How Wireless Charging Transfers Energy Through Changing Magnetic Flux

Your phone can gain energy while sitting on a charging pad, even though no metal contacts connect the pad to the phone. Predict what happens if the current in the pad’s transmitter coil is switched on and then held perfectly constant: will the phone keep receiving energy?

Not through sustained electromagnetic induction. Switching on the current changes the magnetic field for a moment, so a voltage can be induced. Once the current and field settle, the magnetic flux through a stationary receiver stops changing, and the induced EMF falls to zero.

A wireless charger avoids this problem by driving alternating current through its transmitter coil. The changing current produces a changing magnetic field, which allows energy to be transferred to a receiver coil inside the phone.

01From transmitter current to receiver voltage

Start with something familiar: move a magnet towards a loop of wire, and a voltage can be induced in the wire. Hold the magnet still, and the induced voltage disappears, even though the magnetic field is still there.

Wireless charging follows the same principle without needing a moving magnet. Instead, the transmitter uses changing electric current to produce a changing magnetic field.

The analogy has a limit. A wireless charger uses two nearby coils and rapidly alternating current, rather than a hand moving a permanent magnet. The underlying induction principle, however, is the same.

Why changing current matters

A current flowing through a wire produces a magnetic field. If the current increases, decreases, or reverses direction, the magnetic field changes too.

In a wireless charger, the transmitter coil is positioned so that some of its changing magnetic field passes through the receiver coil. The amount of magnetic field passing through the receiver changes over time.

This is the key idea: induction depends on changing magnetic flux, not simply on the presence of a magnetic field.

A steady direct current can produce a magnetic field, but if the current and the relative positions of the coils remain constant, the magnetic flux through the receiver also remains constant. There is no sustained induced EMF.

A brief EMF can still occur when the current is switched on or off, because the magnetic field changes during those transitions.

Changing current is one way to change magnetic flux, but it is not the only way. A changing magnetic field, a changing coil area, or a change in the angle between the field and the coil can also change the flux.

In a charging pad, the transmitter current provides the changing field while the phone remains stationary.

02Magnetic flux: measuring how much field passes through a coil

Magnetic flux is a measure of how much magnetic field passes through a surface. For a uniform magnetic field passing through a flat coil, the flux through one turn is

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

where:

• \(\Phi_B\) is the magnetic flux through one turn, measured in webers (Wb).
• \(B\) is the magnetic field strength, measured in teslas (T).
• \(A\) is the area enclosed by one turn, measured in square metres (\(\mathrm{m^2}\)).
• \(\theta\) is the angle between the magnetic field and the direction perpendicular to the coil’s plane.

The angle is worth checking carefully. When the magnetic field is perpendicular to the coil’s plane, \(\theta=0^\circ\), so the flux is greatest. When the field lies parallel to the plane, \(\theta=90^\circ\), so the flux is zero.

For wireless charging, the receiver’s position and orientation affect how much of the transmitter’s changing magnetic field passes through it.

Faraday’s law of electromagnetic induction

Now we can describe the process mathematically.

Faraday’s law states that the magnitude of the average induced EMF depends on how quickly the magnetic flux through the coil changes and how many turns the coil contains.

\[
\boxed{
|\mathcal{E}_{\mathrm{avg}}|
=N\frac{|\Delta\Phi_B|}{\Delta t}
}
\]

Here:

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

The signed form includes a negative sign:

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

That sign represents the direction of the induced EMF, as described by Lenz’s law. When calculating its magnitude, we use the absolute value.

The equation gives us three useful predictions:

• More turns produce a greater induced EMF for the same flux change and time.
• A larger change in magnetic flux produces a greater induced EMF over the same time.
• The same flux change occurring over a shorter time produces a greater induced EMF.

Notice that the last point is especially relevant to wireless charging. The transmitter uses rapidly changing current so the magnetic flux can change quickly, producing an induced voltage in the receiver.

