Why a Stationary Magnet Does Not Induce Continuous Current

Push a bar magnet into a coil connected to a galvanometer. The needle swings. Now stop the magnet halfway inside the coil and hold it there. After a brief transient, the needle returns to zero.

The magnet is still inside the coil. Its magnetic field hasn’t disappeared, and magnetic flux is still passing through the coil. So why has the current stopped?

Predict what would happen if you left the magnet there for another ten seconds. Would the galvanometer remain deflected?

No. A stationary magnet does not continuously induce current when the coil is also stationary relative to it. The magnetic flux through the coil is no longer changing, so the induced electromotive force (EMF) becomes zero. Without an induced EMF, there is no continuing induced current.

This is the central idea behind electromagnetic induction: a magnetic field passing through a coil is not enough. The magnetic flux through the coil must change.

01The important quantity is magnetic flux, not simply the magnetic field

Before predicting whether a current will be induced, identify two things:

• The particular conducting loop you’re considering.
• How the magnetic field passing through the area enclosed by that loop changes.

This matters because electromagnetic induction depends on the magnetic flux through a specific loop, not simply on whether a magnet is nearby.

What is magnetic flux?

Magnetic flux measures how much magnetic field passes through the area enclosed by a loop.

For a flat coil in a uniform magnetic field:

\[ \Phi=BA\cos\theta \]

where:

• \(\Phi\) is the magnetic flux through one turn, measured in webers (Wb).
• \(B\) is magnetic field strength, measured in teslas (T).
• \(A\) is the area enclosed by the loop, measured in square metres (\(\text{m}^2\)).
• \(\theta\) is the angle between the magnetic field and the normal to the coil.

The normal is an imaginary line perpendicular to the flat surface of the coil.

If the coil contains \(N\) turns, its magnetic flux linkage is:

\[ N\Phi \]

This is the quantity that appears in Faraday’s law.

Why does a stationary magnet produce no continuous induced EMF?

Consider a bar magnet placed next to a coil.

Initially, the magnet might produce a strong magnetic field through the coil. The magnetic flux could be quite large.

However, suppose:

• The magnet is stationary.
• The coil is stationary.
• The distance between them is constant.
• The orientation of the coil is unchanged.
• The magnetic field itself is not changing.

The magnetic field distribution through the coil stays the same.

Consequently, \(B\), \(A\), and \(\theta\) remain constant, meaning the magnetic flux is also constant.

A useful way to think about this is to compare magnetic flux with the water level in a tank.

A tank can contain plenty of water without its water level increasing. The amount of water present and the rate at which that amount changes are different things.

Similarly, a coil can have substantial magnetic flux passing through it without producing an induced EMF.

The analogy has a limit: magnetic fields are not fluids, and magnetic flux has a direction as well as a magnitude. However, it helps separate the amount of flux from its rate of change.

02Faraday’s law explains exactly why the current stops

Michael Faraday’s experiments established that an EMF is induced when the magnetic flux linkage through a circuit changes.

This relationship is described by Faraday’s law:

\[ \boxed{\mathcal{E}_{\text{avg}} =-N\frac{\Delta\Phi}{\Delta t}} \]

where:

• \(\mathcal{E}_{\text{avg}}\) is the average induced EMF, measured in volts (V).
• \(N\) is the number of turns in the coil.
• \(\Delta\Phi\) is the change in magnetic flux through one turn, measured in webers (Wb).
• \(\Delta t\) is the time taken for that change, measured in seconds (s).

The negative sign represents Lenz’s law: the induced EMF produces a current whose magnetic field opposes the change in magnetic flux.

For a stationary magnet and a stationary coil:

\[ \Delta\Phi=0 \]

Therefore:

\[ \mathcal{E}_{\text{avg}} =-N\frac{0}{\Delta t} \]

\[ \boxed{\mathcal{E}=0\text{ V}} \]

The magnetic field can still be present. The magnetic flux can still be non-zero. However, because the flux is not changing, there is no induced EMF.

