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Physics Lesson 16.10.3 - Induced Emf

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Welcome to our Physics lesson on Induced Emf, this is the third lesson of our suite of physics lessons covering the topic of Induction and Energy Transfers, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Induced Emf

Obviously, since there is a current induced in the coil, there is an induced emf in it as well. We must apply the Faraday's Law for to find this induced emf. If we denote the length of the loop by l, and its width by w, we have:

ΦM = B ∙ A
= B ∙ l ∙ w

Let's denote the part of coil's length inside the magnetic field by a as shown in the figure. The value of a varies from 0 to l depending on the position of the coil relative to the magnetic field.

Physics Tutorials: This image provides visual information for the physics tutorial Induction and Energy Transfers

As we move the coil due right, the part of its length inside the field (a therefore) decreases. As a result, the magnetic flux through the coil decreases too, as stated earlier. From Faraday's Law, it is known that any change in flux is accompanied by the induction of an emf in the coil. The magnitude of this induced emf is

εi = ∆ΦM/Δt
= B ∙ w ∙ ∆l/∆t
= B ∙ w ∙ v

where v is the moving speed of the coil.

Example 1

A rectangular coil (as the one shown above in theory) is 20 cm long and 15 cm wide. The coil is pulled at a constant force and speed and as a result, it takes 5 seconds to the coil to move out from the 0.2 T magnetic field, the lines of which are perpendicular to the coil's plane.

Physics Tutorials: This image provides visual information for the physics tutorial Induction and Energy Transfers
  1. What is the emf induced in the coil?
  2. What is the power delivered by the magnetic force if the pulling force is 4N?

Solution 1

Clues:

l = 20 cm = 0.20 m
w = 15 cm = 0.15 m
B = 0.2 T
Δt = 5 s
F = 4 N

  1. The induced emf in the coil is
    εi = ∆ΦM/Δt
    = B ∙ w ∙ ∆l/∆t
    = (0.2 T) ∙ (0.15 m) ∙ (0.20 m-0 m)/5 s
    = 0.0012 V
  2. We must work out the moving velocity of coil first in order to calculate the power delivered by the magnetic force. Thus, since the coil is moving at constant speed, we have
    v = ∆l/∆t
    = 0.20 m/5 s
    = 0.04 m/s
    Yet, since the coil moves at constant speed, the pulling force is balanced by the magnetic force. Hence, we have
    PM = FM ∙ v
    = F ∙ v
    = (4 N) ∙ (0.04 m/s)
    = 0.16 W

You have reached the end of Physics lesson 16.10.3 Induced Emf. There are 6 lessons in this physics tutorial covering Induction and Energy Transfers, you can access all the lessons from this tutorial below.

More Induction and Energy Transfers Lessons and Learning Resources

Magnetism Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
16.10Induction and Energy Transfers
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
16.10.1Induction and Energy Transfer
16.10.2Rate of Work (Power)
16.10.3Induced Emf
16.10.4Induced Current
16.10.5Power and Energy
16.10.6Eddy Currents

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