How to Use Lenz’s Law to Predict Induced Current Direction

A bar magnet moves towards a wire loop and a galvanometer needle deflects. Pull the magnet away, and the needle deflects in the opposite direction. It is tempting to conclude that the induced current simply opposes the magnet’s motion.

That shortcut will eventually give you the wrong answer.

A loop can move through a magnetic field without producing an induced current. A stationary loop can also produce an induced current when its magnetic field changes. The important quantity is not the direction of motion. It is the change in magnetic flux through the loop.

The reliable method is to identify the loop, establish the external magnetic field direction, determine whether the magnetic flux is increasing or decreasing, and only then predict the induced field and current.

Before going further, consider this situation: a magnetic field points into the page, but its strength is decreasing. Should the induced field point out of the page because it opposes the external field?

No. It points into the page.

The loop is trying to maintain the magnetic flux that is disappearing. Lenz’s law opposes the change in flux, not necessarily the original magnetic field.

01Start with the loop, not the motion

Before predicting a current, establish exactly which conducting loop you are analysing.

This sounds obvious, but it prevents several common mistakes.

For example, in a sliding-rod experiment, the loop includes the moving rod, both rails, and the resistor connecting the rails. It is not just the moving rod.

Once you have identified the complete loop, establish three things:

• Which side of the loop are you looking from?
• Which direction does the external magnetic field pass through the loop?
• Is the magnetic flux through the loop increasing or decreasing?

Choose a viewing direction

For most HSC questions, the loop is drawn flat on the page. Imagine looking directly at its surface.

Choose the direction perpendicular to the loop as your normal. For the examples in this article, we will choose the normal pointing out of the page.

This gives us a consistent sign convention:

• Magnetic field out of the page: positive flux.
• Magnetic field into the page: negative flux.
• Anticlockwise conventional current: positive current.
• Clockwise conventional current: negative current.

The right-hand grip rule connects the current direction to the magnetic field produced by the loop.

Curl your right-hand fingers in the direction of conventional current. Your thumb points in the direction of the magnetic field through the centre of the loop.

Therefore, when viewed from the front:

• Anticlockwise current produces a magnetic field out of the page.
• Clockwise current produces a magnetic field into the page.

Separate the field direction from the field change

Now consider two situations involving exactly the same external magnetic field direction.

Situation A: The magnetic field into the page is increasing.

More magnetic flux is entering the page. The loop responds by producing a magnetic field out of the page.

The induced field opposes the increase.

Situation B: The magnetic field into the page is decreasing.

Less magnetic flux is entering the page. The loop responds by producing a magnetic field into the page.

The induced field tries to maintain the existing flux.

Notice the important difference: the external field points in the same direction in both situations, but the induced fields point in opposite directions.

This is why the statement “the induced field always opposes the external field” is incorrect.

Sometimes the induced field opposes the external field. Sometimes it points in the same direction.

What matters is whether the external magnetic flux is increasing or decreasing.

02Why motion does not determine current direction

Motion is often involved in electromagnetic induction, but it is only one possible cause of changing magnetic flux.

Consider three situations.

A magnet approaching a loop

When a magnet approaches a conducting loop, the magnetic field through the loop generally increases in magnitude.

The induced magnetic field opposes that increase.

When the magnet moves away, the magnetic flux generally decreases, so the induced field reverses.

This explains why the galvanometer deflects in opposite directions when the magnet approaches and retreats.

However, the magnet’s motion is not the fundamental rule. It simply causes the magnetic flux to change.

A loop entering a magnetic field

Imagine pulling a rectangular conducting loop into a region where a uniform magnetic field points into the page.

As more of the loop enters the field, the area exposed to the magnetic field increases.

Consequently, the magnetic flux into the page increases, even though the magnetic field strength remains constant.

The induced field must therefore point out of the page.

The important change is the area of the loop exposed to the magnetic field.

A loop moving entirely inside a uniform magnetic field

Now comes the contrasting case.