Compare the three situations

Suppose a bar magnet has its north pole facing a coil. Choose the positive normal to point in the direction of the magnetic field through the coil.

Notice something important.

The magnet produces a magnetic field in all three situations. What changes is the magnetic flux through the coil.

When the magnet approaches, the flux changes in one direction. When it withdraws, the flux changes in the opposite direction.

When it stops, the flux stops changing.

This is why the induced EMF disappears.

What if the magnet is stationary but the coil moves?

This is where the wording of the question becomes important.

A stationary magnet does not automatically mean there can be no electromagnetic induction.

Suppose the magnet remains fixed, but you rotate the coil. The angle \(\theta\) between the magnetic field and the coil’s normal changes.

Because

\[ \Phi=BA\cos\theta \]

the magnetic flux changes even though the magnet is stationary.

The same thing can happen if the coil changes area or moves through a non-uniform magnetic field.

The correct principle is therefore:

An induced EMF depends on a change in magnetic flux linkage, not simply on whether the magnet moves.

03What Faraday’s experiments actually demonstrated

The distinction between a magnetic field and changing magnetic flux emerged from Faraday’s experimental investigations in 1831.

His problem was straightforward: could a magnetic field produce an electric current without a battery directly connected to the circuit?

The moving magnet and galvanometer experiment

Consider an insulated wire coil connected by two wires to a galvanometer, an instrument that detects small electric currents.

A bar magnet is positioned along the axis of the coil, with its north pole facing the opening.

The magnet can be moved into and out of the coil without touching the wire.

What should happen if simply having a magnet near a coil produces current?

The galvanometer should remain deflected while the magnet is held inside the coil. After all, the magnetic field is still there.

But that isn’t what happens.

The observations are:

Stage 1: The magnet enters the coil.

The magnetic flux through the coil changes. The galvanometer briefly deflects.

Stage 2: The magnet is held stationary.

The magnetic flux becomes constant. The galvanometer returns to zero.

Stage 3: The magnet is withdrawn.

The magnetic flux changes in the opposite direction. The galvanometer deflects the other way.

This observation supports Faraday’s central conclusion: changing magnetic flux induces an EMF.

It also demonstrates why the current reverses when the magnet’s direction of motion reverses.

A second experiment removes an important ambiguity

The moving magnet experiment demonstrates induction, but there is another question worth asking.

Is physical movement of the magnet essential, or is changing magnetic flux the actual requirement?

Faraday investigated this using an iron ring with two separate insulated wire coils wound around it.

The primary coil was connected to a battery and switch. The secondary coil was connected to a galvanometer.

The two circuits were electrically separate. They were linked through the magnetic field in the iron ring.

What did Faraday observe?

• When the primary circuit was switched on, the galvanometer in the secondary circuit briefly deflected.
• When the primary current became steady, the galvanometer returned to zero.
• When the primary circuit was switched off, the galvanometer briefly deflected in the opposite direction.

The iron ring and both coils remained stationary throughout.

The changing primary current produced a changing magnetic field, which changed the magnetic flux through the secondary coil.

Once the primary current became steady, the magnetic flux stopped changing, and the induced EMF disappeared.

This experiment demonstrated that the magnet itself does not need to move. A changing magnetic field can also induce an EMF.

It also helped distinguish physical movement from the actual condition required for electromagnetic induction: changing magnetic flux linkage.

04Induced EMF and induced current are not the same thing

There is another distinction that often causes confusion in HSC Physics questions.

An induced EMF is not the same as an induced current.

An EMF is the energy transferred per unit charge around a circuit. It is measured in volts.

An electric current is the rate at which electric charge flows through a conducting path. It is measured in amperes.

For a simple closed circuit with total resistance \(R\), Ohm’s law gives:

\[ \boxed{I=\frac{\mathcal{E}}{R}} \]

where:

• \(I\) is the induced current in amperes (A).
• \(\mathcal{E}\) is the induced EMF in volts (V).
• \(R\) is the total circuit resistance in ohms (\(\Omega\)).

The distinction becomes particularly important when the circuit is open.

