6. Electromagnetic Induction

Big idea: a change in magnetic linkage can transfer mechanical or electrical energy into another circuit.

Learn to calculate magnitudes and explain directions separately. Every clockwise answer must name its viewing side.

In the nineteenth century, Michael Faraday and Joseph Henry independently investigated induction. Their experiments connected changing magnetic conditions with electrical effects and helped make generators and transformers possible.

A magnet can remain beside a coil without driving a current. What matters is how the magnetic flux changes.

Electromagnetic induction produces an electromotive force, or emf, when magnetic flux linked with a circuit changes. Emf means work done per unit charge around the circuit; it is measured in volts, not newtons. An induced emf may exist even if a break in the circuit prevents a sustained current.

The circuit can experience changing flux because the field changes, its area changes, or its orientation changes. Motion is one possible cause; a stationary coil can also have an induced emf. Conversely, motion that leaves flux unchanged need not induce a net emf around a simple closed loop.

A rigid closed coil moves sideways through a large uniform steady field while keeping its area and orientation fixed. Is its net induced emf necessarily nonzero?

Compare its flux before and after the motion. Field, area and orientation have not changed.

Result: In this ideal situation the net emf around the loop is zero. Individual charges can still experience magnetic forces.

1. Can an open coil have an induced emf?
2. Name three quantities whose change can change uniform-field flux.

Answers

1. Yes. A closed conducting path is needed for a sustained induced current, not for emf.
2. Field magnitude, loop area and the angle of its normal to the field.

Use the same coil in several experiments and ask which change produces the galvanometer deflection.

Connect a coil to a sensitive current detector. Bring a magnet towards it: the detector deflects during motion. Stop the magnet: the transient deflection disappears after the circuit settles. Withdraw it: the direction reverses. Faster flux change generally produces a greater emf; the current also depends on the circuit resistance and inductance.

Relative motion can also be produced by moving the coil while the magnet stays fixed. Compare the actual flux histories rather than assuming that every motion gives identical current. Reversing the approaching magnetic pole reverses the relevant field direction and the induced response.

Place two coils nearby. Connect one to a switch and source, and the other to a detector. Switching the first current on or off changes its magnetic field, so the second coil shows a transient response without either coil moving. A steady primary current produces no continuing emf in the secondary after the transient.

The primary current rises slowly, stays steady, then falls rapidly. Compare the secondary emf during the three intervals.

Flux linkage changes during the rise and fall. It is constant in the middle. A faster linkage change gives larger emf magnitude.

Result: Nonzero during rise, zero in the steady interval, and oppositely directed during the fall; the rapid fall can give the larger magnitude.

1. Why does switching a nearby coil matter?
2. Why does reversing the magnet motion reverse the response?

Answers

1. Switching changes its current and hence the magnetic flux through the detector coil.
2. It reverses the sign of the flux change for the same chosen area normal.

Flux measures how a magnetic field passes through an oriented surface.

For a small flat element, $d\Phi_B=\mathbf B\cdot d\mathbf A$. The vector $d\mathbf A$ is perpendicular to the element. Add the contributions over the surface: $\Phi_B=\int\mathbf B\cdot d\mathbf A$. For a uniform field and a flat area, $\Phi_B=BA\cos\theta$. The angle is measured from the normal.

Flux has unit weber, equal to tesla square metre. It is a signed scalar once an area orientation is chosen. A parallel field along the plane contributes zero; a perpendicular field gives the greatest magnitude for a fixed field and area. For a curved surface or nonuniform field, use the integral rather than multiplying arbitrary average values.

A flat loop has area $0.015\,\mathrm{m^2}$ in a $0.32\,\mathrm T$ uniform field. Its normal makes $60^\circ$ with the field. Find the flux.

Use the cosine of the angle to the normal. $\Phi_B=0.32(0.015)\cos60^\circ$.

Result: $\Phi_B=0.0024\,\mathrm{Wb}$.