Imagine the same rectangular loop moving horizontally through a large region of uniform magnetic field. The entire loop remains inside the field.

The loop is moving, but its area, orientation, and the magnetic field strength through it remain constant.

Therefore, the magnetic flux does not change.

There is no induced emf around the loop and no induced current.

This is the key distinction:

Movement can cause a change in magnetic flux, but movement does not automatically mean magnetic flux changes.

For a uniform magnetic field passing through a flat loop, magnetic flux is calculated using:

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

where:

• \(\Phi_B\) is magnetic flux, 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 (m²).
• \(\theta\) is the angle between the magnetic field and the normal to the loop.

Flux can change because the magnetic field strength changes, the loop’s area changes, or the loop rotates and changes its orientation.

If several of these happen simultaneously, consider their combined effect. Do not assume that the direction of motion tells you the answer.

03The experiment that separates magnetic change from motion

Why are we so confident that motion is not the fundamental requirement?

An important experiment from the development of electromagnetic induction provides the answer.

The original problem

A moving-magnet demonstration produces a galvanometer deflection.

But what actually causes it?

One possible explanation is that moving the magnet through the coil generates an electrical effect.

Another is that the movement changes the magnetic flux through the coil, producing an induced emf.

The two explanations are difficult to distinguish when the magnet and the coil are moving relative to each other.

A better investigation keeps the coils stationary and changes the magnetic field instead.

The apparatus

Consider two separate insulated wire coils wound around a shared soft-iron ring.

The first coil, called the primary coil, is connected to a battery and a switch.

The second coil, called the secondary coil, is connected to a centre-zero galvanometer.

There is no direct electrical connection between the two coils. However, the iron ring provides a common magnetic path.

Predict what should happen

Suppose a steady magnetic field is enough to produce an induced current.

If that explanation is correct, the galvanometer should remain deflected for as long as the primary coil carries current.

But if a changing magnetic flux is necessary, the galvanometer should only deflect while the magnetic field through the secondary coil is changing.

Let’s follow the experiment.

Stage 1: Close the switch.

Current begins increasing in the primary coil.

Its magnetic field increases, causing magnetic flux through the secondary coil to increase.

The galvanometer briefly deflects.

Stage 2: Keep the switch closed.

The primary current reaches a steady value.

The magnetic field and magnetic flux through the secondary coil become constant.

The galvanometer returns to zero.

Stage 3: Open the switch.

The primary current decreases towards zero.

The magnetic flux through the secondary coil decreases, and the galvanometer briefly deflects in the opposite direction.

The observations can be summarised as follows:

These are the characteristic observations of Faraday’s 1831 iron-ring experiment.

The result is important for two reasons.

First, neither coil needs to move. A changing magnetic field can induce an emf in a stationary coil.

Second, a steady magnetic field does not produce a sustained induced current in this arrangement.

The experiment demonstrates that changing magnetic flux is the important condition, not mechanical movement.

Faraday’s historical experiment established electromagnetic induction. Heinrich Lenz’s 1834 work subsequently established the direction rule that we now use to predict the induced response.

The original apparatus is described by the Royal Institution, and Lenz’s original direction-rule paper is available through Annalen der Physik.

What the experiment does not tell you

The galvanometer tells you that the induced current reverses when the primary magnetic flux changes in the opposite direction.

However, a galvanometer deflection to the left or right does not automatically tell you whether the current is clockwise or anticlockwise.

To determine that, you must know the secondary coil’s winding direction, its electrical connections, and the viewing direction.

This is exactly why choosing the loop and its orientation before predicting current is so important.

04A reliable method for HSC Lenz’s law questions

You can now use a consistent sequence for almost every direction question.

Step 1: Identify the complete loop

Trace the entire conducting path.

If a rod slides along rails, include the rails and connecting resistor. If a coil is connected to a galvanometer, include the connecting wires.

Choose the side from which you are viewing the loop.

Step 2: Identify the external magnetic field

State its direction through the loop.