Can an open coil have an induced EMF?

Yes.

Imagine moving a magnet towards a coil whose circuit has a gap.

The magnetic flux through the coil changes, so an induced EMF can still exist.

However, because the conducting path is incomplete, there is no sustained current.

A high-resistance voltmeter connected across the gap could measure the induced voltage.

Compare the situations:

This gives you two separate conditions to check:

1. Is the magnetic flux changing? If yes, an induced EMF exists.
2. Is there a complete conducting path? If yes, an induced current can flow.

In the simple HSC circuit model, we generally neglect self-inductance unless the question specifies otherwise. In a real coil, a small current may take time to decay after the magnet stops because the coil’s own changing magnetic field produces a self-induced EMF. This is a brief transient, not a continuous current driven by the stationary magnet.

05Worked examples

06A reliable method for deciding whether current is induced

When you encounter a question involving a magnet and a coil, avoid immediately asking whether the magnet is moving.

Instead, work through these decisions:

Step 1: Identify the loop.

Which conducting path are you considering? Identify the area enclosed by that loop.

Step 2: Identify the magnetic flux.

Consider the magnetic field strength, the loop’s area, and its orientation.

For a uniform field:

\[ \Phi=BA\cos\theta \]

Step 3: Determine whether the flux changes.

Check whether any of the following changes:

• Magnetic field strength, \(B\).
• Loop area, \(A\).
• Angle between the field and the normal, \(\theta\).
• Position of the loop within a non-uniform magnetic field.

Step 4: Calculate the induced EMF.

Use Faraday’s law:

\[ \mathcal{E}_{\text{avg}} =-N\frac{\Delta\Phi}{\Delta t} \]

Step 5: Check whether current can flow.

If the circuit is closed, use the total resistance to calculate the induced current.

\[ I=\frac{\mathcal{E}}{R} \]

This method prevents you from confusing the presence of a magnetic field with a changing magnetic flux.

If you need to determine the direction of the induced current, the next step is to apply Lenz’s law. The tutorgum guide to using Lenz’s law to predict induced current direction develops that reasoning.

07Questions and solutions

Question 1

A 100-turn coil is connected to a resistor with resistance \(5.0\,\Omega\).

A bar magnet is clamped next to the coil, with its north pole facing the coil. The magnet and coil remain stationary relative to each other for \(3.0\,\text{s}\).

The magnetic flux through each turn is \(+0.42\,\text{mWb}\) at the beginning and end of this interval.

Calculate:

(a) The average induced EMF.

(b) The induced current.

(c) Explain why the coil experiences no induced current despite having magnetic flux passing through it.

Solution 1

The induced EMF is \(0\,\text{V}\), and the induced current is \(0\,\text{A}\). The magnet and coil remain stationary relative to each other, so the magnetic flux does not change.

(a) Calculate the change in magnetic flux.

\[ \Delta\Phi=0.42-0.42 \]

\[ \Delta\Phi=0\,\text{Wb} \]

Apply Faraday’s law:

\[ \mathcal{E}_{\text{avg}} =-N\frac{\Delta\Phi}{\Delta t} \]

\[ =-100\frac{0}{3.0} \]

\[ \boxed{\mathcal{E}_{\text{avg}}=0\,\text{V}} \]

(b) Apply Ohm’s law.

\[ I=\frac{\mathcal{E}}{R} \]

\[ I=\frac{0}{5.0} \]

\[ \boxed{I=0\,\text{A}} \]

(c) The magnetic field is still passing through the coil, but its strength and orientation relative to the coil remain unchanged.

Therefore, the magnetic flux linkage is constant.

The tempting mistake is to assume that a magnetic field automatically produces a current. It does not. A changing magnetic flux is required to induce an EMF.

Question 2

A 120-turn coil is connected to a resistor with resistance \(6.0\,\Omega\).

A bar magnet is moved into the coil, held stationary, and then withdrawn.

The magnetic flux per turn is recorded at different times:

Choose the positive normal to point in the direction of the magnetic field produced by the north pole through the coil. Positive current produces a magnetic field in the direction of this normal.

