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Induced Electric Fields

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Magnetism Learning Material
Tutorial ID	Title	Tutorial	Video Tutorial	Revision Notes	Revision Questions
16.11	Induced Electric Fields

In this Physics tutorial, you will learn:

What happens when we place a copper ring inside a magnetic field?
How the electric and magnetic fields are related to each other? Is there any mutual interaction between them or not?
What condition must a magnetic field meet to induce a current in a coil?
How to express the Faraday's law in terms of the induced electric field?
Why it is meaningless to talk about electric potential induced in a coil?

Introduction

Do you think there is any electric field produced due to induction? Where do you base your opinion?

Do you think a change in electric flux brings the induction of an electric field?

Is there any work done when moving a charge inside a magnetic field?

Can a uniform/non uniform magnetic field produce electricity on static/dynamic charges? Explain your opinion.

In this tutorial, we will discuss about induced electric fields, which are very similar to electric field produced by static charges. However, they have some differences we will explain in the following paragraphs.

Induced Electric Fields

We have seen in the previous tutorials that any change in the electric flux when a loop is placed inside a magnetic field results in the induction of a current and electromotive force in the loop. Now, let's see whether this phenomenon is somehow related to electric field. If such a relationship is proven, the true statement "an electric field generates a magnetic field around it" will have its true corresponding counter-statement "a changing magnetic field on a loop generates an electric field in it". In this way, we will have the logical relationship

E ⟹ B and B ⟹ E so E ⟺ B

To prove the above hypothesis, let's insert a copper ring of radius r inside a uniform magnetic field B as shown in the figure. The external magnetic field fills a cylindrical volume of radius R.

If we increase uniformly the strength of the external magnetic field, the magnetic flux through the ring will increase at a steady rate and as a result, a current (and emf) is induced in the ring based on the Faraday's Law. We can determine the direction of these two induced quantities (both of them are anticlockwise here) considering the Lentz Law. Thus, since the flux increases, it is like approaching the north pole of a bar magnet towards the ring in the inwards (onto the page) direction. As a result, a new magnetic field is induced in the opposite direction (out of page). As a result, we find out using the curled right hand rule that the direction of the induced current and emf is anticlockwise.

Remark! To determine the direction of the induced current, you must consider only the direction of the induced magnetic field inside the ring when applying the curled right hand rule.

Moreover, the presence of the induced current in the copper ring implies the presence of an electric field because no current can exist without an electric field around it (an electric field is required to do the necessary work for making the electrons flow around the loop). Since the electric field always has the direction of positive changes (and therefore the direction of current), we appoint an anticlockwise direction to it, just like the other two quantities discussed earlier (induced current and emf). This induced electric field is similar to that produced by static charges, as discussed in Section 14. Both fields will exert an electric force F = Q₀ ∙ E on a positive test charge Q₀.

From the situation described above, we reach in a very important conclusion, which represents another version of the Faraday's Law:

"A changing magnetic field in a coil induces an electric field in it."

In this way, the hypothesis provided at the beginning of this tutorial is confirmed as true. We write

E ⟹ B and B ⟹ E so E ⟺ B

The electric field induced due to the change in the magnetic flux (and field) occurs even when no copper ring is present. This means this phenomenon can occur in every medium, even in vacuum. If the copper ring is replaced with a whatever circular path of radius r, everything said above is still valid.

Due to the increase in the rate of magnet's motion described earlier, the magnetic field inside the circular path increases at a rate of ΔB/Δt. The electric field produced due to this change is tangent to the circular path; this occurs for all similar circular paths, regardless their distance from the centre of loop. Therefore, we obtain some concentric circles representing the induced electric field in various distances from the centre as shown in the figure below.

The electric field will exist as long as the magnetic flux through the loop is changing. This means when the magnetic flux is constant, no induced electric field is present around the loop. Hence, for a constant magnetic flux the circular field lines will not appear anymore in the figure.

If the magnetic flux through the circular path decreases due to any decrease of magnetic field, the electric field will reappear again but this time the field lines are in the opposite direction to before (here clockwise).

Thus, we conclude that an induced electric field is produced every time the magnetic flux (caused by a change in the magnetic field) is changing by time.

Another Version of Faraday's Law

Let's consider a positive trial charge Q₀ moving around a circular path as the one shown in the figure below.

The work done by the electric field E on the test charge Q₀ to make it move in a circular path as the one shown above is

W = Q₀ ∙ ε_i

where εi is the emf induced in the circular loop.