Use dots for out-of-page magnetic fields and crosses for into-page magnetic fields.

If the field is perpendicular to the loop, its direction is particularly easy to establish.

Step 3: Determine how magnetic flux changes

Ask whether the flux through the loop is increasing or decreasing.

Do not jump straight from motion to current.

If the question involves changing magnetic field strength, changing area, or rotation, consider the combined effect using:

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

If the problem provides initial and final values, calculate the signed flux at both points.

Step 4: Apply Lenz’s law

Determine which induced magnetic field would oppose the change in magnetic flux.

Remember:

• Increasing flux into the page produces an induced field out of the page.
• Decreasing flux into the page produces an induced field into the page.
• Increasing flux out of the page produces an induced field into the page.
• Decreasing flux out of the page produces an induced field out of the page.

Step 5: Convert the induced field into current direction

Use the right-hand grip rule.

Keep the reasoning in this order:

\[ \text{Flux change} \rightarrow \text{Induced field} \rightarrow \text{Induced current} \]

Do not skip directly from the direction of movement to the direction of current.

Remember: induced emf is not the same as induced current

There is one more distinction worth getting right.

A changing magnetic flux produces an induced electromotive force (emf). The induced emf is the electrical driving effect around the circuit.

An induced current requires a complete conducting path.

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

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

where:

• \(\varepsilon_{\text{avg}}\) is the average induced emf in volts (V).
• \(N\) is the number of turns in the coil.
• \(\Delta\Phi_B\) is the change in magnetic flux through one turn, in webers (Wb).
• \(\Delta t\) is the time interval in seconds (s).

The negative sign represents Lenz’s law. It tells you that the induced emf acts in the direction that opposes the change in magnetic flux, using your chosen sign convention.

For a simple closed, purely resistive circuit:

\[ I=\frac{\varepsilon}{R} \]

where \(I\) is current in amperes (A), \(\varepsilon\) is induced emf in volts, and \(R\) is total circuit resistance in ohms (Ω).

If the circuit is open, an induced emf can still exist, but there is no sustained current in the ideal circuit model.

This distinction matters in experiments. A high-resistance voltmeter can measure an induced potential difference across an open circuit, even though the circuit does not carry a significant current.

For the calculations below, assume uniform magnetic fields perpendicular to the loop, equal flux through each turn, and negligible self-inductance.

05Worked examples

06Common mistakes to avoid

Before attempting the practice questions, check that you can recognise these mistakes.

07Questions and solutions

The following questions progress from standard HSC calculations to situations where multiple changes in magnetic flux compete.

Question 1

A single-turn rectangular conducting loop has an area of \(0.012\text{ m}^2\).

It is placed in a uniform magnetic field perpendicular to its surface. The field points out of the page and increases from \(0.15\text{ T}\) to \(0.45\text{ T}\) over \(0.10\text{ s}\).

The total resistance of the loop is \(2.0\ \Omega\).

Determine:

a) The direction of the induced magnetic field.

b) The direction of the induced conventional current, viewed from the front.

c) The average induced emf.

d) The induced current.

Solution 1

The induced magnetic field points into the page, the conventional current is clockwise, the average induced emf is \(0.036\text{ V}\), and the current is \(0.018\text{ A}\).

a) Induced magnetic field

The external field points out of the page and increases in strength.

Therefore, the magnetic flux out of the page increases.

According to Lenz’s law, the induced magnetic field must oppose this increase.

It therefore points into the page.

b) Current direction

Using the right-hand grip rule, a magnetic field into the page is produced by a clockwise conventional current.

c) Average induced emf

Choose the normal pointing out of the page.

