Physics Lesson 16.11.2 - Another Version of Faraday's Law

Welcome to our Physics lesson on Another Version of Faraday's Law, this is the second lesson of our suite of physics lessons covering the topic of Induced Electric Fields, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

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 ∙ εi

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 ∙ εi = (Q0 ∙ E) ∙ (2π ∙ r)

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

εi = 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:

εi = E ds

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

εi = -M/dt

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

E ds = -M/dt

Example 1

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.
2. 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 = ?

1. Since the path is regular, we can apply the simplified (not the integral) formula
εi = 2π ∙ r ∙ E
to calculate the electric field produced by the trial charge when moving inside the uniform magnetic field. Thus, we obtain
E = εi/2π ∙ r
= 4000 V/2 ∙ 3.14 ∙ 0.02 m
= 31847 V/m
2. The work done by the electric field on the test charge during one complete rotation is
W = Q0 ∙ εi
= (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.

You have reached the end of Physics lesson 16.11.2 Another Version of Faraday's Law. There are 3 lessons in this physics tutorial covering Induced Electric Fields, you can access all the lessons from this tutorial below.

More Induced Electric Fields Lessons and Learning Resources

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16.11Induced Electric Fields
Lesson IDPhysics Lesson TitleLessonVideo
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16.11.1Induced Electric Fields
16.11.3A New Approach on Electric Potential

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