Why an Open Circuit Can Have Induced EMF but No Current

A bar magnet moves towards a coil of wire, and a voltmeter connected across the coil's separated ends registers an induced voltage. Yet there is no complete conducting path through the coil. How can an induced EMF exist without an induced current?

The answer is that **an induced EMF and an induced current are two different things**. A changing magnetic flux can produce an induced EMF even when the circuit is open. However, a sustained conduction current requires a closed path through which charge can flow. In an ideal open circuit, the induced EMF is non-zero while the current is zero.

This becomes easier to understand when you separate what the changing magnetic field does from what the circuit allows charges to do.

## 1. Start with the magnetic change, then check the circuit

Imagine a coil of wire with a small gap between its two terminals. A bar magnet approaches the coil, with its north pole facing the coil.

As the magnet moves closer, the magnetic field passing through the coil changes. This means the magnetic flux through the coil changes, producing an induced EMF.

But look at the circuit. The wire ends are not connected. There is no complete conducting loop.

[IMAGE: A side-on scientific illustration of a bar magnet approaching a circular wire coil with its north pole facing the coil. Show the magnetic field passing through the coil, the two separated wire terminals, and the changing magnet-to-coil distance. Clearly distinguish the open gap from the magnetic field lines.]

The important distinction is:

- **Induced EMF:** The changing magnetic flux produces an EMF around the coil.
- **Induced current:** Charges flow continuously around the circuit when a conducting path is available.

The changing magnetic flux does not need a closed circuit to produce an EMF. The circuit does need to be closed for a sustained conduction current to flow.

This is why an open circuit can have an induced EMF but no current.

## 2. What actually happens to the charges?

You might reasonably predict that if an EMF is induced, charges should immediately start flowing around the coil.

That prediction misses one important detail: the charges have nowhere to go once they reach the open ends.

When the magnet approaches the coil, the induced electric field exerts forces on charges in the conducting wire. Charges redistribute along the wire and accumulate at the separated terminals.

One terminal becomes relatively positive, and the other becomes relatively negative. This charge separation creates an electric potential difference across the gap.

[IMAGE: Enlarged scientific illustration of the two separated terminals of an open conducting coil. Show positive charge accumulating at one terminal and negative charge at the other, with an electric field across the gap. Include a small coil section connected to both terminals so the charge separation has a clear physical context.]

The charge redistribution continues until the resulting electric field balances the driving effect along the conducting path, under the idealised steady open-circuit conditions.

The result is a potential difference between the terminals, even though there is no sustained current through the open circuit.

Think of the terminals as two ends of a queue separated by a locked gate. People can move towards the gate and accumulate on either side, but they cannot complete a circuit through it. The analogy helps explain charge accumulation, although electric charges respond to electromagnetic forces rather than behaving like people in a queue.

There may be a brief transient current while the charges redistribute. Once the charge distribution settles, the ideal open circuit has zero steady current.

03How can you measure an induced EMF in an open circuit?

There is a practical problem here. If an open circuit has no current, how can an experiment demonstrate that an induced EMF exists?

The answer depends on choosing the right measuring instrument.

Consider an experiment with four main components:

• A 250-turn coil mounted on a bench.
• A bar magnet that can move towards or away from the coil.
• A switch, a resistor, and an ammeter connected in series with the coil.
• A high-resistance voltmeter connected across the open switch.

The magnet moves along the axis of the coil, with its north pole facing the coil. The voltmeter measures the potential difference across the switch, while the ammeter measures current in the main circuit.

What should the experiment show?

Before running the experiment, make two predictions.

1. If the switch is open while the magnet moves, should the voltmeter register a voltage?
2. If the switch is open, should the ammeter register a sustained current?

The expected observations are:

• The voltmeter registers a non-zero voltage because the magnetic flux through the coil is changing.
• The ammeter registers approximately zero because there is no complete low-resistance conducting path.

Now close the switch and repeat the magnet’s movement.

The resistor and ammeter are now part of a complete conducting loop. The changing magnetic flux produces an EMF, which can drive current around the circuit.

The ammeter now registers a current.

The failed measurement: using only a galvanometer

Suppose you initially connect a galvanometer in series with the coil but leave the switch open.

You move the magnet towards the coil, but the galvanometer shows no deflection.

It would be tempting to conclude that no EMF was induced.