The change in magnetic flux is:

\[ \begin{aligned} \Delta\Phi_B &=A(B_f-B_i)\\ &=0.012(0.45-0.15)\\ &=0.0036\text{ Wb} \end{aligned} \]

Using Faraday’s law:

\[ \begin{aligned} \varepsilon_{\text{avg}} &=-\frac{\Delta\Phi_B}{\Delta t}\\ &=-\frac{0.0036}{0.10}\\ &=-0.036\text{ V} \end{aligned} \]

The magnitude of the induced emf is:

\[ |\varepsilon|=0.036\text{ V} \]

The negative sign confirms the clockwise direction.

d) Induced current

Using Ohm’s law:

\[ I=\frac{0.036}{2.0} \]

\[ I=0.018\text{ A} \]

The current is \(18\text{ mA}\), clockwise.

Question 2

A stationary coil contains 40 turns, each with an area of \(2.5\times10^{-3}\text{ m}^2\).

Its plane is parallel to the page, and its normal points out of the page.

A magnetic field perpendicular to the coil changes according to the following data:

The total resistance of the coil is \(5.0\ \Omega\).

For each time interval:

a) Calculate the average induced emf.

b) Determine the direction of the induced current.

c) Calculate the average current.

d) Explain whether the current reverses when the external magnetic field changes from pointing into the page to pointing out of the page.

Solution 2

The induced current is clockwise during both intervals. The average emf is \(0.15\text{ V}\) during the first interval and approximately \(0.0667\text{ V}\) during the second interval. The current does not reverse when the external field changes direction.

The key is to track signed magnetic flux rather than simply memorising the direction of the external field.

a) First interval: 0.00-0.20 s

The external magnetic field points into the page.

Using the normal pointing out of the page:

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

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

At \(0.20\text{ s}\):

\[ \Phi_f=-(0.10)(2.5\times10^{-3}) \]

\[ \Phi_f=-0.25\times10^{-3}\text{ Wb} \]

Therefore:

\[ \Delta\Phi_B=+0.75\times10^{-3}\text{ Wb} \]

The average induced emf is:

\[ \begin{aligned} \varepsilon_{\text{avg}} &=-40\frac{0.75\times10^{-3}}{0.20}\\ &=-0.15\text{ V} \end{aligned} \]

The negative sign indicates a clockwise current.

The average current is:

\[ I_{\text{avg}}=\frac{0.15}{5.0} \]

\[ I_{\text{avg}}=0.030\text{ A} \]

b) Second interval: 0.20-0.50 s

The external magnetic field changes from into the page to out of the page.

The initial flux is:

\[ \Phi_i=-0.25\times10^{-3}\text{ Wb} \]

The final flux is:

\[ \Phi_f=+(0.10)(2.5\times10^{-3}) \]

\[ \Phi_f=+0.25\times10^{-3}\text{ Wb} \]

Therefore:

\[ \Delta\Phi_B=+0.50\times10^{-3}\text{ Wb} \]

The average induced emf is:

\[ \begin{aligned} \varepsilon_{\text{avg}} &=-40\frac{0.50\times10^{-3}}{0.30}\\ &=-0.0667\text{ V} \end{aligned} \]

Again, the negative sign indicates a clockwise current.

The average current is:

\[ I_{\text{avg}}=\frac{0.0667}{5.0} \]

\[ I_{\text{avg}}=0.0133\text{ A} \]

c) Compare the two intervals

d) Why does the current not reverse?

This is the important conceptual part.

During the first interval, the external magnetic field points into the page but weakens. The signed magnetic flux becomes less negative.

During the second interval, the field changes from into the page to out of the page. The signed magnetic flux continues increasing.

In both intervals, the change in flux is in the positive, out-of-page direction.

Therefore, the induced field points into the page throughout both intervals.

The induced current remains clockwise, even though the external magnetic field eventually reverses direction.

A student who simply reverses the induced current whenever the external field reverses would get this question wrong.

Question 3

A rectangular conducting loop has a width of \(0.20\text{ m}\) in the direction of travel and a height of \(0.15\text{ m}\).

It moves horizontally to the right at a constant speed of \(0.50\text{ m s}^{-1}\) through a region of uniform magnetic field directed into the page.

The loop is completely inside the magnetic field region for the first \(0.40\text{ s}\). At the end of this interval, its right edge reaches the boundary of the field region.

