Induced EMF vs Current: Avoid a Common HSC Physics Mistake

You calculate a non-zero induced electromotive force (emf), then immediately write \(I=\mathcal E/R\) to find the current. The arithmetic looks reasonable-until the question tells you the coil is open. If the terminals aren’t connected, where is the current supposed to go?

The mistake isn’t necessarily your use of Faraday’s law. It’s treating induced emf and current as interchangeable quantities. A changing magnetic flux can produce an induced emf even when no sustained current flows.

Before reading on, predict what happens when the magnetic flux through a coil changes while its terminals are disconnected. Does an ideal voltmeter read zero? Does a sustained current flow?

The voltmeter can read a non-zero voltage, while the sustained current is zero. Understanding why will help you avoid a common error in HSC Physics questions on electromagnetic induction.

01Keep induced emf and current separate

Think of a pump pushing water against a closed valve. The pump can establish a pressure difference even though water isn’t flowing through the closed valve.

In this analogy, the pressure difference represents the electrical driving effect, while the flow rate represents current. The analogy has limits: electrical charge responds to electric fields, not to water pressure, but the distinction between a driving effect and actual flow is useful.

What induced emf tells you

Electromotive force, or emf, is the energy supplied per unit charge by a source. Despite its name, it isn’t a force measured in newtons.

For electromagnetic induction, a changing magnetic flux produces an induced emf. Magnetic flux describes how much magnetic field passes through a surface.

For a uniform magnetic field passing through a flat surface,

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

where:

• \(\Phi_B\) is the magnetic flux, measured in webers (Wb).
• \(B\) is the magnetic field strength, measured in teslas (T).
• \(A\) is the surface area, measured in square metres (\(\text{m}^2\)).
• \(\theta\) is the angle between the magnetic field and the surface’s normal.

Flux can change because the magnetic field changes, the area changes, or the surface rotates relative to the field.

Faraday’s law gives the average induced emf over a time interval:

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

Here, \(N\) is the number of turns in the coil, \(\Delta\Phi_B\) is the change in magnetic flux per turn, and \(\Delta t\) is the time taken for that change. The emf is measured in volts (V).

If you’re calculating only the magnitude, use

\[ |\mathcal E_{\text{avg}}| =\frac{N|\Delta\Phi_B|}{\Delta t} \]

The minus sign in Faraday’s law describes the direction of the induced emf, as explained by Lenz’s law. It doesn’t mean that the emf is negative in every situation.

Notice what the equation actually depends on: the rate of change of magnetic flux, not simply the amount of flux present.

For example, a coil sitting in a strong but steady magnetic field can have substantial magnetic flux and still have zero induced emf. If the flux isn’t changing, Faraday’s law gives zero induced emf.

What current tells you

Electric current measures the rate at which charge passes a point:

\[ I=\frac{\Delta Q}{\Delta t} \]

where \(I\) is current in amperes (A), \(\Delta Q\) is the charge transferred in coulombs (C), and \(\Delta t\) is the time interval in seconds (s).

For a sustained current to flow through an ordinary conducting circuit, there must be a complete conducting path. The electric field drives charge around that path.

An induced emf can provide the driving effect, but it doesn’t guarantee that charge can circulate.

This gives you the key distinction:

• Induced emf: describes the energy supplied per unit charge by induction.
• Current: describes the rate of charge flow through a circuit.

You can have an induced emf without a sustained current. You can also have current driven by a battery without any induced emf from a changing magnetic flux.

02What happens when the coil is open?

Imagine a coil exposed to a changing magnetic field. Its two terminals aren’t connected to an external load, so there is no complete external conducting path.

The changing flux induces an electric field along the winding. Charges respond to this field and redistribute, producing a potential difference between the terminals.

Once the charge distribution has settled, there is no sustained current through the open circuit in the ideal model. However, the terminal voltage can remain non-zero while the magnetic flux continues to change.

This is why an open circuit can have a voltage across its terminals without carrying a sustained current.

What the voltmeter measures

A voltmeter is connected in parallel across the two terminals whose voltage you want to measure. It has a very high resistance, so it draws very little current.

For an open coil, an ideal voltmeter reads the induced open-circuit voltage. In the standard HSC model, this is approximately equal to the magnitude of the induced emf, provided the meter leads don’t enclose additional changing magnetic flux.

A real voltmeter has finite resistance, so it technically provides a very weak conducting path and draws a tiny current. We usually neglect this effect in HSC calculations unless the question provides the meter’s resistance.

What the ammeter measures

An ammeter is connected in series with the circuit. It measures the current passing through that path.

If a switch elsewhere in the circuit remains open, the ammeter reads zero because the circuit is incomplete.

Be careful, though: connecting an ammeter directly across the two open terminals can bridge the gap. Because an ammeter has very low resistance, doing this can create a substantial current and change the circuit you’re trying to measure.

An ammeter isn’t a passive observer of an open circuit. Its connection matters.

03Open circuit versus closed circuit

The same coil can behave differently when you change only the circuit connection.

When the circuit is closed, the induced emf can drive current through the coil and any connected load. In a simple resistive model, neglecting inductive transients, the current magnitude is

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

where \(R_{\text{total}}\) is the total resistance of the closed circuit, measured in ohms (\(\Omega\)).

The total resistance matters. It can include the coil’s resistance, the external resistor, and other circuit components.

A second detail is easy to miss: the induced emf and the terminal voltage under load aren’t always equal. If the coil has internal resistance, some of the emf is associated with the voltage drop across that resistance.

For a simple source model,

\[ V_{\text{terminal}}=|\mathcal E|-Ir \]

where \(r\) is the coil’s internal resistance. This expression assumes the current direction is consistent with the source supplying energy to the load.

So a voltmeter and an ammeter answer different questions. The voltmeter tells you about the potential difference between two points; the ammeter tells you how much charge is flowing per second through a particular path.

04Three shortcuts that fail in exams

The last shortcut is especially important. If the flux rises and then falls back to its original value, the induced emf reverses direction. A zero average over the entire interval doesn’t mean the emf was zero throughout.

If the circuit is closed, current can flow in both directions during those intervals. If the circuit is open during one of them, the induced emf can still exist during that interval without producing a sustained current.

05A decision method you can reuse

When an HSC question mentions an induced emf and asks about current, work through these steps in order.

Step 1: Identify the change in magnetic flux.

Find the initial and final flux per turn. Calculate \(\Delta\Phi_B=\Phi_{B,f}-\Phi_{B,i}\). Keep the signs if the flux changes direction.

Step 2: Calculate the induced emf.

Use Faraday’s law, including the number of turns. If the question asks for magnitude, use the absolute value.

Step 3: Check the circuit condition.

Ask whether charge has a complete conducting path during the interval you’re analysing. Don’t assume the circuit is closed just because an induced emf exists.

Step 4: Calculate current only when appropriate.

For a closed, resistance-dominated circuit in the simplified model, use the induced emf and total resistance to calculate current. For an open circuit, the sustained current is zero, even when the induced emf is non-zero.

If the circuit changes state during the question, analyse each interval separately.

Step 5: Determine the direction if asked.

Use Lenz’s law. The induced current produces a magnetic field that opposes the change in magnetic flux, not necessarily the magnetic field itself.

06Worked examples

07Questions and solutions

Question 1

A 30-turn coil experiences a decrease in magnetic flux per turn from \(0.020\ \text{Wb}\) to \(0.008\ \text{Wb}\) over \(0.12\ \text{s}\).

The coil’s terminals are open. Calculate the magnitude of the average induced emf, and state the ideal voltmeter reading and the ammeter reading if an ammeter is in series with a separate open switch.

Solution 1

The average induced emf has a magnitude of \(3.0\ \text{V}\), the ideal voltmeter reads approximately \(3.0\ \text{V}\), and the ammeter reads \(0\ \text{A}\).

First, calculate the change in flux per turn:

\[ \Delta\Phi_B=0.008-0.020 =-0.012\ \text{Wb} \]

Apply Faraday’s law:

\[ \begin{aligned} |\mathcal E_{\text{avg}}| &=\frac{N|\Delta\Phi_B|}{\Delta t}\\ &=\frac{30(0.012)}{0.12}\\ &=3.0\ \text{V} \end{aligned} \]

The changing flux produces an induced emf despite the open circuit. The voltmeter reads the open-circuit voltage, while the open switch prevents sustained current through the ammeter.