That conclusion is incorrect. A galvanometer detects current, not EMF directly. With the switch open, there is no complete conducting path for a sustained current.

The experiment has failed to distinguish between two possibilities:

• No induced EMF was produced.
• An induced EMF was produced, but the circuit was open.

The original measurement cannot tell you which explanation is correct.

Revising the experiment

Replace the current-only measurement with a high-resistance voltmeter across the open switch.

A voltmeter is designed to measure potential difference. Its very high resistance means it draws very little current.

This allows you to detect the induced open-circuit voltage without significantly loading the circuit.

There is one important qualification: a real voltmeter has finite resistance. Connecting it across the gap technically creates a very high-resistance conducting path. The ideal open-circuit model assumes an infinitely high resistance, so its current is exactly zero.

For HSC calculations, treat the voltmeter as ideal unless its resistance is provided.

What the measurements tell us

The following values are illustrative predictions for the experimental arrangement. They are not measurements from an actual experiment.

Assume:

• Induced EMF magnitude: \(0.50\ \text{V}\).
• Coil resistance: \(5\ \Omega\).
• Voltmeter resistance: \(10\ \text{M}\Omega\).
• External resistor: \(20\ \Omega\).

The small current drawn by the voltmeter in the first experiment is approximately \(50\ \text{nA}\). This is negligible for the idealised HSC model.

Notice something important: the induced EMF has not disappeared simply because the circuit is open. Changing the circuit connection changes whether a sustained current can flow.

There is also a practical measurement rule worth remembering: an ammeter must be connected in series, not directly across an open switch. An ammeter has very low resistance, so placing it across the gap can effectively close the circuit and produce a current that changes the situation being measured.

04Using Faraday’s law without assuming a current exists

Now that the physical distinction is clear, we can put numbers to it.

The induced EMF depends on the rate of change of magnetic flux linkage, not on whether the circuit is open or closed.

For a coil with \(N\) turns, Faraday’s law gives:

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

where:

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

For a uniform magnetic field passing through a flat coil, the flux through one turn is:

\[ \Phi_B=BA\cos\theta \]

Here, \(B\) is magnetic field strength in teslas, \(A\) is the area of one turn in square metres, and \(\theta\) is the angle between the magnetic field and the normal to the coil.

The negative sign in Faraday’s law represents Lenz’s law. It tells us that the induced effect opposes the change in magnetic flux.

For an open circuit, the sign still describes the induced EMF’s polarity relative to your chosen reference direction. It does not mean a sustained current is flowing.

For a refresher on calculating flux changes and applying the equation, see Faraday’s Law for HSC Physics: Calculating Induced EMF.

The three decisions you should make in an HSC question

A reliable approach is to separate the magnetic situation from the electrical connection.

The most common mistake is jumping straight from Faraday’s law to Ohm’s law.

Calculating an induced EMF does not automatically give you an induced current.

For a simple closed resistive circuit:

\[ I=\frac{|\mathcal E|}{R_{\text{total}}} \]

Here, \(I\) is current in amperes and \(R_{\text{total}}\) is the total circuit resistance in ohms.

If the circuit is ideally open, the current is zero. You cannot use the coil’s resistance alone to calculate a current because the open switch prevents a complete conducting path.

05Common mistakes to avoid

Before attempting the questions, check these four predictions.

One final distinction matters when you are asked about direction.

Lenz’s law tells you the direction of the induced effect relative to the change in magnetic flux. If the circuit is closed, you can use this to predict the induced current direction.

If the circuit is open, you can still identify the induced EMF’s polarity, but you should not claim that a sustained current flows.

04Questions and solutions

The following questions progress from a standard Faraday’s law calculation to situations where you must track both the changing magnetic flux and the circuit connection.

Question 1

A 150-turn coil experiences an increase in magnetic flux per turn from \(0.10\ \text{mWb}\) to \(0.34\ \text{mWb}\) over \(0.12\ \text{s}\).

The coil is connected to an external circuit containing an open switch.

(a) Calculate the magnitude of the average induced EMF.

(b) Determine the sustained current in the ideal open-circuit model.

Solution 1

The induced EMF is \(0.30\ \text{V}\), while the ideal open-circuit current is zero.

Part (a): Calculate the induced EMF.