Calculate the average induced EMF and current during each of the following intervals:

(a) \(0.00\) to \(0.20\,\text{s}\).

(b) \(0.20\) to \(0.70\,\text{s}\).

(c) \(0.70\) to \(0.80\,\text{s}\).

(d) Explain why the current during withdrawal has a greater magnitude than the current during entry.

Solution 2

The coil has a negative current during entry, zero current while the magnet is held stationary, and a positive current during withdrawal. The withdrawal produces the largest current because the same flux change occurs in half the time.

(a) Magnet entering the coil

Calculate the change in flux:

\[ \Delta\Phi=0.46-0.10 \]

\[ =+0.36\,\text{mWb} \]

Convert to webers:

\[ \Delta\Phi=+3.6\times10^{-4}\,\text{Wb} \]

The time interval is:

\[ \Delta t=0.20\,\text{s} \]

Apply Faraday’s law:

\[ \mathcal{E}_{\text{avg}} =-120\frac{3.6\times10^{-4}}{0.20} \]

\[ \boxed{\mathcal{E}_{\text{avg}}=-0.216\,\text{V}} \]

Calculate the current:

\[ I_{\text{avg}}=\frac{-0.216}{6.0} \]

\[ \boxed{I_{\text{avg}}=-0.036\,\text{A}} \]

The current is \(36\,\text{mA}\) in the negative direction.

(b) Magnet held stationary

The magnetic flux remains at \(+0.46\,\text{mWb}\).

Therefore:

\[ \Delta\Phi=0 \]

Applying Faraday’s law:

\[ \boxed{\mathcal{E}_{\text{avg}}=0\,\text{V}} \]

And:

\[ \boxed{I=0\,\text{A}} \]

The magnet remains inside the coil, but there is no changing magnetic flux.

(c) Magnet withdrawing

Calculate the change in flux:

\[ \Delta\Phi=0.10-0.46 \]

\[ =-0.36\,\text{mWb} \]

\[ =-3.6\times10^{-4}\,\text{Wb} \]

The time interval is:

\[ \Delta t=0.80-0.70=0.10\,\text{s} \]

Apply Faraday’s law:

\[ \mathcal{E}_{\text{avg}} =-120\frac{-3.6\times10^{-4}}{0.10} \]

\[ \boxed{\mathcal{E}_{\text{avg}}=+0.432\,\text{V}} \]

Calculate the current:

\[ I_{\text{avg}}=\frac{0.432}{6.0} \]

\[ \boxed{I_{\text{avg}}=+0.072\,\text{A}} \]

The current is \(72\,\text{mA}\) in the positive direction.

(d) The same magnitude of magnetic flux change occurs during entry and withdrawal:

\[ |\Delta\Phi|=3.6\times10^{-4}\,\text{Wb} \]

However, withdrawal takes only \(0.10\,\text{s}\), compared with \(0.20\,\text{s}\) during entry.

Faraday’s law tells us that a greater rate of change of magnetic flux produces a greater induced EMF.

Consequently, the withdrawal EMF and current magnitudes are both twice as large.

The important distinction is that current depends on how quickly the flux changes, not just on the total amount of flux change.

Question 3

A 90-turn coil has an open circuit and is connected across an ideal high-resistance voltmeter.

A bar magnet is moved towards the coil. The magnetic flux per turn increases from \(+0.20\,\text{mWb}\) to \(+0.50\,\text{mWb}\) over \(0.15\,\text{s}\).

The positive normal is chosen in the direction of the increasing magnetic flux.

(a) Calculate the average induced EMF.

(b) What current flows through the open circuit while the magnet is moving?

(c) The magnet is then held stationary, and the switch is closed to connect the coil to a \(5.0\,\Omega\) resistor. Calculate the induced EMF and current after the switching transient.

Assume an ideal open circuit, a purely resistive closed circuit, and negligible self-inductance.