Question 2

A 50-turn coil is connected in a closed circuit. The magnetic flux per turn changes from \(-0.0040\ \text{Wb}\) to \(+0.0080\ \text{Wb}\) over \(0.040\ \text{s}\). The coil has a resistance of \(2.0\ \Omega\), and the external resistor has a resistance of \(4.0\ \Omega\).

Calculate the average induced emf magnitude and the current magnitude. Explain why using \(0.0040\ \text{Wb}\) as the change in flux gives an incorrect result.

Assume inductive transients can be neglected.

Solution 2

The induced emf has a magnitude of \(15\ \text{V}\), and the current has a magnitude of \(2.5\ \text{A}\).

The key is to subtract the signed flux values rather than simply compare their magnitudes.

\[ \begin{aligned} \Delta\Phi_B &=0.0080-(-0.0040)\\ &=0.0120\ \text{Wb} \end{aligned} \]

The flux changes from a negative value, through zero, to a positive value. Its total change is therefore \(0.0120\ \text{Wb}\), not \(0.0040\ \text{Wb}\).

Using Faraday’s law,

\[ \begin{aligned} |\mathcal E_{\text{avg}}| &=\frac{50(0.0120)}{0.040}\\ &=15\ \text{V} \end{aligned} \]

The total resistance is

\[ R_{\text{total}}=2.0+4.0 =6.0\ \Omega \]

Since the circuit is closed,

\[ \begin{aligned} I&=\frac{|\mathcal E|}{R_{\text{total}}}\\ &=\frac{15}{6.0}\\ &=2.5\ \text{A} \end{aligned} \]

The induced current flows in the direction that opposes the increase in the chosen positive direction of magnetic flux, as required by Lenz’s law.

Question 3

An 80-turn coil is connected to a closed circuit with a total resistance of \(8.0\ \Omega\). The magnetic flux per turn changes as follows:

• During the first \(0.10\ \text{s}\), it increases linearly from \(+0.010\ \text{Wb}\) to \(+0.030\ \text{Wb}\).
• During the next \(0.10\ \text{s}\), it decreases linearly back to \(+0.010\ \text{Wb}\).

Ignore inductive transients.

Calculate the induced emf magnitude and current magnitude during each interval. Determine the average induced emf over the full \(0.20\ \text{s}\), and explain whether a zero average emf means that no current flowed.

Solution 3

The induced emf has a magnitude of \(16\ \text{V}\) during both intervals, and a current of magnitude \(2.0\ \text{A}\) flows in opposite directions during the two intervals. The average emf over the full interval is zero, but current flows during both halves.

First interval

The flux change per turn is

\[ \Delta\Phi_B=0.030-0.010 =0.020\ \text{Wb} \]

Therefore,

\[ \begin{aligned} |\mathcal E_1| &=\frac{80(0.020)}{0.10}\\ &=16\ \text{V} \end{aligned} \]

The current magnitude is

\[ I_1=\frac{16}{8.0} =2.0\ \text{A} \]

Second interval

The flux change per turn is

\[ \Delta\Phi_B=0.010-0.030 =-0.020\ \text{Wb} \]

The flux changes at the same rate magnitude but in the opposite direction. Consequently, the induced emf reverses direction and retains its \(16\ \text{V}\) magnitude. The current also reverses direction, retaining its \(2.0\ \text{A}\) magnitude.

Average emf over the full interval

The initial and final flux values are equal. Hence,

\[ \begin{aligned} \mathcal E_{\text{avg}} &=-N\frac{\Phi_{B,f}-\Phi_{B,i}}{\Delta t}\\ &=-80\frac{0.010-0.010}{0.20}\\ &=0\ \text{V} \end{aligned} \]

This is the average emf over the whole interval, not the emf at every instant. The flux first increases and then decreases, so the instantaneous induced emf is non-zero during both halves.

The signed charge transferred during the first interval is \(2.0(0.10)=0.20\ \text{C}\) in one direction. The same amount passes in the opposite direction during the second interval. The net signed charge is zero, but \(0.40\ \text{C}\) passes a point when both directions of transfer are counted.

