How to Calculate Induced EMF When a Coil’s Area Shrinks

A rectangular coil sits in a steady magnetic field. You squeeze the coil so that the area enclosed by its wire becomes smaller. The magnetic field strength stays exactly the same. Will an induced EMF still be produced?

Yes. Changing the area changes the magnetic flux through the coil, even when the magnetic field itself remains constant.

To find the average induced EMF when a coil’s area shrinks, use:

\[ |\mathcal E_{\text{avg}}| =\frac{NB|\cos\theta|(A_i-A_f)}{\Delta t} \]

This simplified formula applies when the magnetic field strength and the coil’s orientation remain constant.

Here:

• \(N\) is the number of turns in the coil.
• \(B\) is the magnetic field strength, in teslas (T).
• \(\theta\) is the angle between the magnetic field and the normal to the coil.
• \(A_i\) is the initial area enclosed by each turn, in \(\text{m}^2\).
• \(A_f\) is the final enclosed area, in \(\text{m}^2\).
• \(\Delta t\) is the time taken for the area to shrink, in seconds (s).

If the magnetic field is perpendicular to the plane of the coil, \(\theta=0^\circ\), so the equation becomes:

\[ |\mathcal E_{\text{avg}}| =\frac{NB(A_i-A_f)}{\Delta t} \]

The important part is understanding why this works, when you can use the shortcut, and how to determine the direction of the induced current separately.

01Why shrinking the area produces an induced EMF

Imagine holding a circular hoop in steady rain. If you make the hoop smaller, less rain passes through its opening.

The magnetic equivalent is called magnetic flux, represented by \(\Phi\). It describes the magnetic field passing through the area enclosed by one turn of the coil, taking its orientation into account.

For a uniform magnetic field:

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

The area \(A\) means the area enclosed by the wire, not the surface area of the wire itself.

The angle \(\theta\) is measured between the magnetic field and the normal to the coil. The normal is an imaginary line perpendicular to the plane of the coil.

Consider three possible orientations:

This reveals an important detail.

Shrinking the area does not automatically mean that magnetic flux decreases. The area must contribute to the magnetic flux in the first place.

If the magnetic field is parallel to the coil’s plane, then \(\cos90^\circ=0\). Shrinking the area while maintaining this orientation produces no change in magnetic flux and therefore no induced EMF.

For the usual HSC question, however, the magnetic field is perpendicular to the coil’s plane. This makes the calculation much simpler.

02An experiment that isolates the effect of area

How could we demonstrate that changing the area alone can produce an induced EMF?

The key is to keep the magnetic field constant while changing the size of the loop.

Setting up a variable-area loop

Imagine a rectangular conducting loop placed between the pole pieces of an electromagnet.

The loop consists of two parallel conducting rails, joined at one end. A conducting bar slides along the rails, completing the rectangular circuit. A sensitive galvanometer is connected in series with the loop.

The magnetic field points into the page and is perpendicular to the loop.

The distance between the rails is \(L\), and the distance from the fixed end to the sliding bar is \(x\).

Therefore, the enclosed area is:

\[ A=Lx \]

What should happen when the bar moves?

Before looking at the result, predict what the galvanometer should show.

As the sliding bar moves towards the fixed end, \(x\) decreases. Since \(A=Lx\), the enclosed area also decreases.

The magnetic field strength stays constant, and the coil’s orientation does not change.

Therefore, the magnetic flux decreases. Faraday’s law predicts an induced EMF, and because the circuit is closed, an induced current should flow.

If the bar stops moving, the area stops changing. The galvanometer should return to zero.

If the bar moves outwards again, increasing the area, the galvanometer should deflect in the opposite direction.

These are the predicted observations for this experimental arrangement.

What could make the experiment unreliable?

The original setup needs to control more than just the bar’s movement.

For example, if the bar moves outside the uniform region of the magnetic field, the field strength through the loop may change as well. We could no longer attribute the entire flux change to the changing area.

