Lesson Explainer: Inductance | Nagwa Lesson Explainer: Inductance | Nagwa

Lesson Explainer: Inductance Physics

In this explainer, we will learn how to calculate the self-inductance of a conducting loop and the mutual inductance of a pair of conducting coils.

Current-carrying wires create magnetic fields around themselves. If the current in a wire changes over time, so does the magnetic field it produces.

When a conducting loop is exposed to a changing magnetic field, a potential difference is induced in the loop. This process is known as electromagnetic induction, or simply induction.

Definition: Inductance

Inductance is the capacity of a conductor to experience a change in current due to a changing magnetic field. The greater that capacity, the more inductance the conductor is said to have.

Consider a solenoid that carries current.

When the current is steady, a constant magnetic field is created within the solenoid loops, as shown below.

Imagine then that we increase the potential difference across the solenoid by some fixed amount.

Doing so will create the following sequence of changes in the solenoid.

Increasing the potential difference across the solenoid will increase the current in the solenoid. This change in current will increase the magnetic field strength around the solenoid.

The increasing magnetic field induces a current in the opposite direction to the original current in the solenoid. This opposing current is of lesser magnitude than the original current.

The small opposing current changes the magnetic field around the solenoid, reducing the rate of increase of the increasing magnetic field.

The magnetic field continues to change at the reduced rate, however. The still-changing magnetic field continues to induce a current. The current induced is now of lesser magnitude than the earlier induced current.

The smaller current generates a magnetic field that is smaller than the magnetic field that induced it. The effect of this magnetic field is to continue to reduce the rate of increase of the magnetic field.

This process continues, with each subsequent induced current value being lesser in magnitude than the previous value. After sufficient repetitions of the process, the current change becomes negligible, and the increase in the magnetic field strength becomes negligible.

At this time, the current change and magnetic field strength change are effectively zero. We can then consider there to be a new constant value of current in the solenoid and a new constant value of maximum magnetic field strength for the solenoid.

An important thing to understand about this sequence of changes is that it reduces the rate of change in current and magnetic field strength to zero but does not make them negative. This means that increasing the potential difference across a solenoid

  • does not result in the current in the solenoid increasing without limit,
  • does not result in the current in the solenoid decreasing to less than its initial value,
  • does not result in the current around the solenoid oscillating.

Rather, the current in the solenoid increases at a continually decreasing rate until it eventually ceases to increase. This change in the current occurs over a time interval. This is also true for the magnetic field strength around the solenoid.

This phenomenon is known as self-inductance—the changing of current in a conductor due to a time-varying magnetic field created by a changing current already in the conductor.

Equation: Inductance

If 𝜀 is the potential difference induced in a conductor, Δ𝐼Δ𝑡 is the rate at which current in the conductor changes over time, and 𝐿 is the conductor’s inductance, then 𝜀=𝐿×Δ𝐼Δ𝑡.

Mathematically, the inductance of a conductor is the constant of proportionality between the potential difference induced in the conductor and the rate at which current in the conductor changes over time.

The minus sign in the equation indicates the polarity of the induced potential difference. This voltage tends to generate current that opposes the change in current over time, Δ𝐼Δ𝑡.

Example 1: Using Self-Inductance to Determine the Time Needed for Current to Change a Set Amount

A changing current in a loop of wire induces a potential difference of 1.2 V across it. The loop has a self-inductance of 125 mH. How much time is required for the loop to increase the current through it by 0.25 A? Give your answer to two decimal places.

Answer

We relate the potential difference across the loop, its self-inductance, and the change in current over time using the following relationship: 𝜀=𝐿×Δ𝐼Δ𝑡.

We want to solve for the change in time, Δ𝑡. Rearranging, Δ𝑡=(𝐿×Δ𝐼)𝜀.

Since Δ𝑡 will not be negative, we can assume the potential difference has a negative polarity, making it equal to 1.2 V.

Knowing that the self-inductance of the loop, 𝐿, is 125 mH and the change in current, Δ𝐼, is 0.25 A, Δ𝑡=125×10×(0.25)(1.2)=((0.125)×(0.25))(1.2)=0.026041̇6.HAVHAVs

Rounding this result to two decimal places, the time needed to increase the current in the loop by 0.25 A is 0.03 s.

We can combine the equation for inductance with Faraday’s law, which states that the voltage induced in a conductor is proportional to the change in magnetic flux over time experienced by the conductor. Specifically, 𝜀=𝑁×ΔΦΔ𝑡, where 𝑁 is the number of turns in the conductor and ΔΦ is the change in magnetic flux through the conductor over a time Δ𝑡.

Equating Faraday’s law with the earlier expression for inductance, 𝜀=𝑁×ΔΦΔ𝑡=𝐿×Δ𝐼Δ𝑡.

Therefore, 𝑁×ΔΦ=𝐿×Δ𝐼.

Or, equivalently, 𝐿=𝑁×ΔΦΔ𝐼.

This result indicates that inductance equals the number of turns in a conductor multiplied by the change in magnetic flux it experiences, divided by the change in current in the conductor.

Example 2: Calculating Self-Inductance Using Changes in Current and Magnetic Flux

A loop of wire increases the current that it carries by 180 mA. The change in magnetic flux produced by the change in current is 0.77 Wb. What is the self-inductance of the loop? Give your answer to one decimal place.

Answer

The self-inductance 𝐿 of the loop is given by 𝐿=𝑁×ΔΦΔ𝐼, where 𝑁 is the number of turns in the loop, ΔΦ is the change in magnetic flux through it, and Δ𝐼 is the change in current through the loop.

Here, our loop consists of a single turn of wire, so 𝑁 is one.

Magnetic flux and current are given in units of webers (Wb) and milliamperes (mA) respectively.