A zero average emf does not mean that no current flowed.

Question 4

A changing magnetic flux induces an emf of \(12\ \text{V}\) in a coil with an internal resistance of \(3.0\ \Omega\). Model the coil as an ideal emf source in series with this resistance, and neglect inductive transients.

(a) A voltmeter with a resistance of \(9.0\ \text{M}\Omega\) is connected across the otherwise open terminals. Estimate the voltmeter reading and the current through the meter.

(b) The coil is then connected to a \(3.0\ \Omega\) external resistor, with an ideal ammeter in series. Calculate the current and the voltage across the external resistor.

(c) Explain why connecting the ammeter directly across the open terminals would not measure the original open-circuit current.

Solution 4

The voltmeter reads approximately \(12.0\ \text{V}\) while drawing a tiny current of about \(1.3\ \mu\text{A}\). With the external resistor connected, the current is \(2.0\ \text{A}\) and the load voltage is \(6.0\ \text{V}\). An ammeter connected directly across the open terminals would change the circuit by providing a low-resistance path.

(a) Voltmeter connected

The voltmeter technically completes a high-resistance circuit. The total resistance is approximately

\[ R_{\text{total}}=9.0\times10^6+3.0 \approx9.0\times10^6\ \Omega \]

The current through the meter is

\[ \begin{aligned} I&=\frac{\mathcal E}{R_{\text{total}}}\\ &\approx\frac{12}{9.0\times10^6}\\ &=1.33\times10^{-6}\ \text{A}\\ &=1.33\ \mu\text{A} \end{aligned} \]

The voltage across the voltmeter is approximately

\[ \begin{aligned} V_{\text{meter}} &=IR_{\text{meter}}\\ &\approx(1.33\times10^{-6})(9.0\times10^6)\\ &\approx12.0\ \text{V} \end{aligned} \]

The current is tiny because the voltmeter’s resistance is very high. This is why the voltmeter gives a good approximation to the open-circuit voltage.

(b) External resistor connected

The total resistance becomes

\[ R_{\text{total}}=3.0+3.0 =6.0\ \Omega \]

The current is

\[ I=\frac{12}{6.0} =2.0\ \text{A} \]

The voltage across the external resistor is

\[ V_R=IR=(2.0)(3.0) =6.0\ \text{V} \]

The other \(6.0\ \text{V}\) is the voltage drop across the coil’s internal resistance in this model.

(c) Ammeter connected directly across the terminals

An ammeter has very low resistance. Connecting it directly across the open terminals would bridge the gap and create a new, low-resistance circuit.

In the stated model, the current would approach

\[ I\approx\frac{12}{3.0} =4.0\ \text{A} \]

This is not the original open-circuit condition. The ammeter would create the path that allows current to flow, rather than passively measuring a current that existed before it was connected.

Question 5

An 80-turn coil is connected to a \(4.0\ \Omega\) total resistive circuit containing a switch and an ammeter. A high-resistance voltmeter is connected across the coil terminals.

The magnetic flux per turn changes linearly as follows:

• From \(t=0.00\ \text{s}\) to \(t=0.10\ \text{s}\), the flux increases from \(+0.010\ \text{Wb}\) to \(+0.030\ \text{Wb}\). The switch is closed.
• From \(t=0.10\ \text{s}\) to \(t=0.20\ \text{s}\), the flux decreases from \(+0.030\ \text{Wb}\) back to \(+0.010\ \text{Wb}\). The switch is open.

Ignore coil inductance and switching transients, and treat the voltmeter as ideal. Use a fixed sign convention in which positive current corresponds to positive induced emf.

(a) Calculate the signed induced emf during each interval.

(b) Determine the ammeter reading during each interval.

(c) State the signed voltmeter reading during the second interval.

(d) Calculate the total signed charge passing through the ammeter over the full \(0.20\ \text{s}\).

A student argues that because the initial and final magnetic fluxes are equal, both the total induced emf and the total transferred charge must be zero. Evaluate this claim.