Step 1: Calculate the change in magnetic flux per turn.

\[ \begin{aligned} \Delta\Phi_B &=0.34-0.10\\\ &=0.24\ \text{mWb}\\\ &=2.4\times10^{-4}\ \text{Wb} \end{aligned} \]

Step 2: Apply Faraday’s law.

\[ |\mathcal E| = \frac{N|\Delta\Phi_B|}{\Delta t} \]

Substitute:

\[ |\mathcal E| = \frac{150(2.4\times10^{-4})}{0.12} \]

\[ \boxed{|\mathcal E|=0.30\ \text{V}} \]

Part (b): Determine the current.

The switch remains open, so there is no complete conducting path.

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

The changing magnetic flux produces an induced EMF, but the open circuit prevents a sustained current.

Question 2

A 200-turn coil experiences a decrease in magnetic flux per turn from \(+0.80\ \text{mWb}\) to \(+0.20\ \text{mWb}\) over \(0.15\ \text{s}\).

The coil’s circuit is open.

A positive magnetic flux is defined as pointing out of the page, towards the observer. A positive EMF is defined as acting anticlockwise when viewed from the front of the coil.

(a) Calculate the signed average induced EMF.

(b) Explain what the sign tells you.

(c) Would a sustained anticlockwise current flow in the open circuit?

Solution 2

The signed average induced EMF is \(+0.80\ \text{V}\). It corresponds to the anticlockwise reference direction, but no sustained current flows because the circuit is open.

Part (a): Calculate the signed EMF.

Step 1: Calculate the change in flux.

\[ \begin{aligned} \Delta\Phi_B &=0.20-0.80\\\ &=-0.60\ \text{mWb}\\\ &=-6.0\times10^{-4}\ \text{Wb} \end{aligned} \]

Step 2: Substitute into Faraday’s law.

\[ \mathcal E = -\frac{N\Delta\Phi_B}{\Delta t} \]

\[ \mathcal E = -\frac{200(-6.0\times10^{-4})}{0.15} \]

\[ \boxed{\mathcal E=+0.80\ \text{V}} \]

Part (b): Interpret the sign.

The positive sign means the induced EMF acts in the defined positive direction.

In this question, that direction is anticlockwise.

The positive magnetic flux is decreasing. According to Lenz’s law, the induced effect opposes this decrease by tending to maintain the original outward magnetic flux.

Part (c): Determine whether current flows.

No sustained current flows.

The positive EMF identifies the polarity and reference direction of the induced electrical effect. However, the open switch prevents charges from circulating continuously around the coil.

If the switch were closed, the induced EMF would drive an anticlockwise current under the stated assumptions.

Question 3

A student investigates electromagnetic induction using a coil, a moving bar magnet, a switch, and a galvanometer.

The galvanometer is connected in series with the coil. The switch is left open while the magnet moves towards the coil.

The galvanometer shows no deflection.

The student concludes:

“There was no induced EMF because the galvanometer detected no current.”

(a) Explain why this conclusion is not justified.

(b) Describe how the student could modify the apparatus to detect an induced EMF while keeping the main circuit open.

(c) Explain why the modified experiment does not produce a perfectly zero current in a real circuit.

Solution 3

The student’s conclusion is incorrect because a galvanometer measures current, not induced EMF. The experiment needs a high-resistance voltmeter across the open switch to detect the induced potential difference.

Part (a): Explain the failed conclusion.

The moving magnet changes the magnetic flux through the coil.

According to Faraday’s law, this produces an induced EMF.

However, the switch is open, so the circuit is incomplete. There is no continuous conducting path for a sustained current.

The galvanometer therefore shows no significant deflection, even though an induced EMF exists.

The student has confused the absence of current with the absence of EMF.

Part (b): Modify the apparatus.

Connect a high-resistance voltmeter in parallel across the two open switch contacts.

When the magnet moves towards the coil, the changing magnetic flux produces an induced EMF. The voltmeter can detect the resulting potential difference across the gap.

The main circuit remains effectively open because the voltmeter has a very high resistance.

Part (c): Explain the small real current.

A real voltmeter has finite resistance rather than infinite resistance.

Consequently, connecting it across the switch creates a high-resistance path through which a tiny current can flow.

This current is usually negligible compared with the current in a closed circuit containing an ordinary load resistor.

The ideal open-circuit model treats this current as zero.

Question 4

A 300-turn coil has an internal resistance of \(4\ \Omega\). Its magnetic flux per turn increases from \(0.20\ \text{mWb}\) to \(0.50\ \text{mWb}\) over \(0.15\ \text{s}\).