A coil with identical turns has flux linkage N times the flux through one turn. Do not confuse flux per turn with total linkage. If different turns enclose different fluxes, add their individual linkages.

1. What is the flux when B lies in the loop plane?
2. Does reversing the chosen normal change physical induction?

Answers

1. Zero for a uniform field and flat loop.
2. It reverses the signed flux and the associated positive circulation convention together; the physical result stays the same.

Faraday’s law connects the rate of change of linkage to a measurable emf.

Choose a positive area normal and use the right-hand rule to define positive circulation. Faraday’s law is $\mathcal E=-d\Phi_B/dt$ for one turn and $\mathcal E=-N\,d\Phi_B/dt$ for identical turns. For a finite interval the average emf is $-N\Delta\Phi_B/\Delta t$. The derivative gives an instantaneous value.

The minus sign carries directional information. First compute the magnitude of the rate; then decide the direction using a consistent circulation or Lenz’s law. Induced current equals emf divided by resistance only when a purely resistive model is justified. A coil’s own inductance can delay current changes.

A stationary $160$-turn coil of area $0.012\,\mathrm{m^2}$ has its normal parallel to B. The field rises uniformly by $0.45\,\mathrm T$ in $0.30\,\mathrm s$. Find the emf magnitude.

For fixed area and orientation, $|\mathcal E|=NA|\Delta B|/\Delta t$.

Result: $|\mathcal E|=2.88\,\mathrm V$. Its induced field opposes the increase.

A changing magnetic field creates a circulating electric field. Its line integral around a stationary loop need not vanish. Unlike electrostatic potential differences, this induced emf cannot generally be described by a single-valued electrostatic potential alone.

1. Which emf uses a finite flux difference divided by time?
2. Does zero instantaneous flux imply zero emf?

Answers

1. The average emf over that interval.
2. No. Flux may pass through zero while its rate of change is nonzero.

The induced response resists the change that causes it, which explains where the electrical energy comes from.

Lenz’s law says the induced current produces a magnetic effect opposing the change in linked flux. If an outward field grows, the induced field points inward. If the outward field weakens, the induced field points outward. It is the change that is opposed, not always the original field itself.

• Choose a viewing side and state whether the external flux points towards or away from you.
• Decide whether that flux is increasing or decreasing.
• Choose the induced field that resists that change.
• Use the curled fingers of your right hand for current and thumb for the induced field.

If a moving magnet induced a current that pulled it faster in the same direction without any energy supply, both mechanical energy and electrical heating could grow without a source. The opposing effect prevents that. An external agent must supply work to maintain the motion against the induced magnetic force.

Viewed from above, a loop’s upward magnetic flux is decreasing. Determine the induced current direction.

The loop tries to support the upward flux. A right-hand thumb pointing upward requires anticlockwise conventional current as seen from above.

Result: Anticlockwise from above. Looking from below reverses the apparent clockwise label.

1. Does Lenz’s law always make the induced field opposite to the external field?
2. What supplies heating energy when a magnet is pushed steadily into a resistive coil?

Answers

1. No. During decreasing external flux it can point in the same direction.
2. The mechanical work supplied by the external agent.

A moving conductor can separate charges even in a steady magnetic field.

Positive charges moving with a rod experience $q\mathbf v\times\mathbf B$. For a straight rod of length l with rod, velocity and uniform field mutually perpendicular, the magnitude of its motional emf is $Blv$. Charge separation builds an electric field until the open-circuit electric and magnetic forces balance.

In a closed sliding-rod circuit, the emf can drive a current. Determine polarity from the force on a positive test charge; electron motion is opposite. The simple expression needs the perpendicular geometry. In general, use the line integral of the magnetic force per unit charge along the moving conductor.

A $0.40\,\mathrm m$ rod moves at $2.5\,\mathrm{m\,s^{-1}}$ in a $0.60\,\mathrm T$ perpendicular field. The closed circuit resistance is $3.0\,\Omega$. Ignore inductive transients. Find emf, current and mechanical power.

$\mathcal E=Blv$, $I=\mathcal E/R$, and the opposing force is $BIl$.

Result: $\mathcal E=0.6\,\mathrm V$, $I=0.2\,\mathrm A$, $F=0.048\,\mathrm N$ and $Fv=0.12\,\mathrm W=I^2R$.

One can derive the sliding-rod result from flux: if the rod moves by dx, loop area changes by l dx, so $|d\Phi_B/dt|=Bl\,dx/dt=Blv$. This agrees with the magnetic-force argument. With speed maintained and inductive effects negligible, $I=Blv/R$, opposing force $BIl$, and external power $Fv=I\mathcal E=I^2R$.

A rotating rod in a perpendicular field has different speeds at different radii. For a rod extending from its rotation axis to radius l, integrate the local emf contribution Bωr dr to obtain a magnitude Bωl²/2. Do not use the speed of its outer end for every point on the rod.

1. Which rule determines the positive end of a moving isolated rod?
2. Can a moving rod have emf without current?

Answers

1. The direction of v cross B for positive charge.
2. Yes, if its circuit is open.

A circuit stores magnetic energy, so it resists rapid changes of its current.

For fixed linear magnetic geometry, the linkage of a secondary coil caused by primary current is $N_2\Phi_{21}=MI_1$. Thus $\mathcal E_2=-M\,dI_1/dt$, with sign depending on the chosen winding conventions. M depends on turns, geometry, separation and core material. For ideal coaxial long solenoids, $M=\mu_0N_1N_2A/l$ when their linked area is A.

Two fixed coils have $M=0.006\,\mathrm H$. The primary current changes by $4.0\,\mathrm A$ in $0.050\,\mathrm s$. Find the average secondary emf magnitude.

Use $|\mathcal E_2|=M|\Delta I_1|/\Delta t$.

Result: $0.48\,\mathrm V$.

A coil’s own linkage is $LI$ in the same linear approximation. Its self-induced emf is $-L\,dI/dt$. Inductance has unit henry. A long air-core solenoid has $L=\mu_0N^2A/l$ when end effects are negligible.

The external source must do work while building the current. With $dW=LI\,dI$, integration from zero current gives $U=LI^2/2$. In a uniform vacuum field the magnetic energy density is $B^2/(2\mu_0)$. Real resistive coils also dissipate heat; not all supplied energy is stored.

An $0.080\,\mathrm H$ coil current rises from zero to $3.0\,\mathrm A$ in $0.020\,\mathrm s$. Find the average back-emf magnitude and final stored energy.

Use $L\Delta I/\Delta t$ for magnitude and $LI^2/2$ for energy.

Result: $12\,\mathrm V$ and $0.36\,\mathrm J$.

1. Does an ideal inductor oppose a steady DC current through an induced emf?
2. Which ideal solenoid quantity grows as the square of turn count?

Answers

1. No. Its self-induced emf is zero once its current is steady.
2. Its self-inductance, for unchanged area and length.

A generator converts controlled mechanical rotation into a periodic emf.

Let the area normal of a uniform N-turn coil rotate at angular speed ω in uniform B. If its normal is parallel to B at t=0, $\Phi_B=BA\cos\omega t$ per turn. Differentiation gives $\mathcal E=NBA\omega\sin\omega t$, so peak emf is $\mathcal E_0=NBA\omega$. A different initial orientation changes phase, not amplitude.

The emf is greatest when the flux passes through zero because its slope is then greatest. Emf vanishes when flux is maximum or minimum. Slip rings connect the rotating ends to a stationary external circuit without changing their identity; brushes maintain contact. This ideal model ignores winding resistance, friction and detailed field geometry.

A $120$-turn coil of area $0.020\,\mathrm{m^2}$ rotates at $40\,\mathrm{rad\,s^{-1}}$ in $0.25\,\mathrm T$. Find the emf amplitude.

Multiply $NBA\omega$.

Result: $\mathcal E_0=24\,\mathrm V$.

When the generator supplies a load, its current produces a torque opposing the rotation. Maintaining speed requires mechanical power. Raising speed raises the ideal emf amplitude and electrical frequency together; merely adding turns raises amplitude without changing rotation frequency.

1. What fixes the frequency in this simple two-pole generator model?
2. Why does a loaded generator require more driving torque?

Answers

1. The coil’s rotation frequency.
2. Its current produces an opposing magnetic torque, so additional mechanical work supplies the output energy.

• $\Phi_B=\int\mathbf B\cdot d\mathbf A$; uniform flat case $BA\cos\theta$.
• $\mathcal E=-d(N\Phi_B)/dt$; Lenz’s law gives the physical direction.
• Perpendicular moving rod: $|\mathcal E|=Blv$.
• Linear inductors: $\mathcal E_L=-L\,dI/dt$ and $U=LI^2/2$.
• Rotating coil: $\mathcal E_0=NBA\omega$.

• Induction concerns flux change; a large constant field need not cause emf.
• A circuit can have emf without current.
• Treat the normal and the positive loop circulation as a linked sign convention.

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

6.1 A coil and a magnet remain fixed relative to one another. Must the detector show a continuing current?

6.2 Distinguish emf from current.

6.3 Find flux for $B=0.24\,\mathrm T$, $A=0.025\,\mathrm{m^2}$, normal parallel to B.

6.4 A flat loop rotates until its plane is parallel to B. What is its final flux?

6.5 A $75$-turn coil loses $0.004\,\mathrm{Wb}$ per turn in $0.20\,\mathrm s$. Find average emf magnitude.

6.6 Upward flux increases. Give induced current viewed from above.

6.7 Upward flux decreases. Give induced current viewed from above.

6.8 A $0.30\,\mathrm m$ rod moves at $4.0\,\mathrm{m\,s^{-1}}$ in a perpendicular $0.50\,\mathrm T$ field. Find emf magnitude.

6.9 Why does the direction phrase “clockwise” need a viewing side?

6.10 An $0.15\,\mathrm H$ inductor current falls at $8.0\,\mathrm{A\,s^{-1}}$. Find induced emf magnitude.

6.11 Find energy in a $0.12\,\mathrm H$ coil carrying $2.5\,\mathrm A$.

6.12 A long solenoid keeps its dimensions but doubles its turns. How does L change?

6.13 For $M=0.025\,\mathrm H$ and primary slope $12\,\mathrm{A\,s^{-1}}$, find secondary emf magnitude.

6.14 A generator flux is maximum. Is its emf maximum at that instant?

6.15 A generator has $N=80$, $B=0.30\,\mathrm T$, $A=0.015\,\mathrm{m^2}$ and $\omega=50\,\mathrm{rad\,s^{-1}}$. Find peak emf.

6.16 Explain why a generator delivering electrical power does not violate energy conservation.

Answers

6.1 No. After transients, unchanged linked flux gives no induced emf.

6.2 Emf is work supplied per unit charge; current is charge flow per unit time. An open circuit can have emf.

6.3 $0.006\,\mathrm{Wb}$.

6.4 Zero in a uniform field.

6.5 $1.5\,\mathrm V$.

6.6 Clockwise; it produces a downward field.

6.7 Anticlockwise; it produces an upward field.

6.8 $0.6\,\mathrm V$.

6.9 The same physical circulation looks clockwise from one side and anticlockwise from the other.

6.10 $1.2\,\mathrm V$; its direction tends to maintain the current.

6.11 $0.375\,\mathrm J$.

6.12 It increases by a factor of four in the same linear-core approximation.

6.13 $0.3\,\mathrm V$.

6.14 No. For sinusoidal rotation the flux slope, and hence emf, is zero then.

6.15 $18\,\mathrm V$.

6.16 External mechanical work drives rotation against the induced magnetic torque, supplying output energy and losses.