Solution 3

The average induced EMF while the magnet moves is \(-0.18\,\text{V}\), but the open circuit carries no current. After the magnet stops, closing the circuit produces zero induced EMF and zero induced current.

(a) Calculate the change in flux:

\[ \Delta\Phi=0.50-0.20 \]

\[ =+0.30\,\text{mWb} \]

Convert to webers:

\[ \Delta\Phi=+3.0\times10^{-4}\,\text{Wb} \]

Apply Faraday’s law:

\[ \mathcal{E}_{\text{avg}} =-90\frac{3.0\times10^{-4}}{0.15} \]

\[ \boxed{\mathcal{E}_{\text{avg}}=-0.18\,\text{V}} \]

The negative sign indicates that the induced EMF acts opposite to the chosen positive direction.

(b) Even though an induced EMF exists, the circuit is open.

There is no complete conducting path through which charge can continuously flow.

Therefore, in the ideal open-circuit model:

\[ \boxed{I=0\,\text{A}} \]

The voltmeter can still detect the induced potential difference across the open terminals.

This demonstrates that induced EMF and induced current are separate quantities.

(c) Once the magnet is held stationary, the magnetic flux becomes constant.

Therefore:

\[ \Delta\Phi=0 \]

And Faraday’s law gives:

\[ \boxed{\mathcal{E}=0\,\text{V}} \]

When the switch is closed:

\[ I=\frac{\mathcal{E}}{R} \]

\[ I=\frac{0}{5.0} \]

\[ \boxed{I=0\,\text{A}} \]

Closing the circuit does not automatically create an induced current. There must still be a changing magnetic flux to produce the induced EMF.

Question 4

A stationary magnet assembly produces a broad region of approximately uniform magnetic field.

An 80-turn rectangular coil is connected to a \(4.0\,\Omega\) resistor.

The coil is moved sideways through the magnetic field without changing its area or orientation.

During the first \(0.10\,\text{s}\), the entire coil remains inside the uniform-field region. Its magnetic flux per turn stays constant at \(+0.60\,\text{mWb}\).

The coil then moves across the edge of the field. Its flux per turn decreases from \(+0.60\,\text{mWb}\) to \(+0.15\,\text{mWb}\) over \(0.12\,\text{s}\).

Calculate:

(a) The average induced EMF and current while the coil moves through the uniform-field region.

(b) The average induced EMF and current while the coil crosses the edge of the magnetic field.

(c) Explain why the coil can move through a magnetic field without producing a net induced EMF.

Solution 4

There is no induced EMF while the coil moves entirely within the uniform magnetic field. An average EMF of \(+0.30\,\text{V}\) and current of \(+0.075\,\text{A}\) are produced while it crosses the edge of the field.

(a) During the first interval, the magnetic flux remains constant:

\[ \Delta\Phi=0 \]

Therefore:

\[ \mathcal{E}_{\text{avg}}=0\,\text{V} \]

Using Ohm’s law:

\[ I=\frac{0}{4.0} \]

\[ \boxed{I=0\,\text{A}} \]

Although the coil is moving, its area, orientation, and magnetic field strength remain unchanged.

Consequently, the magnetic flux through it does not change.

(b) While the coil crosses the field boundary, the flux decreases.

Calculate the change in flux:

\[ \Delta\Phi=0.15-0.60 \]

\[ =-0.45\,\text{mWb} \]

Convert to webers:

\[ \Delta\Phi=-4.5\times10^{-4}\,\text{Wb} \]

Apply Faraday’s law:

\[ \mathcal{E}_{\text{avg}} =-80\frac{-4.5\times10^{-4}}{0.12} \]

\[ \boxed{\mathcal{E}_{\text{avg}}=+0.30\,\text{V}} \]

Calculate the current:

\[ I_{\text{avg}}=\frac{0.30}{4.0} \]

\[ \boxed{I_{\text{avg}}=+0.075\,\text{A}} \]

(c) The tempting assumption is that moving a conductor through a magnetic field must always produce an induced current.

However, the relevant quantity is the net magnetic flux through the entire loop.

