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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 Q_{0} 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 Q_{0} to make it move in a circular path as the one shown above is

W = Q_{0} ∙ 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 = (Q_{0} ∙ 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

Q_{0} ∙ ε_{i} = (Q_{0} ∙ E) ∙ (2π ∙ r)

Simplifying Q_{0} 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

=__∮__Q_{0} Eds

= Q_{0} __∮__E ds

=

= Q

(the circle in each integral is the symbol that indicates a closed path). Therefore, since the work W = Q_{0} ∙ 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

- 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?

We have the following clues in this problem:

Q_{0} = 2 μC = 2 × 10^{-6} C

r = 2 cm = 0.02 m

εi = 4 kV = 0.004 V = 4 × 10^{3} V

(π = 3.14)

a) E = ?

b) W = ?

- Since the path is regular, we can apply the simplified (not the integral) formula εto calculate the electric field produced by the trial charge when moving inside the uniform magnetic field. Thus, we obtain
_{i}= 2π ∙ r ∙ EE =*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 = QAs 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.
_{0}∙ ε_{i}

= (2 × 10^{-6}C) ∙ (4 × 10^{3}V)

= 8 × 10^{-3}J

= 0.008 J

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.

Enjoy the "Another Version of Faraday's Law" physics lesson? People who liked the "Induced Electric Fields lesson found the following resources useful:

- Faradays Law Feedback. Helps other - Leave a rating for this faradays law (see below)
- Magnetism Physics tutorial: Induced Electric Fields. Read the Induced Electric Fields physics tutorial and build your physics knowledge of Magnetism
- Magnetism Revision Notes: Induced Electric Fields. Print the notes so you can revise the key points covered in the physics tutorial for Induced Electric Fields
- Magnetism Practice Questions: Induced Electric Fields. Test and improve your knowledge of Induced Electric Fields with example questins and answers
- Check your calculations for Magnetism questions with our excellent Magnetism calculators which contain full equations and calculations clearly displayed line by line. See the Magnetism Calculators by iCalculator™ below.
- Continuing learning magnetism - read our next physics tutorial: RL Circuits

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