two things are true at once:

• Something external must apply a force equal and opposite to the magnetic drag, since the loop is not accelerating. That force does work at a rate $P_{\text{mech}} = F v$.
• The circuit dissipates energy in its resistance at a rate $P_{\text{elec}} = \mathcal{E}I = I^2 R$.

Substitute the chain and the two are identically equal:

No approximation anywhere. Every joule of electrical energy in the circuit was pushed in by whoever is pulling the loop.

That identity is what makes Lenz's law non-negotiable rather than a convention. If the force helped the motion instead of opposing it, the loop would accelerate on its own while also heating its own resistance, and there would be no source for either. The minus sign in Faraday's law is conservation of energy wearing a sign.

When the loop is not moving at constant speed the audit still closes, it just includes kinetic energy. A loop given a shove and then left alone slows down, and the kinetic energy it loses appears as heat in its resistance. If it never leaves the field boundary, all of its initial kinetic energy ends up dissipated:

There is one quantity in this situation that does not depend on the speed at all, and it is a favourite. The total charge that flows while the flux changes is

The time cancels out of the integral, so pulling a loop out of a field quickly and pulling it out slowly move exactly the same amount of charge. What changes with speed is the current, the force, the power and the energy dissipated. The charge does not. The third worked example uses this.

Working the force integral honestly

The sheet prints the force as an integral, and there are cases where you have to treat it as one.

Read the pieces. $d\vec{\ell}$ is an element of the conductor pointing along the conventional current. The cross product means the force on each element is perpendicular to both the element and the local field, with the magnitude carrying a sine of the angle between them. The integral runs over the conductor.

Three observations make it usable.

The current comes out. In a series loop the current is the same everywhere, so $I$ is a constant of the integration and the integral is over geometry alone.

A straight segment in a uniform field collapses immediately. With $\vec{B}$ constant and $d\vec{\ell}$ all in one direction, $\int d\vec{\ell} = \vec{\ell}$ and the force is $I\vec{\ell} \times \vec{B}$, of magnitude $I\ell B\sin\theta$. Almost every exam case is this one, which is why the shortcut is worth having and worth labelling as a shortcut.

A curved or partially immersed conductor needs the integral. For a curved wire in a uniform field, $\int d\vec{\ell}$ between two endpoints is the straight vector joining them, so a semicircular wire in a uniform field feels the same force as a straight wire connecting its ends. For a conductor in a non-uniform field, $\vec{B}$ stays inside the integral and you integrate properly.

Directions come from the right-hand rule, and the CED lists skill 1.A here precisely because the reliable way to get a cross product right is to draw it. Point the fingers along the current, curl toward the field, and the thumb gives the force. For an induced current you never actually need to do this, because Lenz's law tells you the answer in advance: the force opposes the motion. Use the right-hand rule as the check, not as the method. If the two disagree, you have the current direction backwards.

If you landed here from AP Physics 2

There is no AP Physics 2 topic called Induced Currents and Magnetic Forces. The algebra-based course splits this content between two topics and never assembles the chain.

• The force on a current-carrying wire lives in AP Physics 2 Topic 12.3, Magnetism and Current-Carrying Wires, as $F_B = I\ell B\sin\theta$.
• The induced current lives in AP Physics 2 Topic 12.4, Electromagnetic Induction and Faraday's Law, which prints the sliding rod as its worked case.

Those pages are for students in AP Physics 2. This page is for students in AP Physics C: Electricity and Magnetism. If you are in the algebra-based course, the two equations above and Lenz's law are the whole of it, and the drag on a sliding rod is as far as the course goes. If you are in Physics C, three things are added: the force arrives as an integral rather than a product, the CED explicitly asks you to apply Newton's second law to the moving loop, which turns the situation into a differential equation, and functional dependence on four separate quantities is a listed skill for the topic.

The AP Physics 1 topic on Newton's second law is the other half of the prerequisite here, and it is worth revisiting if setting up a velocity-dependent force equation feels unfamiliar. The mathematics is the same as any resistive-force problem: a constant driving term, a term proportional to speed, and a terminal state where they balance.

How Topic 13.3 is tested

The patterns are narrow and they repeat.

1. Run the chain and report a force (skill 2.B). Flux rate, emf, current, force, in that order, with units carried.
2. Give a factor of change (skill 2.D). Doubling the field quadruples the force. Doubling the resistance halves it. This is the signature question of the topic.
3. Argue a direction and justify it (skill 3.B). The force opposes the motion, and you say why using Lenz's law rather than naming the right-hand rule.
4. Set up Newton's second law for the loop and either find a terminal velocity or reason from the equation without solving it (skills 2.B, 3.B).
5. Close an energy audit, showing that mechanical power in equals electrical power dissipated.
6. Sketch force, current or velocity against time or position for a loop crossing a field boundary (skill 1.A), remembering the three regions.

On justification, the CED's science-practices page for this unit is direct: simply being able to solve for a final answer is insufficient, and students may benefit from practice crafting clear, concise arguments, derivations, and calculations that follow a logical pathway. In this topic that means writing the causal chain as sentences. "The loop moves, so the enclosed area falls, so the flux falls, so an emf is induced, so a current flows, so the field exerts a force on the current-carrying side inside the field, and by Lenz's law that force opposes the loop's motion." Every clause in that sentence is a step a scoring guideline can recognise.

One last thing worth having ready. The CED's exam-preparation note for Unit 13 says students may be asked to determine the changes in an induced current when the magnetic flux through the loop is changing or if the magnetic field is turned on and off. Turning a field on or off is the version of this topic with no motion at all: the flux changes because the field does, a current is induced, and the force appears on whatever segments happen to be inside the field while it is changing.