While the coil translates through the uniform region, the magnetic field strength, enclosed area, and orientation remain constant.

There is no change in flux linkage, so there is no net induced EMF around the closed loop.

When the coil crosses the field boundary, the amount of magnetic field passing through its enclosed area changes. An induced EMF is then produced.

This is an important distinction: movement can occur without electromagnetic induction.

Question 5

Two identical circular coils, A and B, are placed in a region where a stationary magnet produces an approximately uniform magnetic field of \(0.40\,\text{T}\).

Each coil has:

• 100 turns.
• An area of \(25\,\text{cm}^2\) per turn.
• A total circuit resistance of \(5.0\,\Omega\).

Both coils are initially positioned with their normals parallel to the magnetic field.

Choose the positive normal to point along the magnetic field. Positive current is defined as the direction that produces a magnetic field along this normal.

The coils undergo different movements.

Coil A: The circular coil spins through \(180^\circ\) about its own normal. Its centre and the orientation of its plane remain unchanged.

Coil B: The coil is tilted about a diameter. Its normal moves from \(0^\circ\) to \(60^\circ\) relative to the magnetic field over \(0.25\,\text{s}\). It is then held at \(60^\circ\) for \(0.50\,\text{s}\), before returning to \(0^\circ\) over another \(0.25\,\text{s}\).

Assume both circuits are closed, purely resistive, and have negligible self-inductance.

Answer the following:

(a) Does coil A experience an induced EMF? Explain your reasoning, even though the wire itself is moving.

(b) Calculate the average induced EMF and current in coil B during the initial tilt.

(c) Determine the induced EMF and current while coil B is held at \(60^\circ\).

(d) Calculate the average induced EMF and current during the return movement.

(e) Calculate the signed charge transferred through coil B’s resistor during each movement. Hence, determine the net signed charge transferred over the complete cycle.

(f) Explain why a zero average EMF over the complete cycle does not mean that no electromagnetic induction occurred.

Solution 5

Coil A experiences no induced EMF, while coil B experiences two periods of induced EMF in opposite directions. Coil B’s initial and final magnetic fluxes are equal, but its flux changes during the intermediate movement.

This distinction is the key to the entire problem.

(a) Coil A: spinning about its own normal

Coil A rotates, but its orientation relative to the magnetic field does not change.

Its area remains constant, and its normal remains parallel to the magnetic field.

Therefore:

\[ \Phi=BA\cos0^\circ \]

remains constant throughout the rotation.

Consequently:

\[ \boxed{\mathcal{E}=0\,\text{V}} \]

and:

\[ \boxed{I=0\,\text{A}} \]

The wire is physically moving, but the magnetic flux through the loop is unchanged.

This is why movement alone is not sufficient to produce a net induced EMF around a closed loop.

(b) Coil B: initial tilt from \(0^\circ\) to \(60^\circ\)

First, convert the coil area into square metres.

\[ A=25\times10^{-4} \]

\[ A=2.5\times10^{-3}\,\text{m}^2 \]

Calculate the initial magnetic flux per turn:

\[ \Phi_i=BA\cos0^\circ \]

\[ =0.40(2.5\times10^{-3}) \]

\[ \Phi_i=1.0\times10^{-3}\,\text{Wb} \]

Now calculate the flux at \(60^\circ\):

\[ \Phi_f=BA\cos60^\circ \]

\[ =0.40(2.5\times10^{-3})(0.5) \]

\[ \Phi_f=5.0\times10^{-4}\,\text{Wb} \]

The change in flux is:

\[ \Delta\Phi=\Phi_f-\Phi_i \]

\[ =-5.0\times10^{-4}\,\text{Wb} \]

Apply Faraday’s law:

\[ \mathcal{E}_{\text{avg}} =-100\frac{-5.0\times10^{-4}}{0.25} \]

\[ \boxed{\mathcal{E}_{\text{avg}}=+0.20\,\text{V}} \]

Calculate the average current:

\[ I_{\text{avg}}=\frac{0.20}{5.0} \]

\[ \boxed{I_{\text{avg}}=+0.040\,\text{A}} \]

The average current is \(40\,\text{mA}\) in the positive direction.