The graph reveals something subtle. The induced EMF is not greatest when the magnetic flux is greatest. It is greatest when the flux is changing most rapidly.

At the maximum and minimum of a smooth sinusoidal flux curve, the flux briefly stops increasing or decreasing before reversing direction. The induced EMF is zero at those instants.

03How the receiver actually obtains electrical energy

So far, we have explained how a voltage appears in the receiver coil. But a voltage by itself does not tell us how much energy reaches the phone.

An electromotive force is energy transferred per unit charge. It is measured in volts, but it is not a force in the mechanical sense, and it does not guarantee that current will flow.

Consider two situations.

If the receiver coil is disconnected from a load, an induced voltage can exist across its terminals. However, there is no continuous path for current through the load, so useful power delivery is essentially zero.

If the receiver coil is connected to a load, current can flow. The electrical energy delivered to that load comes from the transmitter’s power supply.

Lenz’s law explains where the energy comes from

Faraday’s law tells us the magnitude of the induced EMF. Lenz’s law tells us its direction.

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

The distinction is important. The induced effect does not always oppose the original magnetic field. It opposes the change in flux.

• If the external flux through the receiver is increasing, the induced magnetic effect acts to oppose that increase.
• If the external flux is decreasing, the induced magnetic effect acts to oppose that decrease.

When the receiver supplies current to a load, that current produces its own magnetic field. Its effect on the transmitter means the transmitter’s power supply must provide additional energy to maintain the changing current and deliver power to the receiver.

This is how energy conservation appears in electromagnetic induction. The receiver is not creating energy from the magnetic field. Energy supplied to the transmitter is transferred through the coupled electromagnetic fields and delivered to the receiver circuit.

Without a load drawing power, there is no sustained transfer of useful electrical power to that load, even if an induced voltage is present.

04The role of the charging electronics

Induction is the mechanism that produces the receiver’s voltage. Additional electronics make that energy suitable for charging a battery.

A typical wireless charging system works as follows.

1. Transmitter electronics: The power supply provides electrical energy. Switching electronics drive alternating current through the transmitter coil at a selected frequency. The supply may provide DC initially, which the electronics convert into the required alternating current.
2. Electromagnetic induction: The changing transmitter current produces a changing magnetic field. Changing magnetic flux through the receiver coil induces an alternating voltage.
3. Rectification: The receiver’s rectifier converts the induced alternating current into a unidirectional current suitable for producing DC.
4. Voltage and current regulation: The receiver’s power electronics and charging controller adjust the output to suit the battery and its charging limits.
5. Charging control: The system can adjust power delivery in response to factors such as battery state, temperature, and the quality of the connection between the coils.

These are separate jobs. Faraday’s law explains why a voltage is induced; rectification and regulation explain how the receiver turns that output into controlled electrical power for the battery.

The current in the receiver coil is therefore not necessarily the same as the current entering the battery. The electronics can transform the voltage and current between those points.

05Where energy is lost in a real wireless charger

Not all the electrical energy entering a charging pad reaches the battery.

Several processes account for the difference.

• Resistance in the coils and wiring: Electrical resistance converts some electrical energy into heat. For a resistive component, the power dissipated as heat is \(P=I^2R\), where \(I\) is the current in amperes and \(R\) is the resistance in ohms.
• Switching and conversion losses: Transmitter electronics, rectifiers, and voltage regulators are not perfectly efficient. Some energy is dissipated during switching and conversion.
• Imperfect magnetic coupling: Not all the transmitter’s magnetic flux links the receiver coil. Separation and misalignment can reduce the amount of flux passing through the receiver.
• Magnetic and eddy-current losses: Magnetic materials can dissipate energy, while changing magnetic fields can induce currents in nearby conductive materials. These currents may produce unwanted heating.

The overall charging efficiency is the percentage of input electrical power that reaches the battery as useful charging power.

\[
\eta=
\frac{P_{\mathrm{battery}}}
{P_{\mathrm{input}}}\times100\%
\]

Here, \(\eta\) is efficiency, \(P_{\mathrm{battery}}\) is the power delivered to the battery, and \(P_{\mathrm{input}}\) is the electrical power entering the charging system. Both powers are measured in watts (W).