The coil is connected to an \(11\ \Omega\) external resistor through a switch.

(a) Calculate the magnitude of the induced EMF.

(b) Calculate the current when the switch is closed.

(c) Calculate the potential difference across the external resistor.

(d) The switch is now opened while the magnetic flux is still changing at the same rate. Determine the new current and explain whether the induced EMF changes.

Solution 4

The induced EMF is \(0.60\ \text{V}\). With the switch closed, the current is \(0.040\ \text{A}\), and the external resistor has \(0.44\ \text{V}\) across it. Opening the switch reduces the ideal sustained current to zero without removing the induced EMF.

Part (a): Calculate the induced EMF.

Step 1: Calculate the change in flux.

\[ \begin{aligned} \Delta\Phi_B &=0.50-0.20\\\ &=0.30\ \text{mWb}\\\ &=3.0\times10^{-4}\ \text{Wb} \end{aligned} \]

Step 2: Apply Faraday’s law.

\[ |\mathcal E| = \frac{N|\Delta\Phi_B|}{\Delta t} \]

\[ |\mathcal E| = \frac{300(3.0\times10^{-4})}{0.15} \]

\[ \boxed{|\mathcal E|=0.60\ \text{V}} \]

Part (b): Calculate the current.

The switch is closed, so the circuit has a complete conducting path.

The total resistance includes both the coil’s internal resistance and the external resistor:

\[ R_{\text{total}}=4+11 \]

\[ R_{\text{total}}=15\ \Omega \]

Using Ohm’s law:

\[ I=\frac{|\mathcal E|}{R_{\text{total}}} \]

\[ I=\frac{0.60}{15} \]

\[ \boxed{I=0.040\ \text{A}} \]

Part (c): Calculate the voltage across the external resistor.

\[ V_R=IR \]

\[ V_R=0.040(11) \]

\[ \boxed{V_R=0.44\ \text{V}} \]

The remaining \(0.16\ \text{V}\) is the voltage drop across the coil’s internal resistance:

\[ V_{\text{internal}}=0.040(4)=0.16\ \text{V} \]

The two voltage drops add to the induced EMF:

\[ 0.44+0.16=0.60\ \text{V} \]

Part (d): Open the switch.

The induced EMF remains \(0.60\ \text{V}\) because the magnetic flux is still changing at the same rate.

However, opening the switch breaks the conducting path.

Therefore:

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

The induced EMF and the current have changed differently. The EMF remains non-zero, but the ideal sustained current becomes zero.

Question 5

A 100-turn coil is exposed to a changing external magnetic field. The magnetic flux per turn follows this sequence:

A switch connects the coil to a \(10\ \Omega\) total resistive circuit.

The switch is:

• Open from \(0.00\) to \(0.20\ \text{s}\).
• Closed from \(0.20\) to \(0.50\ \text{s}\).
• Open again from \(0.50\) to \(0.70\ \text{s}\).

Assume the stated flux is externally imposed, the total resistance remains constant when the switch is closed, and self-inductance, stored electrical energy, and switching transients are negligible.

(a) Identify every interval during which an induced EMF exists.

(b) Identify every interval during which a sustained current flows through the resistor.

(c) A student argues that the current must be greatest between \(0.20\) and \(0.50\ \text{s}\), because the magnetic flux is at its maximum and the switch is closed. Explain the error.

(d) Calculate the magnitude of the current that would flow if the switch were instead closed throughout the first interval.

(e) Now consider a revised experiment in which the switch is closed during both changing-flux intervals and open during the constant-flux interval. Calculate the charge passing a fixed point in the circuit during each changing-flux interval. Explain why the net signed charge transferred over the full experiment is zero, even though charge has flowed through the resistor.

Solution 5

Under the original switching schedule, no sustained current flows at any point. The induced EMF exists during the first and final intervals, but the circuit is open during both. The circuit is closed only when the magnetic flux is constant, so the induced EMF is zero.

The revised experiment produces two equal-magnitude current pulses in opposite directions. Their signed charge transfers cancel, but the total amount of charge that passes the fixed point is non-zero.

Part (a): Identify the intervals with induced EMF.

Faraday’s law depends on the rate of change of magnetic flux.

During the first interval, the flux increases from zero to \(+0.60\ \text{mWb}\). Therefore, an induced EMF exists.