The weber is the SI unit of magnetic flux. To use units of electric current that agree with the weber, we must convert our current units to amperes.

1‎ ‎000 milliamperes equals one ampere, so if 𝑌 is a number of milliamperes, then 𝑌=𝑌×10.mAA

Thus, 180 milliamperes equals 0.180 amperes.

Substituting this value for current and the given magnetic flux into the equation for 𝐿, we get 𝐿=(1)×(0.77)(0.180)=4.2̇7.WbAH

Rounding the answer to one decimal place, the self-inductance of the loop is 4.3 henries.

A conductor producing a changing magnetic field may induce current in itself, and it may also induce current in a separate conductor. When two conductors interact this way, they exhibit what is called mutual inductance.

Consider two conductors, numbered one and two. Conductor one carries current 𝐼, producing a changing magnetic flux through conductor two, which induces a voltage 𝜀.

Similarly to the earlier equation for inductance, these quantities are related as follows: 𝜀=𝑀×Δ𝐼Δ𝑡.

The voltage induced in conductor two due to the current in conductor one depends on the mutual inductance from conductor one to conductor two (𝑀).

In general, the mutual inductance from conductor one to conductor two is not equal to the mutual inductance of conductor two to conductor one. That is, 𝑀𝑀.

Nonetheless, it is not unusual to see the equation for voltage, mutual inductance, and current written: 𝜀=𝑀×Δ𝐼Δ𝑡.

Even in this unspecified form, we are still considering current changing in one conductor to induce voltage in the other through their mutual inductance.

Example 3: Determining Potential Difference Using Mutual Inductance

A transformer with an iron core has a primary coil that has 75 turns and a secondary coil that also has 75 turns. The coils have a mutual inductance of 15 H. The current in the primary coil increases the current in the secondary coil at a rate of 1.25 A/s. What is the potential difference across the coils? Give your answer to one decimal place.

Answer

The change in current in the secondary transformer coil is due ultimately to a change in current in the primary coil.

We recall that 𝜀=𝑀×Δ𝐼Δ𝑡, where 𝜀 is the voltage induced by a change in current over time Δ𝐼Δ𝑡.

We can solve for this voltage using the given rate of current change (1.25 A/s) and the mutual inductance between the coils (15 H). Substituting in these values, 𝜀=(15)×(1.25/).HAs

The negative sign in this equation reflects the fact that induced current generates a magnetic field opposing the magnetic field originally causing it. Regarding voltage, this sign indicates a change in polarity, but not magnitude—we can disregard the negative sign while solving for voltage: 𝜀=(15)×(1.25/)=18.75.HAsV

Rounding this result to one decimal place, we find the potential difference across the coils is 18.8 volts.

Example 4: Computing the Number of Turns in a Transformer Coil

A transformer consists of a primary coil and a secondary coil, each with the same number of turns, wrapped around an iron core. The coils have a mutual inductance of 32 mH. A current in the primary coil increases the magnetic flux through the core by 4.48 mWb. The current induced in the secondary coil is 1.4 A. How many turns does the coil have?

Answer

Faraday’s law relates voltage induced in a conductor to a change in magnetic flux over time: 𝜀=𝑁×ΔΦΔ𝑡, where 𝜀 is the induced potential difference, 𝑁 is the number of turns in the conductor, and ΔΦΔ𝑡 equals the change in magnetic flux over time experienced by the conductor.

There is another expression for induced voltage, involving the mutual inductance between two conductors such as the coils of a transformer: 𝜀=𝑀×Δ𝐼Δ𝑡.

These equations may be combined to yield a third equation: 𝑁×ΔΦΔ𝑡=𝑀×Δ𝐼Δ𝑡.

Note that the negative signs and fractions of 1Δ𝑡 appear on both sides, therefore cancelling out: 𝑁×(ΔΦ)=𝑀×(Δ𝐼).

Dividing both sides by ΔΦ, 𝑁=𝑀×(Δ𝐼)(ΔΦ).

We can now solve for the number of turns in the secondary transformer coil by substituting in the given values of mutual inductance (32 mH), change in current (1.4 A), and change in magnetic flux (4.48 mWb).

First, however, we convert the mutual inductance to a value in units of henries and the change in magnetic flux to possess units of webers: 1000=1mHH and 1000=1.mWbWb

So 32=32×10=0.032mHHH and 4.48=4.48×10=0.00448.mWbWbWb

Therefore, 𝑁=(0.032)×(1.4)(0.00448)=10.HAWb

The secondary coil in the transformer has 10 turns.

Let us now summarize what we have learned in this explainer.

Key Points

  • The self-inductance of an inductor determines the rate at which the current in the inductor can be changed.
  • 𝜀=𝐿×Δ𝐼Δ𝑡, where 𝜀 is the potential difference across a conductor, Δ𝐼Δ𝑡 is the change in current over time, and 𝐿 is the inductance of the conductor.
  • Variable current in one conductor may also induce current in a second conductor, a phenomenon known as mutual inductance.
  • 𝜀=𝑀×Δ𝐼Δ𝑡, where 𝜀 is the potential difference induced across one conductor, 𝑀 is the mutual inductance between two conductors, and Δ𝐼Δ𝑡 is the change in current over time.
  • Both inductance equations may be combined with Faraday’s law to yield new equations: 𝐿=𝑁×ΔΦΔ𝐼 and 𝑀=𝑁×ΔΦΔ𝐼. Here, 𝐿 and 𝑀 are inductance and mutual inductance, respectively, 𝑁 is the number of turns in a given conductor, and ΔΦΔ𝐼 is the change in magnetic field strength divided by the change in current.

Download the Nagwa Classes App

Attend sessions, chat with your teacher and class, and access class-specific questions. Download the Nagwa Classes app today!

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy