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  4. Electromagnetic induction
  5. Induced electric fields

13.4 Induced electric fields  (Page 6/4)

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By the end of this section, you will be able to:
  • Connect the relationship between an induced emf from Faraday’s law to an electric field, thereby showing that a changing magnetic flux creates an electric field
  • Solve for the electric field based on a changing magnetic flux in time

The fact that emfs are induced in circuits implies that work is being done on the conduction electrons in the wires. What can possibly be the source of this work? We know that it’s neither a battery nor a magnetic field, for a battery does not have to be present in a circuit where current is induced, and magnetic fields never do work on moving charges. The answer is that the source of the work is an electric field E → that is induced in the wires. The work done by E → in moving a unit charge completely around a circuit is the induced emf ε; that is,

ε = ∮ E → · d l → ,

where ∮ represents the line integral around the circuit. Faraday’s law can be written in terms of the induced electric field    as

∮ E → · d l → = − d Φ m d t .

There is an important distinction between the electric field induced by a changing magnetic field and the electrostatic field produced by a fixed charge distribution. Specifically, the induced electric field is nonconservative because it does net work in moving a charge over a closed path, whereas the electrostatic field is conservative and does no net work over a closed path. Hence, electric potential can be associated with the electrostatic field, but not with the induced field. The following equations represent the distinction between the two types of electric field:

∮ E → · d l → ≠ 0 ( induced ) ; ∮ E → · d l → = 0 ( electrostatic ) .

Our results can be summarized by combining these equations:

ε = ∮ E → · d l → = − d Φ m d t .

Induced electric field in a circular coil

What is the induced electric field in the circular coil of [link] (and [link] ) at the three times indicated?

Strategy

Using cylindrical symmetry, the electric field integral simplifies into the electric field times the circumference of a circle. Since we already know the induced emf, we can connect these two expressions by Faraday’s law to solve for the induced electric field.

Solution

The induced electric field in the coil is constant in magnitude over the cylindrical surface, similar to how Ampere’s law problems with cylinders are solved. Since E → is tangent to the coil,

∮ E → · d l → = ∮ E d l = 2 π r E .

When combined with [link] , this gives

E = ε 2 π r .

The direction of ε is counterclockwise, and E → circulates in the same direction around the coil. The values of E are

E ( t 1 ) = 6.0 V 2 π ( 0.50 m ) = 1.9 V/m; E ( t 2 ) = 4.7 V 2 π ( 0.50 m ) = 1.5 V/m; E ( t 3 ) = 0.040 V 2 π ( 0.50 m ) = 0.013 V/m .

Significance

When the magnetic flux through a circuit changes, a nonconservative electric field is induced, which drives current through the circuit. But what happens if d B / d t ≠ 0 in free space where there isn’t a conducting path? The answer is that this case can be treated as if a conducting path were present ; that is, nonconservative electric fields are induced wherever d B / d t ≠ 0 , whether or not there is a conducting path present.

These nonconservative electric fields always satisfy [link] . For example, if the circular coil of [link] were removed, an electric field in free space at r = 0.50 m would still be directed counterclockwise, and its magnitude would still be 1.9 V/m at t = 0 , 1.5 V/m at t = 5.0 × 1 0 − 2 s , etc. The existence of induced electric fields is certainly not restricted to wires in circuits.

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Questions & Answers

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Joy Reply
Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
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Kate Reply
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Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
Maria Reply
please solve
Sharon
8m/s²
Aishat
What is Thermodynamics
Muordit
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
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50 m/s due south east
Someone
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Someone
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Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
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ALIYU
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Nassze Reply
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Esrael
Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
JALLAH Reply
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Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
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Raymond
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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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