The induced current produces a magnetic field that opposes the decrease in the original magnetic flux.

(c) Coil B: held stationary at \(60^\circ\)

Once the coil stops moving, its orientation becomes constant.

The magnetic flux therefore remains at:

\[ \Phi=5.0\times10^{-4}\,\text{Wb} \]

There is no change in magnetic flux:

\[ \Delta\Phi=0 \]

Therefore:

\[ \boxed{\mathcal{E}=0\,\text{V}} \]

And:

\[ \boxed{I=0\,\text{A}} \]

The stationary magnet is still producing its magnetic field, but there is no continuing induced current.

(d) Coil B: returning from \(60^\circ\) to \(0^\circ\)

The magnetic flux increases from:

\[ 5.0\times10^{-4}\,\text{Wb} \]

to:

\[ 1.0\times10^{-3}\,\text{Wb} \]

Therefore:

\[ \Delta\Phi=+5.0\times10^{-4}\,\text{Wb} \]

The return movement also takes \(0.25\,\text{s}\).

Apply Faraday’s law:

\[ \mathcal{E}_{\text{avg}} =-100\frac{5.0\times10^{-4}}{0.25} \]

\[ \boxed{\mathcal{E}_{\text{avg}}=-0.20\,\text{V}} \]

Calculate the average current:

\[ I_{\text{avg}}=\frac{-0.20}{5.0} \]

\[ \boxed{I_{\text{avg}}=-0.040\,\text{A}} \]

The average current is now \(40\,\text{mA}\) in the negative direction.

The current reverses because the magnetic flux is now increasing rather than decreasing.

(e) Calculate the charge transferred

The average current is related to the charge transferred by:

\[ q=I_{\text{avg}}\Delta t \]

During the initial tilt:

\[ q_1=(0.040)(0.25) \]

\[ \boxed{q_1=+0.010\,\text{C}} \]

During the return movement:

\[ q_2=(-0.040)(0.25) \]

\[ \boxed{q_2=-0.010\,\text{C}} \]

The total signed charge is:

\[ q_{\text{net}}=q_1+q_2 \]

\[ =0.010-0.010 \]

\[ \boxed{q_{\text{net}}=0\,\text{C}} \]

The net signed charge is zero because equal amounts of charge pass through the resistor in opposite directions.

However, the total magnitude of charge transferred is:

\[ |q_1|+|q_2|=0.020\,\text{C} \]

(f) Why doesn’t zero average EMF mean zero induction?

This is the most important conceptual part of the question.

A student might notice that the coil starts and finishes with the same orientation and conclude that no induced EMF was produced.

That conclusion is incorrect.

The magnetic flux changes during the initial tilt. It then remains constant while the coil is held. Finally, it changes in the opposite direction during the return movement.

The complete sequence is:

Over the complete cycle, the final magnetic flux equals the initial magnetic flux.

Therefore, the average EMF over the entire \(1.0\,\text{s}\) interval is:

\[ \mathcal{E}_{\text{avg}} =-N\frac{\Phi_f-\Phi_i}{\Delta t} \]

\[ \boxed{\mathcal{E}_{\text{avg}}=0\,\text{V}} \]

But this does not mean the instantaneous EMF was zero throughout the cycle.

The coil experienced two periods of induction, producing opposite currents and transferring charge in both directions.

This is the subtle distinction: equal initial and final magnetic flux does not mean magnetic flux remained constant in between.

08Where this concept leads next

A stationary magnet does not continuously induce current in a stationary coil because the magnetic flux linkage remains constant.

Once you understand this, the next useful question is how to deliberately keep changing magnetic flux to generate electricity.

A DC generator does exactly that. Its coil rotates in a magnetic field, continuously changing its orientation and therefore its magnetic flux. The resulting induced EMF can then be converted into pulsating direct current using a split-ring commutator.

You can explore that application in tutorgum’s guide to DC generators and how rotating coils produce direct current.