For a system operating steadily, with stored energy not changing appreciably, the input power is divided between useful output and losses:

\[
P_{\mathrm{input}}
=P_{\mathrm{battery}}+P_{\mathrm{losses}}
\]

This explains why increasing the induced EMF does not automatically make charging faster. The receiver must be able to draw useful current, and the electronics must be able to convert and deliver that power within their operating limits.

Misalignment may reduce the flux linking the receiver. A controller can respond by increasing the transmitter’s drive to maintain charging power, but this may increase heating and other losses. If the system reaches its operating limit, the battery’s charging power may fall.

Resonant circuits and magnetic materials can improve the transfer of energy in a designed charging system. They help the system operate efficiently, but they do not replace the underlying induction mechanism.

06Questions and solutions

Question 1

A stationary receiver coil has 100 turns. The magnetic flux through each turn changes from \(2.0\times10^{-5}\,\mathrm{Wb}\) to \(5.0\times10^{-5}\,\mathrm{Wb}\) in \(1.5\,\mathrm{ms}\).

Calculate the magnitude of the average induced EMF.

Solution 1

The magnitude of the average induced EMF is \(2.0\,\mathrm{V}\).

First, calculate the change in flux per turn:

\[
\begin{aligned}
|\Delta\Phi_B|
&=(5.0-2.0)\times10^{-5}\\
&=3.0\times10^{-5}\,\mathrm{Wb}
\end{aligned}
\]

Convert the time interval to seconds:

\[
\Delta t=1.5\times10^{-3}\,\mathrm{s}
\]

Using Faraday’s law,

\[
\begin{aligned}
|\mathcal{E}_{\mathrm{avg}}|
&=N\frac{|\Delta\Phi_B|}{\Delta t}\\[4pt]
&=\frac{100(3.0\times10^{-5})}
{1.5\times10^{-3}}\\[4pt]
&=2.0\,\mathrm{V}
\end{aligned}
\]

The coil has an average induced EMF of \(2.0\,\mathrm{V}\). This result gives the induced voltage, not the power delivered to a battery.

Question 2

Two operating modes use the same receiver coil and alignment.

In Mode A, the magnetic flux per turn varies from \(+\Phi_0\) to \(-\Phi_0\) during each half-cycle at a frequency of \(100\,\mathrm{kHz}\).

In Mode B, the transmitter current amplitude is halved. Assume linear magnetic coupling, so the flux amplitude is also halved. The flux now varies from \(+\Phi_0/2\) to \(-\Phi_0/2\), but the frequency is doubled to \(200\,\mathrm{kHz}\).

Compare the magnitudes of the average induced EMF over a half-cycle in the two modes. Does your result establish that both modes deliver the same charging power to the battery? Explain.

Solution 2

The average induced EMF magnitudes are equal under the stated assumptions, but this does not establish that the battery charging powers are equal.

In Mode A, the flux change during a half-cycle is

\[
|\Delta\Phi_A|=2\Phi_0
\]

The half-cycle duration is

\[
\Delta t_A=\frac{1}{2f_A}
=\frac{1}{2(100\,000)}
=5.0\times10^{-6}\,\mathrm{s}
\]

Therefore,

\[
|\mathcal{E}_A|
=N\frac{2\Phi_0}{5.0\times10^{-6}}
\]

In Mode B, the flux change is half as large:

\[
|\Delta\Phi_B|=\Phi_0
\]

However, the half-cycle duration is also halved:

\[
\Delta t_B=\frac{1}{2(200\,000)}
=2.5\times10^{-6}\,\mathrm{s}
\]

Hence,

\[
|\mathcal{E}_B|
=N\frac{\Phi_0}{2.5\times10^{-6}}
\]

The two expressions are equal because

\[
\frac{2\Phi_0}{5.0\times10^{-6}}
=
\frac{\Phi_0}{2.5\times10^{-6}}
\]

The smaller flux change in Mode B occurs over a proportionally shorter time, so the average induced EMF is unchanged.

However, equal induced EMF does not guarantee equal charging power. The operating frequency affects the receiver’s electrical behaviour, while coil impedance, resonance, conversion losses, and charging control affect how much power reaches the battery. Those factors would need to be considered before comparing the charging rates.

Question 3

A phone is moved sideways on a charging pad, reducing the changing magnetic flux that links its receiver coil.

The charging controller responds by increasing the transmitter’s drive to keep the battery’s charging power approximately constant. The pad becomes warmer during this process.

Explain, in order, why the misalignment initially reduces the induced EMF, how the controller can compensate, and why heating can increase even though the battery charging power remains nearly constant. What happens if the controller reaches its maximum operating limit?

Solution 3

Misalignment initially reduces the induced EMF because less changing magnetic flux links the receiver coil. The controller can compensate by increasing the transmitter’s drive, but this can increase losses and heating. If the controller reaches its limit, the charging power may fall.

The reasoning has three stages.

First, consider the flux change. Moving the phone changes the magnetic coupling between the coils. For the same transmitter current waveform, a smaller change in flux through the receiver generally produces a smaller induced EMF, according to Faraday’s law.

Second, consider the controller. The controller can increase the transmitter’s drive to produce a larger changing magnetic field. This can restore the rate of change of flux through the receiver, allowing the receiver electronics to maintain approximately the same battery charging power.

Third, consider the energy losses. Higher transmitter current can increase resistive heating because the power dissipated in a resistance is proportional to \(I^2R\). Switching and other conversion losses may also increase. Maintaining the same useful output can therefore require greater input power.

If the controller reaches its maximum operating limit, it can no longer compensate sufficiently for the reduced coupling. The induced output may then become insufficient to maintain the previous battery charging power.

There is no violation of energy conservation. The battery can receive approximately the same power while the system draws more input power, because the difference is accounted for by increased losses.

Question 4

During a wireless charging test, the following real electrical powers are measured:

• Input power to the charging pad: \(15.0\,\mathrm{W}\).
• Power delivered from the receiver coil terminals into the receiver’s rectifier: \(11.0\,\mathrm{W}\).
• Power delivered to the battery: \(8.8\,\mathrm{W}\).

Calculate:

(a) the overall charging efficiency;

(b) the power not delivered to the receiver coil terminals; and

(c) the power lost between the receiver coil terminals and the battery.

A student claims that all the power not reaching the battery must be heating in the transmitter coil. Explain why this conclusion is unjustified.

Solution 4

The overall charging efficiency is \(58.7\%\). The power not delivered to the receiver coil terminals is \(4.0\,\mathrm{W}\), and the power lost between those terminals and the battery is \(2.2\,\mathrm{W}\).

(a) Calculate the overall efficiency.

Use the ratio of useful battery power to input power:

\[
\begin{aligned}
\eta
&=\frac{P_{\mathrm{battery}}}
{P_{\mathrm{input}}}\times100\%\\[4pt]
&=\frac{8.8}{15.0}\times100\%\\[4pt]
&=58.7\%
\end{aligned}
\]

The battery receives \(58.7\%\) of the input electrical power.

(b) Calculate the power not delivered to the receiver coil terminals.

\[
\begin{aligned}
P_{\mathrm{loss,\,before}}
&=15.0-11.0\\
&=4.0\,\mathrm{W}
\end{aligned}
\]

This difference accounts for losses in the transmitter electronics, transmitter coil, and imperfect energy transfer, among other possible losses before the measured receiver output.

(c) Calculate the power lost in the receiver electronics.

\[
\begin{aligned}
P_{\mathrm{loss,\,receiver}}
&=11.0-8.8\\
&=2.2\,\mathrm{W}
\end{aligned}
\]

The total power not reaching the battery is

\[
\begin{aligned}
P_{\mathrm{loss,total}}
&=15.0-8.8\\
&=6.2\,\mathrm{W}
\end{aligned}
\]

The student’s conclusion is unjustified because the measurements identify losses across two different parts of the system. The \(2.2\,\mathrm{W}\) difference occurs after the receiver coil terminals, while the \(4.0\,\mathrm{W}\) difference occurs before that point and cannot be assigned entirely to the transmitter coil.

The missing useful power can be dissipated in multiple components or lost from the intended transfer pathway. Measuring the total difference alone does not reveal exactly where each loss occurs.

Question 5

A transmitter is controlled so that its current waveform and frequency remain fixed. An aligned receiver coil is initially connected to a resistive load, and useful power is being transferred.

The load is then disconnected, leaving the receiver coil open-circuit. The transmitter controller continues to maintain the same current waveform, and the coils do not move. Assume linear coupling and ignore standby losses for the main comparison.

A student argues:

“The transmitter current has not changed, so the induced EMF has not changed. Therefore, the receiver must still be getting the same energy per second.”

Assess this argument by explaining:

(a) what happens to the open-circuit induced EMF;

(b) what happens to the receiver current and useful power;

(c) what happens to the transmitter’s real input power; and

(d) how Lenz’s law and energy conservation explain the result.

Finally, explain why the induced EMF alone cannot determine a real phone’s battery charging rate.

Solution 5

The student’s conclusion is wrong: an induced EMF can remain across the open receiver terminals while the useful power delivered to the load falls to zero. Maintaining the same transmitter current does not require the transmitter to supply the same real input power.

(a) The open-circuit induced EMF

With the transmitter current waveform, frequency, and coil positions unchanged, the changing magnetic flux produced by the transmitter remains unchanged under the stated model.

Faraday’s law therefore predicts the same open-circuit induced EMF as before disconnection. However, this does not mean that the loaded terminal voltage must have been identical to the open-circuit voltage. When current flows, the receiver coil’s own electrical behaviour affects the voltage available across its load.

(b) The receiver current and useful power

After disconnection, there is no closed path through the load. The receiver current falls to zero, so the load receives no useful electrical power.

This is the crucial distinction: an induced EMF describes a voltage, whereas power delivery requires current as well. A voltage across open terminals is not evidence that energy is continuing to flow into a load at the previous rate.

(c) The transmitter’s real input power

The real input power associated with transferring energy to the receiver decreases when the load is disconnected. The transmitter still needs to supply any remaining resistive and electronic losses, but the component of power previously delivered to the receiver is no longer required.

The controller is maintaining the current waveform, not necessarily the input power. It can adjust its output voltage, and the real power drawn from the supply can change even when the current waveform is fixed.

This is the point the student’s argument misses. An unchanged current waveform does not mean that every aspect of the system’s energy transfer must remain unchanged.

(d) Lenz’s law and energy conservation

Before disconnection, current flows in the receiver coil. That current produces a magnetic field whose effect opposes the change in flux that induced it. This interaction acts back on the transmitter, which must supply additional energy to maintain its prescribed current waveform while delivering power to the load.

After disconnection, the receiver current disappears. Its associated magnetic effect disappears too, and the transmitter no longer needs to provide the additional real power that was being transferred to the load.

If the receiver could continue delivering the same power without current flowing or energy being supplied by the transmitter, energy would effectively be created from nowhere. Lenz’s law and the energy balance of the coupled system rule out that result.

Why this matters for a real phone: the induced EMF alone cannot determine the battery charging rate. The receiver must draw current, its electronics must rectify and regulate the output, and the system must supply enough power after accounting for losses. Battery state, temperature, coil alignment, and the controller’s operating limits also affect the charging rate.

This is the same conceptual bridge used when studying transformers and AC generators: Faraday’s law predicts the induced EMF, while the connected circuit determines how much current and useful power can actually be delivered.