Solution 5

The induced emf is \(-16\ \text{V}\) during the first interval and \(+16\ \text{V}\) during the second. The ammeter reads a current of magnitude \(4.0\ \text{A}\) in the first interval and zero in the second. The voltmeter reads approximately \(+16\ \text{V}\) during the second interval, and the total signed charge through the ammeter is \(-0.40\ \text{C}\) under the stated sign convention.

The student’s claim is incorrect about the transferred charge. Equal initial and final fluxes give zero average emf over the whole interval, but the circuit is open during the second half, so the opposite emf cannot drive a corresponding current.

(a) Signed induced emf

For the first interval,

\[ \begin{aligned} \Delta\Phi_{B,1} &=0.030-0.010\\ &=+0.020\ \text{Wb} \end{aligned} \]

Applying Faraday’s law,

\[ \begin{aligned} \mathcal E_1 &=-80\frac{+0.020}{0.10}\\ &=-16\ \text{V} \end{aligned} \]

For the second interval,

\[ \begin{aligned} \Delta\Phi_{B,2} &=0.010-0.030\\ &=-0.020\ \text{Wb} \end{aligned} \]

Therefore,

\[ \begin{aligned} \mathcal E_2 &=-80\frac{-0.020}{0.10}\\ &=+16\ \text{V} \end{aligned} \]

The emf reverses sign because the change in magnetic flux reverses direction.

(b) Ammeter readings

During the first interval, the switch is closed, so current can flow:

\[ \begin{aligned} I_1&=\frac{\mathcal E_1}{R_{\text{total}}}\\ &=\frac{-16}{4.0}\\ &=-4.0\ \text{A} \end{aligned} \]

The ammeter reads \(4.0\ \text{A}\) in the direction opposite to the defined positive direction.

During the second interval, the switch is open. The induced emf still exists, but the conducting path is incomplete. The ideal ammeter therefore reads

\[ I_2=0\ \text{A} \]

The crucial point is that the switch changes the current condition without removing the changing magnetic flux.

(c) Voltmeter reading during the second interval

The flux is decreasing during the second interval, producing a positive induced emf under the chosen sign convention.

With the switch open and the voltmeter ideal, the voltmeter reads approximately

\[ V_{\text{meter}}=+16\ \text{V} \]

There is a non-zero terminal voltage despite the absence of sustained circuit current.

(d) Total signed charge

The signed charge transferred is the current integrated over time. Here, the current is constant during each interval, so we can calculate the charge in each one separately.

During the first interval,

\[ \begin{aligned} Q_1&=I_1\Delta t\\ &=(-4.0)(0.10)\\ &=-0.40\ \text{C} \end{aligned} \]

During the second interval,

\[ Q_2=I_2\Delta t=0(0.10)=0\ \text{C} \]

Hence,

\[ \begin{aligned} Q_{\text{total}}&=Q_1+Q_2\\ &=-0.40+0\\ &=-0.40\ \text{C} \end{aligned} \]

The total signed charge is \(-0.40\ \text{C}\), or \(0.40\ \text{C}\) in the first-interval current direction.

The student has identified a real result: because the initial and final fluxes are equal, the average induced emf over the full interval is zero.

However, that doesn’t mean the emf was zero throughout. It was \(-16\ \text{V}\) in the first interval and \(+16\ \text{V}\) in the second.

If the circuit had remained closed with constant resistance throughout both intervals, the oppositely directed currents would have transferred equal amounts of charge in opposite directions under this simplified model. But the circuit was open during the second interval. The second emf could not drive a corresponding current, so it could not cancel the charge transferred in the first interval.

This is the deeper lesson: you must track the changing flux and the circuit condition over the same time intervals. The total change in flux alone doesn’t determine the total charge transferred when the circuit connection changes during the process.

08A quick exam self-check

Before submitting a response about electromagnetic induction, run through these five points:

• Have I calculated the change in magnetic flux, rather than using the magnetic field strength alone?
• Have I used Faraday’s law to find the induced emf?
• Have I checked whether a complete conducting path exists during each interval?
• Have I used the total resistance only when calculating current in a closed circuit?
• Have I distinguished what the voltmeter reads from what the ammeter measures?

Once you’ve established whether current can flow, the next useful step is applying Lenz’s law to determine its direction. Remember that the induced current opposes the change in magnetic flux, not necessarily the magnetic field itself.