Similarly, poor electrical contacts could change the circuit resistance and affect the galvanometer reading.

A better design would use:

• An electromagnet with constant current to maintain a steady magnetic field.
• Pole pieces that produce a reasonably uniform field over the loop’s movement.
• Guides that keep the sliding bar parallel to the fixed conductor.
• Measurements of the rail separation, bar position, and elapsed time.
• A calibrated circuit resistance, or a suitable high-resistance voltmeter, to distinguish induced EMF from induced current.

The experiment would then isolate the effect of changing area much more reliably.

A small experimental calculation

Suppose the illustrative apparatus has the following values:

The initial and final areas are:

\[ A_i=Lx_i=0.20(0.15)=0.030\,\text{m}^2 \]

\[ A_f=Lx_f=0.20(0.10)=0.020\,\text{m}^2 \]

The magnitude of the average induced EMF is:

\[ \begin{aligned} |\mathcal E| &=\frac{NB(A_i-A_f)}{\Delta t}\\ &=\frac{1(0.40)(0.030-0.020)}{0.25}\\ &=0.016\,\text{V} \end{aligned} \]

The predicted average induced EMF is \(0.016\,\text{V}\), or \(16\,\text{mV}\).

The important experimental conclusion is that an EMF can be induced by changing the area of a loop, without changing the magnetic field strength.

03Calculating induced EMF when the area shrinks

Now let’s turn that physical idea into a reliable HSC calculation method.

Start with Faraday’s law:

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

The change in magnetic flux is always:

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

Substituting the magnetic flux equation gives:

\[ \mathcal E_{\text{avg}} =-N\frac{B A_f\cos\theta-B A_i\cos\theta}{\Delta t} \]

If \(B\) and \(\theta\) remain constant:

\[ \mathcal E_{\text{avg}} =-NB\cos\theta \frac{A_f-A_i}{\Delta t} \]

For the magnitude of the induced EMF:

\[ |\mathcal E_{\text{avg}}| =NB|\cos\theta| \frac{A_i-A_f}{\Delta t} \]

Notice that we subtract the final area from the initial area when calculating the magnitude.

This is because the area is shrinking, so \(A_i>A_f\).

04When a side slides, use the rate of area change

Sometimes an HSC question does not give you the initial and final areas directly.

Instead, it describes a rectangular loop with one side moving inwards at a particular speed.

You can still use the same method.

Consider a rectangular loop with:

• A fixed length \(L\).
• A movable side at distance \(x\) from the opposite side.
• A uniform magnetic field perpendicular to the loop.

The enclosed area is:

\[ A=Lx \]

If the movable side travels inwards at speed \(v\), the width decreases.

The magnitude of the rate of area change is:

\[ \frac{|\Delta A|}{\Delta t}=Lv \]

This gives a useful alternative form of Faraday’s law:

\[ |\mathcal E|=NBLv \]

This simplified version assumes that the magnetic field is perpendicular to the coil and remains constant.

If the field makes an angle \(\theta\) with the coil’s normal, include the orientation factor:

\[ |\mathcal E|=NBLv|\cos\theta| \]

This is particularly useful for questions involving a moving conductor or a changing rectangular loop.

05Determining the direction of the induced current

So far, we have calculated the magnitude of the induced EMF.

But what if an HSC question asks for the direction of the induced current?

This is where Lenz’s law comes in.

Lenz’s law states that the induced current flows in a direction that produces a magnetic field opposing the change in magnetic flux.

It does not necessarily oppose the original magnetic field. It opposes the change in that field’s magnetic flux through the loop.

Case 1: Magnetic field points into the page

Imagine looking directly at a coil with a magnetic field pointing into the page.

The coil’s area decreases, so the magnetic flux into the page decreases.

The induced current must therefore produce its own magnetic field into the page to oppose this reduction.

Using the right-hand grip rule, a clockwise current produces a magnetic field into the page.

Therefore, the induced current is clockwise.

Case 2: Magnetic field points out of the page

Now reverse the magnetic field.

The magnetic flux points out of the page. As the area shrinks, this outward flux decreases.

The induced magnetic field must point out of the page to oppose the decrease.

Using the right-hand grip rule, this requires an anticlockwise current.

The two cases can be summarised as follows:

Remember that these current directions assume the circuit is closed.

Induced EMF is not the same as induced current

This distinction is particularly important in HSC Physics.

An induced EMF can exist even when the circuit is open. However, an open circuit does not provide a complete conducting path for a sustained current.

Think of the two quantities separately:

• Induced EMF: calculated using Faraday’s law.
• Induced current: calculated using Ohm’s law when the circuit is closed and its resistance is known.

For an ideal, purely resistive circuit:

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

In the examples above, we assumed that self-inductance and other transient effects were negligible.

Do not automatically calculate a current just because a question gives you an induced EMF.

06Common mistakes when the coil’s area shrinks

Before attempting exam questions, check these common errors.

One final qualification: the simplified equation assumes a uniform magnetic field through the coil. If the field is non-uniform, you cannot generally calculate flux using \(BA\cos\theta\) with a single field strength. You need to account for how the field varies across the enclosed area.

07Questions and solutions

The following questions build from a straightforward area-change calculation to situations where other changes in the magnetic flux compete with the shrinking area.

Question 1

A 50-turn coil is placed in a uniform magnetic field of \(0.25\,\text{T}\). The magnetic field is perpendicular to the plane of the coil.

The enclosed area decreases from \(0.030\,\text{m}^2\) to \(0.018\,\text{m}^2\) over \(0.20\,\text{s}\).

Calculate the magnitude of the average induced EMF.

Assume that the magnetic field strength and coil orientation remain constant.

Solution 1

The magnitude of the average induced EMF is \(0.75\,\text{V}\).

Because the magnetic field is perpendicular to the coil, \(\theta=0^\circ\), and therefore \(\cos\theta=1\).

Start with:

\[ |\mathcal E| =\frac{NB(A_i-A_f)}{\Delta t} \]

Substitute:

\[ \begin{aligned} |\mathcal E| &=\frac{50(0.25)(0.030-0.018)}{0.20}\\ &=\frac{50(0.25)(0.012)}{0.20}\\ &=\frac{0.15}{0.20}\\ &=0.75\,\text{V} \end{aligned} \]

The induced EMF has a magnitude of \(0.75\,\text{V}\).

Notice that we used the difference between the initial and final areas, not simply the final area.

Question 2

A 120-turn coil is placed in a uniform magnetic field of \(0.35\,\text{T}\).

The enclosed area decreases from \(0.050\,\text{m}^2\) to \(0.032\,\text{m}^2\) over \(0.25\,\text{s}\).

The magnetic field makes an angle of \(60^\circ\) with the plane of the coil. The total resistance of the closed circuit is \(3.0\,\Omega\).

Calculate:

(a) The magnitude of the average induced EMF.

(b) The magnitude of the induced current.

Assume the magnetic field strength and coil orientation remain constant.

Solution 2

The average induced EMF is approximately \(2.6\,\text{V}\), and the induced current has a magnitude of approximately \(0.87\,\text{A}\).

The important detail is the angle.

The question gives the angle between the magnetic field and the plane of the coil, but the magnetic flux equation requires the angle between the field and the normal.

Therefore:

\[ \theta=90^\circ-60^\circ=30^\circ \]

Part (a): Induced EMF

Use:

\[ |\mathcal E| =\frac{NB|\cos\theta|(A_i-A_f)}{\Delta t} \]

Substitute:

\[ \begin{aligned} |\mathcal E| &=\frac{120(0.35)\cos30^\circ(0.050-0.032)} {0.25}\\ &=\frac{120(0.35)(0.866)(0.018)}{0.25}\\ &\approx2.62\,\text{V} \end{aligned} \]

Therefore:

\[ |\mathcal E|\approx2.6\,\text{V} \]

Part (b): Induced current

The circuit is closed, and its total resistance is known.

Apply Ohm’s law:

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

\[ \begin{aligned} I&=\frac{2.62}{3.0}\\ &\approx0.87\,\text{A} \end{aligned} \]

The induced current has a magnitude of \(0.87\,\text{A}\).

The main trap was using \(60^\circ\) directly in the cosine. Always check whether the given angle is measured from the coil’s plane or its normal.

Question 3

A 20-turn square coil has a side length of \(0.20\,\text{m}\). It is reshaped into a rectangle measuring \(0.30\,\text{m}\) by \(0.10\,\text{m}\) over \(0.10\,\text{s}\).

The wire length remains constant throughout the process.

The coil is perpendicular to a uniform magnetic field of \(0.50\,\text{T}\), directed into the page. The total resistance of the closed circuit is \(5.0\,\Omega\).

Calculate:

(a) The magnitude of the average induced EMF.

(b) The magnitude and direction of the induced current.

Explain why the unchanged wire length does not mean that the magnetic flux remains constant.

Solution 3

The induced EMF is \(1.0\,\text{V}\), and the induced current is \(0.20\,\text{A}\) clockwise.

The key is to calculate the enclosed area rather than assume that an unchanged wire length means an unchanged area.

Part (a): Induced EMF

First, calculate the initial area.

The coil starts as a square:

\[ A_i=0.20(0.20) \]

\[ A_i=0.040\,\text{m}^2 \]

Now calculate the final area.

The coil becomes a rectangle:

\[ A_f=0.30(0.10) \]

\[ A_f=0.030\,\text{m}^2 \]

The area has decreased by:

\[ A_i-A_f=0.040-0.030 \]

\[ =0.010\,\text{m}^2 \]

Apply Faraday’s law:

\[ |\mathcal E| =\frac{NB(A_i-A_f)}{\Delta t} \]

\[ \begin{aligned} |\mathcal E| &=\frac{20(0.50)(0.010)}{0.10}\\ &=1.0\,\text{V} \end{aligned} \]

Part (b): Induced current

Apply Ohm’s law:

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

\[ I=\frac{1.0}{5.0} =0.20\,\text{A} \]

To determine the direction, consider the original magnetic field.

The magnetic field points into the page. As the area decreases, the magnetic flux into the page decreases.

According to Lenz’s law, the induced current must produce a magnetic field into the page to oppose this decrease.

Using the right-hand grip rule, this requires a clockwise current.

Therefore, the induced current is \(0.20\,\text{A}\) clockwise.

Why the wire length does not determine the flux

Both shapes have the same perimeter:

\[ 4(0.20)=0.80\,\text{m} \]

\[ 2(0.30+0.10)=0.80\,\text{m} \]

However, their enclosed areas are different.

The wire length has stayed constant, but the area enclosed by the wire has decreased. Therefore, the magnetic flux has decreased and an induced EMF has been produced.

Question 4

A 60-turn coil is placed perpendicular to a magnetic field directed into the page.

Initially, the magnetic field strength is \(0.30\,\text{T}\), and the enclosed area is \(0.040\,\text{m}^2\).

After \(0.20\,\text{s}\), the magnetic field strength has increased to \(0.42\,\text{T}\), while the enclosed area has decreased to \(0.030\,\text{m}^2\).

The total resistance of the closed circuit is \(6.0\,\Omega\).

Calculate:

(a) The magnitude of the average induced EMF.

(b) The magnitude and direction of the induced current.

A student claims that the induced magnetic field must point into the page because the coil’s area is shrinking. Explain why this prediction is incorrect.

Solution 4

The induced EMF is \(0.18\,\text{V}\), and the induced current is \(0.030\,\text{A}\) anticlockwise.

The student’s prediction is incorrect because the magnetic field strength is also changing. We must calculate the total change in magnetic flux rather than consider the area change alone.

Part (a): Induced EMF

The initial magnetic flux through one turn is:

\[ \begin{aligned} \Phi_i&=B_iA_i\\ &=0.30(0.040)\\ &=0.012\,\text{Wb} \end{aligned} \]

The final magnetic flux is:

\[ \begin{aligned} \Phi_f&=B_fA_f\\ &=0.42(0.030)\\ &=0.0126\,\text{Wb} \end{aligned} \]

Now calculate the change in flux:

\[ \begin{aligned} \Delta\Phi &=\Phi_f-\Phi_i\\ &=0.0126-0.012\\ &=0.0006\,\text{Wb} \end{aligned} \]

The magnetic flux into the page has actually increased.

Apply Faraday’s law:

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

\[ \begin{aligned} |\mathcal E| &=\frac{60(0.0006)}{0.20}\\ &=0.18\,\text{V} \end{aligned} \]

Part (b): Induced current and direction

Calculate the current:

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

\[ I=\frac{0.18}{6.0} =0.030\,\text{A} \]

Now apply Lenz’s law.

The original magnetic field points into the page, and the total magnetic flux into the page has increased.

The induced magnetic field must oppose this increase, so it must point out of the page.

Using the right-hand grip rule, an anticlockwise current produces a magnetic field out of the page.

Therefore, the induced current is \(0.030\,\text{A}\) anticlockwise.

Why the student’s prediction fails

The area decreased, which by itself would reduce the magnetic flux.

However, the magnetic field strength increased from \(0.30\,\text{T}\) to \(0.42\,\text{T}\).

That increase was large enough to more than compensate for the reduction in area.

This is why you must calculate the initial and final flux whenever more than one quantity changes.

Question 5

A 100-turn coil is connected in series with a resistor. The total circuit resistance is \(5.0\,\Omega\).

The coil is perpendicular to a uniform magnetic field of \(0.40\,\text{T}\), directed into the page.

Its enclosed area changes in two stages:

• From \(t=0\) to \(t=0.30\,\text{s}\), the area decreases uniformly from \(0.050\,\text{m}^2\) to \(0.020\,\text{m}^2\).
• From \(t=0.30\,\text{s}\) to \(t=0.90\,\text{s}\), the area increases uniformly back to \(0.050\,\text{m}^2\).

Assume the total resistance remains constant and ignore self-inductance.

Define clockwise current as positive when viewed directly at the coil.

Determine:

(a) The signed average induced EMF and signed induced current during each stage.

(b) The signed charge that passes through the resistor over the entire \(0.90\,\text{s}\).

(c) The total amount of charge that passes through the resistor, counting both directions.

(d) A student calculates the average induced EMF over the entire \(0.90\,\text{s}\) and obtains zero. They conclude that no induced current flowed at any point. Explain why their conclusion is wrong.

Solution 5

During compression, the induced EMF is \(+4.0\,\text{V}\) and the current is \(+0.80\,\text{A}\), clockwise.

During expansion, the induced EMF is \(-2.0\,\text{V}\) and the current is \(-0.40\,\text{A}\), anticlockwise.

The net signed charge is zero, but the total amount of charge passing through the resistor is \(0.48\,\text{C}\).

The distinction between signed charge and total charge is the central idea in this question.

Part (a): Signed EMF and current during each stage

The positive current direction is defined as clockwise.

Because the magnetic field points into the page, the positive magnetic flux direction is also into the page.

For the first stage, the area decreases.

Initial flux:

\[ \Phi_i=0.40(0.050)=0.020\,\text{Wb} \]

Final flux:

\[ \Phi_f=0.40(0.020)=0.008\,\text{Wb} \]

The change in flux is:

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

Apply Faraday’s law:

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

\[ \begin{aligned} \mathcal E_1 &=-100\left(\frac{-0.012}{0.30}\right)\\ &=+4.0\,\text{V} \end{aligned} \]

The positive sign indicates clockwise EMF.

Calculate the current:

\[ I_1=\frac{4.0}{5.0} =+0.80\,\text{A} \]

Therefore, during compression:

• Induced EMF: \(+4.0\,\text{V}\).
• Induced current: \(+0.80\,\text{A}\), clockwise.

Now consider the second stage.

The coil expands, so its area increases.

The flux changes from \(0.008\,\text{Wb}\) back to \(0.020\,\text{Wb}\).

Therefore:

\[ \Delta\Phi=0.020-0.008 =+0.012\,\text{Wb} \]

The second interval lasts:

\[ \Delta t_2=0.90-0.30=0.60\,\text{s} \]

Apply Faraday’s law:

\[ \begin{aligned} \mathcal E_2 &=-100\left(\frac{0.012}{0.60}\right)\\ &=-2.0\,\text{V} \end{aligned} \]

The negative sign indicates anticlockwise EMF.

Calculate the current:

\[ I_2=\frac{-2.0}{5.0} =-0.40\,\text{A} \]

Therefore, during expansion:

• Induced EMF: \(-2.0\,\text{V}\).
• Induced current: \(0.40\,\text{A}\), anticlockwise.

Notice that the current is smaller during expansion because the same area change takes twice as long.

Part (b): Net signed charge

The charge passing through a circuit is:

\[ Q=I\Delta t \]

During compression:

\[ \begin{aligned} Q_1&=(+0.80)(0.30)\\ &=+0.24\,\text{C} \end{aligned} \]

During expansion:

\[ \begin{aligned} Q_2&=(-0.40)(0.60)\\ &=-0.24\,\text{C} \end{aligned} \]

The net signed charge is:

\[ \begin{aligned} Q_{\text{net}} &=Q_1+Q_2\\ &=0.24-0.24\\ &=0\,\text{C} \end{aligned} \]

Therefore, the net signed charge passing through the resistor is zero.

Part (c): Total amount of charge

This time, we count the charge passing through the resistor regardless of its direction.

Therefore, we add the magnitudes:

\[ \begin{aligned} Q_{\text{total}} &=|Q_1|+|Q_2|\\ &=0.24+0.24\\ &=0.48\,\text{C} \end{aligned} \]

The total amount of charge passing through the resistor is \(0.48\,\text{C}\).

The zero net charge does not mean that no charge moved. It means that equal amounts of charge passed in opposite directions during the two stages.

Part (d): Why the student’s conclusion is wrong

The student is correct that the average induced EMF over the entire \(0.90\,\text{s}\) is zero.

The initial and final areas are identical, and the magnetic field is constant. Therefore, the initial and final magnetic fluxes are identical.

Using Faraday’s law over the entire interval:

\[ \begin{aligned} \mathcal E_{\text{avg}} &=-N\frac{\Phi_f-\Phi_i}{\Delta t}\\ &=-100\frac{0.020-0.020}{0.90}\\ &=0\,\text{V} \end{aligned} \]

However, this is an average over two stages with opposite changes in flux.

During the first stage, the flux decreases, producing a clockwise current.

During the second stage, the flux increases, producing an anticlockwise current.

The average over the complete process hides these individual changes.

This is the important distinction: zero net change in magnetic flux over an entire interval does not mean zero induced EMF throughout that interval.

You must examine what happens during each stage before drawing conclusions about the induced current.

08Where to go next

Once you can calculate induced EMF from a shrinking coil, the next useful step is recognising the same process when the question gives you magnetic flux rather than the coil’s dimensions.

The underlying calculation is still Faraday’s law. The difference is that you no longer need to calculate \(BA\cos\theta\) yourself.

For a broader explanation of flux linkage, Faraday’s law, and Lenz’s law, read Faraday’s Law for HSC Physics: Calculating Induced EMF.

If the question provides a magnetic-flux-time graph, learn how to calculate the gradient and use it to find induced EMF in How to Read a Magnetic-Flux-Time Graph to Calculate EMF.

The key habit to carry into those questions is the same: work out what happens to magnetic flux first. Then calculate the induced EMF, and only afterwards determine the current.