During the second interval, the flux remains constant. Its rate of change is zero, so:

\[ \mathcal E=0 \]

During the final interval, the flux decreases from \(+0.60\ \text{mWb}\) to zero. Therefore, another induced EMF exists, with the opposite polarity.

The induced EMF exists during:

• \(0.00\)-\(0.20\ \text{s}\).
• \(0.50\)-\(0.70\ \text{s}\).

Part (b): Identify the intervals with current.

Now compare those intervals with the switch positions.

There is no interval in which a non-zero induced EMF and a closed conducting path exist together.

Therefore:

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

Part (c): Explain the student’s error.

The student has confused the magnitude of magnetic flux with its rate of change.

The flux is largest between \(0.20\) and \(0.50\ \text{s}\), but it is also constant during that interval.

A constant magnetic flux produces no induced EMF, even when the circuit is closed.

The first and final intervals have changing flux, so they produce an induced EMF. However, the switch is open during both intervals.

The key is that two conditions must be satisfied for an induced current:

1. There must be an induced EMF.
2. There must be a complete conducting path.

Neither condition alone guarantees a sustained current.

Part (d): Calculate the current if the switch is closed during the first interval.

Step 1: Calculate the induced EMF during the first interval.

The flux changes by:

\[ \Delta\Phi_B=0.60\times10^{-3}\ \text{Wb} \]

The time taken is \(0.20\ \text{s}\).

Using Faraday’s law:

\[ |\mathcal E| = \frac{N|\Delta\Phi_B|}{\Delta t} \]

\[ |\mathcal E| = \frac{100(0.60\times10^{-3})}{0.20} \]

\[ |\mathcal E|=0.30\ \text{V} \]

Step 2: Calculate the current.

The total circuit resistance is \(10\ \Omega\).

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

\[ I=\frac{0.30}{10} \]

\[ \boxed{I=0.030\ \text{A}} \]

The circuit would carry \(30\ \text{mA}\) during the first interval if it were closed.

Part (e): Calculate the charge transferred during both changing-flux intervals.

This is where the distinction between net charge transfer and total charge movement becomes important.

Step 1: Calculate the current during the first interval.

The induced EMF magnitude is \(0.30\ \text{V}\), so:

\[ I_1=\frac{0.30}{10} =0.030\ \text{A} \]

The current flows for \(0.20\ \text{s}\).

Using:

\[ Q=It \]

we obtain:

\[ Q_1=0.030(0.20) \]

\[ \boxed{Q_1=0.0060\ \text{C}} \]

Step 2: Determine the current during the final interval.

The magnetic flux decreases by the same magnitude over the same time.

Therefore, the induced EMF has the same magnitude but opposite polarity.

The current magnitude is still:

\[ I_2=0.030\ \text{A} \]

Its direction is reversed.

The charge transferred in this interval is:

\[ Q_2=0.030(0.20) \]

\[ \boxed{Q_2=0.0060\ \text{C}} \]

Step 3: Compare the two charge transfers.

The first interval transfers \(0.0060\ \text{C}\) in one direction.

The final interval transfers \(0.0060\ \text{C}\) in the opposite direction.

Taking the first direction as positive:

\[ Q_{\text{net}} = +0.0060-0.0060 \]

\[ \boxed{Q_{\text{net}}=0\ \text{C}} \]

However, the total magnitude of charge passing the fixed point is:

\[ Q_{\text{total}} = 0.0060+0.0060 \]

\[ \boxed{Q_{\text{total}}=0.012\ \text{C}} \]

These are two different quantities.

The net signed charge is zero because the current reverses direction. The total charge transferred in magnitude is \(0.012\ \text{C}\), because charge passed the point in both directions.

This is the important result: a zero net charge transfer does not mean that no current flowed.

05Where to go next

When solving an electromagnetic induction question, keep the order consistent:

1. Identify what is changing in the magnetic field and calculate the change in flux.
2. Define the coil and its orientation.
3. Use Faraday’s law to calculate the induced EMF.
4. Check whether the circuit is open or closed.
5. Only calculate current when a complete conducting path exists.

Once you know that current can flow, the next challenge is determining its direction.

That is where Lenz’s law becomes useful. It tells you how the induced current responds to the change in magnetic flux, rather than simply responding to the magnetic field itself.

Continue with How to Use Lenz’s Law to Predict Induced Current Direction.