If we want to calculate the work done by the electric field to make the test charge complete one revolution only, is

W = F ∙ s = (Q₀ ∙ E) ∙ (2π ∙ r)

where the first expression inside the brackets represents the electric force and the second the circumference of the circular path which is equal to the distance travelled by the test charge during one revolution.

Combining the last two equations, we obtain

Q₀ ∙ ε_i = (Q₀ ∙ E) ∙ (2π ∙ r)

Simplifying Q₀ from both sides, we get for the induced emf

ε_i = 2π ∙ r ∙ E

The above approach is a simplified version; it takes place only when the applied force is constant. In the general case, we use the integration method explained in tutorial 16.6. Thus, we have

W = ∮F ds
= ∮Q₀ Eds
= Q₀ ∮E ds

(the circle in each integral is the symbol that indicates a closed path). Therefore, since the work W = Q₀ ∙ εi we obtain for the induced emf in the loop:

ε_i = ∮E ds

When the last expression is combined with the standard form of Faraday's law

ε_i = -dΦ_M/dt

we obtain a new (integral) version of this law, that is

∮E ds = -dΦ_M/dt

Example 1

What is the electric field generated when a 2 μC trial charge moves in a circular path of radius 2 cm inside a uniform magnetic field if the emf induced in the coil during this process is 4 mV? The charges makes one complete rotation around the centre of the path.
What is the work done by the electric field on the trial charge during this process?

Solution 1

We have the following clues in this problem:

Q₀ = 2 μC = 2 × 10^-6 C
r = 2 cm = 0.02 m
εi = 4 kV = 0.004 V = 4 × 10³ V
(π = 3.14)
a) E = ?
b) W = ?

Since the path is regular, we can apply the simplified (not the integral) formula
ε_i = 2π ∙ r ∙ E
to calculate the electric field produced by the trial charge when moving inside the uniform magnetic field. Thus, we obtain
E = ε_i/2π ∙ r
= 4000 V/2 ∙ 3.14 ∙ 0.02 m
= 31847 V/m
The work done by the electric field on the test charge during one complete rotation is
W = Q₀ ∙ ε_i
= (2 × 10^-6 C) ∙ (4 × 10³ V)
= 8 × 10^-3 J
= 0.008 J
As you see, the value of work is very small despite the large value of electric field. This is because the magnitude of the trial charge is very small.

A New Approach on Electric Potential

As stated in the previous paragraph, both the static and dynamic charges produce electric fields around them. However, they have a fundamental difference: the electric field lines of static charges are open (start from positive charges and end at negative ones) while dynamic charges form closed lines due to their circular path around the loop. Since the electric field produced by dynamic charges is uniform, it is meaningless to talk about electric potential (and potential difference) caused by moving charges.

Therefore, the main difference between electric fields produced by static and dynamic charges is as follows:

"Electric potential has meaning only for static charges, not for dynamic ones."

Let's prove the correctness of the above statement by writing the integral form of the potential difference produced by static charges:

∆V = ∫E ds

When moving in a closed loop at the same distance from the centre of field (in equipotential surfaces), the above expression becomes

∆V = ∫E ds = 0

However, when the magnetic flux changes, we have (from the new version of Faraday's Law discussed earlier in this tutorial)

∮E ds = -dΦ_M/dt

This means the integral is not zero, so ΔV ≠ 0. These two different results for the same thing represent a contradiction. Therefore, it is meaningless to discuss about induced electric potential.

Example 2

How many seconds are needed to change the electric flux from 0.3 Wb to 1.5 Wb if a test charge inserted inside a uniform magnetic field is moving according the circular path shown in the figure? The electric field induced during one complete rotation in the loop is 12 V/m and the loop's diameter is 2 cm.

Solution 2

Clues:

Φ₁ = 0.3 Wb
Φ₂ = 1.5 Wb
E = 12 V/m
d = 2 cm = 0.02 m
dt = ?

First, we find the radius of loop. We have

r = d/2
= 0.02 m/2
= 0.01 m

Since the magnetic field is uniform, we can write the integral version of Faraday's Law

∮E ds = -dΦ_M/dt

E ∙ 2πr = -dΦ_M/dt

Ignoring the negative sign in the right side of the equation (it just shows direction), we obtain

dt = dΦ_M/E ∙ 2πr
= Φ₂-Φ₁/E ∙ 2πr
= 1.5 Wb-0.3 Wb/(12 V/m) ∙ 2 ∙ 3.14 ∙ (0.01 m)
= 1.59 s

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Popular Today

Induced Electric Fields

Introduction

Induced Electric Fields

Another Version of Faraday's Law

Example 1

Solution 1

A New Approach on Electric Potential

Example 2

Solution 2

Whats next?

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