It then takes another \(0.40\text{ s}\) to leave the field region completely.

The magnetic field strength is \(0.40\text{ T}\), and the total resistance of the loop is \(3.0\ \Omega\).

Determine:

a) Whether an induced emf exists while the loop is entirely inside the uniform field.

b) The average induced emf while the loop leaves the field region.

c) The direction and magnitude of the induced current while the loop leaves the field region.

d) Whether an induced emf exists after the loop has completely left the field region.

Solution 3

There is no induced emf while the loop is entirely inside the uniform field. While it leaves the field region, the induced emf has an average magnitude of \(0.030\text{ V}\), producing a clockwise current of \(0.010\text{ A}\). After the loop has completely left, the induced emf returns to zero.

a) Loop entirely inside the field

The loop is moving, but its area, orientation, and the magnetic field strength through it remain constant.

Therefore:

\[ \Delta\Phi_B=0 \]

Using Faraday’s law:

\[ \varepsilon=0\text{ V} \]

There is no induced current.

This is a direct example of why movement alone does not determine induction.

b) Loop leaving the field region

The area of the loop exposed to the magnetic field decreases as the loop moves out.

The magnetic flux into the page therefore decreases.

First, calculate the full loop area:

\[ A=0.20\times0.15 \]

\[ A=0.030\text{ m}^2 \]

At the start of the exit:

\[ \Phi_i=-(0.40)(0.030) \]

\[ \Phi_i=-0.012\text{ Wb} \]

Once the loop has completely left the field:

\[ \Phi_f=0\text{ Wb} \]

Therefore:

\[ \Delta\Phi_B=0-(-0.012) \]

\[ \Delta\Phi_B=+0.012\text{ Wb} \]

The average induced emf is:

\[ \begin{aligned} \varepsilon_{\text{avg}} &=-\frac{\Delta\Phi_B}{\Delta t}\\ &=-\frac{0.012}{0.40}\\ &=-0.030\text{ V} \end{aligned} \]

The average emf has a magnitude of \(0.030\text{ V}\).

c) Induced field and current

The external magnetic flux into the page is decreasing.

The induced field must therefore point into the page to oppose the decrease.

Using the right-hand grip rule, the induced current is clockwise.

Its magnitude is:

\[ I=\frac{|\varepsilon|}{R} \]

\[ I=\frac{0.030}{3.0} \]

\[ I=0.010\text{ A} \]

The induced current is \(10\text{ mA}\), clockwise.

d) After leaving the field

The loop is now outside the magnetic field region.

The magnetic flux through it is zero and remains constant.

Therefore, there is no induced emf or current.

Notice that the loop continues moving, but induction only occurs while its magnetic flux changes.

Question 4

A student performs an experiment using a bar magnet and a conducting coil connected to a galvanometer.

When the north pole of the magnet approaches the coil, the galvanometer deflects in one direction. When the magnet moves away, the galvanometer deflects in the opposite direction.

The student concludes:

“The direction of induced current is determined by the direction in which the magnet moves.”

a) Explain why this experiment alone does not establish that motion determines the induced current direction.

b) Propose a modified experiment using two coils, a battery, a switch, and a galvanometer that tests whether mechanical movement is necessary for induction.

c) Describe the expected galvanometer readings when the primary circuit is switched on, held on, and switched off.

d) Explain what would happen if the secondary circuit were open but a high-resistance voltmeter were connected across the gap.

Solution 4

The student’s conclusion is not justified because the magnet’s movement and the change in magnetic flux occur together. A stationary two-coil experiment can demonstrate induction without mechanical movement.

a) Limitation of the original experiment

When the magnet approaches the coil, the magnetic flux through the coil changes.

When it retreats, the magnetic flux changes in the opposite direction.

The experiment shows that reversing the magnet’s motion reverses the induced current in this particular arrangement.

However, it does not establish that movement itself is the fundamental cause.

The movement changes the magnetic flux, and that is the quantity responsible for the induced emf.

b) Modified experimental arrangement

Use two separate coils wound around a shared soft-iron ring.

Connect the primary coil to a battery and switch.

Connect the secondary coil to a centre-zero galvanometer.

Keep both coils stationary.

The battery and switch control the magnetic field produced by the primary coil. The secondary coil detects the induced current.

This arrangement removes mechanical movement as a variable.

c) Expected observations

When the switch is closed, the primary current increases. Magnetic flux through the secondary coil increases, producing a brief galvanometer deflection.

When the switch remains closed, the primary current becomes steady. Magnetic flux stops changing, so the galvanometer returns to zero.

When the switch is opened, the primary current decreases. Magnetic flux decreases, producing a brief deflection in the opposite direction.

The observations demonstrate that mechanical movement is not necessary.

A changing magnetic flux is sufficient to produce an induced emf.

The exact clockwise or anticlockwise direction cannot be determined from the words left and right on the galvanometer alone. You would also need the secondary coil’s winding direction and the meter lead connections.

d) Open secondary circuit

If the secondary circuit is open, a changing magnetic flux can still produce an induced emf.

The high-resistance voltmeter can measure the resulting potential difference across the gap.

However, there is no complete conducting path for a sustained current.

This demonstrates the distinction between induced emf and induced current.

Question 5

A rectangular conducting loop has a width of \(0.20\text{ m}\) and a height of \(0.20\text{ m}\).

It moves horizontally to the right at a constant speed into a large region of uniform magnetic field directed into the page.

At \(t=0\), half of the loop’s width is inside the magnetic field region.

At \(t=0.10\text{ s}\), the entire loop has entered the region. It continues moving to the right and remains completely inside the field until \(t=0.20\text{ s}\).

At the same time, the magnetic field strength decreases linearly:

The field is uniform over the part of the loop inside the region and zero outside it.

The total loop resistance is \(4.0\ \Omega\).

Use an out-of-page normal as your positive direction.

a) Calculate the signed magnetic flux through the loop at each of the three times.

b) Calculate the average induced emf during each \(0.10\text{ s}\) interval.

c) Determine the induced magnetic field and current direction during each interval.

d) Calculate the average current during each interval.

e) A student calculates the change in flux between \(t=0\) and \(t=0.20\text{ s}\), obtains zero, and concludes that no induced current exists at any time. Explain precisely why this reasoning is incorrect.

Solution 5

The average induced emf is \(+0.040\text{ V}\) during the first interval and \(-0.040\text{ V}\) during the second interval. The induced current is anticlockwise during the first interval and clockwise during the second. Both intervals have an average current magnitude of \(0.010\text{ A}\).

The student’s conclusion is incorrect because the total change in flux over the entire \(0.20\text{ s}\) interval does not reveal what happened during the two individual intervals.

This question combines two competing effects: the loop’s exposed area increases while the external magnetic field weakens.

a) Calculate the signed magnetic flux

Remember that the external magnetic field points into the page, so the signed magnetic flux is negative.

At \(t=0\), only half the loop is inside the field.

The area exposed to the field is:

\[ A_0=0.20\times0.10 \]

\[ A_0=0.020\text{ m}^2 \]

Therefore:

\[ \Phi_0=-(0.40)(0.020) \]

\[ \Phi_0=-0.0080\text{ Wb} \]

At \(t=0.10\text{ s}\), the entire loop is inside the field.

\[ A_1=0.20\times0.20 \]

\[ A_1=0.040\text{ m}^2 \]

Therefore:

\[ \Phi_1=-(0.30)(0.040) \]

\[ \Phi_1=-0.0120\text{ Wb} \]

At \(t=0.20\text{ s}\), the loop remains completely inside the field.

Its area is still \(0.040\text{ m}^2\).

Therefore:

\[ \Phi_2=-(0.20)(0.040) \]

\[ \Phi_2=-0.0080\text{ Wb} \]

Our results are:

Notice that the flux first becomes more negative, then becomes less negative.

b) Calculate the average induced emf during each interval

First interval: \(0.00\)-\(0.10\text{ s}\).

The change in flux is:

\[ \Delta\Phi_1=-0.0120-(-0.0080) \]

\[ \Delta\Phi_1=-0.0040\text{ Wb} \]

Using Faraday’s law:

\[ \begin{aligned} \varepsilon_1 &=-\frac{\Delta\Phi_1}{\Delta t}\\ &=-\frac{-0.0040}{0.10}\\ &=+0.040\text{ V} \end{aligned} \]

The average induced emf is \(+0.040\text{ V}\).

Second interval: \(0.10\)-\(0.20\text{ s}\).

The change in flux is:

\[ \Delta\Phi_2=-0.0080-(-0.0120) \]

\[ \Delta\Phi_2=+0.0040\text{ Wb} \]

Therefore:

\[ \begin{aligned} \varepsilon_2 &=-\frac{\Delta\Phi_2}{\Delta t}\\ &=-\frac{0.0040}{0.10}\\ &=-0.040\text{ V} \end{aligned} \]

The average induced emf is \(-0.040\text{ V}\).

These are interval averages. During the first interval, both the field strength and the exposed area change, so the instantaneous emf is not constant.

c) Determine the induced magnetic field and current direction

During the first interval, the magnitude of the magnetic flux into the page increases.

Although the external magnetic field weakens, the area exposed to the field doubles.

The increase in exposed area has the greater effect.

The induced magnetic field must therefore point out of the page to oppose the increasing into-page flux.

Using the right-hand grip rule, the current is anticlockwise.

During the second interval, the loop is completely inside the field, so its exposed area remains constant.

The external magnetic field continues weakening.

The magnetic flux into the page decreases, so the induced field points into the page.

The current is now clockwise.

The current reverses even though the loop continues moving to the right in the same direction.

This is the central result of the question: the direction of movement has remained unchanged, but the direction of the induced current has reversed because the flux change has reversed.

d) Calculate the average current

The total resistance is \(4.0\ \Omega\).

For the first interval:

\[ I_{1,\text{avg}}=\frac{0.040}{4.0} \]

\[ I_{1,\text{avg}}=0.010\text{ A} \]

The average current is \(0.010\text{ A}\), anticlockwise.

For the second interval:

\[ I_{2,\text{avg}}=\frac{0.040}{4.0} \]

\[ I_{2,\text{avg}}=0.010\text{ A} \]

The average current is \(0.010\text{ A}\), clockwise.

These average current values assume a purely resistive circuit with negligible self-inductance.

e) Why the student’s conclusion is wrong

The student calculates the total change in flux:

\[ \Delta\Phi_{\text{total}} =\Phi_2-\Phi_0 \]

\[ =-0.0080-(-0.0080) \]

\[ =0\text{ Wb} \]

Therefore, the average induced emf over the entire \(0.20\text{ s}\) interval is:

\[ \varepsilon_{\text{avg}} =-\frac{0}{0.20} =0\text{ V} \]

That calculation is mathematically correct.

The mistake is interpreting a zero average emf as proof that no emf was induced at any time.

During the first \(0.10\text{ s}\), the flux becomes more negative, producing a positive induced emf.

During the next \(0.10\text{ s}\), the flux becomes less negative, producing a negative induced emf.

The two contributions cancel when averaged over the entire interval.

But they do not cancel at every moment.

The loop experiences induced emf and current in both intervals, with opposite directions.

This is a particularly important distinction in induction questions: a zero net change in magnetic flux over an interval does not necessarily mean zero induction throughout that interval.

08Where to go next

Once you can confidently predict the direction of induced current, the next step is calculating the magnitude of induced emf from the rate of change of magnetic flux.

Practise interpreting the gradient of a magnetic flux-time graph in How to Calculate Induced EMF from a Piecewise Flux Graph.

Keep the two jobs separate: Lenz’s law determines the direction, while Faraday’s law determines the induced emf magnitude.