Physics Lesson 16.10.5 - Power and Energy

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Welcome to our Physics lesson on Power and Energy, this is the fifth 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.

Power and Energy

Again, we use the outcomes of the analysis made in the previous paragraphs to calculate the rate of work (power) delivered during the process discussed earlier. We have:

P = FM ∙ v
= (B2 ∙ w2 ∙ v/R) ∙ v
= B2 ∙ w2 ∙ v2/R
= εi2/R

As for the rate of thermal energy produced in the coils, we have:

P = i2 ∙ R
= (B ∙ w ∙ v/R)2 ∙ R
= B2 ∙ w2 ∙ v2/R

This value is the same as the one obtained earlier for the rate of work done in the coil. This means the work done for pulling the loop through a magnetic field is transferred entirely to the loop in the form of thermal energy.

Example 2

A 30 cm × 25 cm rectangular loop is pulled at constant speed out a 5.0 T uniform magnetic field as shown in the figure.

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

It takes 10 seconds to the pulling force to shift the coil out of the magnetic field. If the average resistance provided by the coil during this process is 2.0 Ω, calculate:

  1. The current induced in the coil
  2. The average pulling force applied in the coil
  3. The emf induced in the coil
  4. The rate at which the external source is doing work on the coil
  5. The total thermal energy transferred to the coil

Solution 2

Clues:

l = 30 cm = 0.30 m
w = 25 cm = 0.25 m
B = 5.0 T
Δt = 10 s
R = 2.0 Ω
a) i = ?
b) F = ?
c) εi = ?
d) P = ?
e) E = ?

  1. First, we must calculate the speed of the coil. Since it is moving at constant speed, we have:
    v = l/∆t
    = (0.30 m)/(10 s)
    = 0.03 m/s
    The current induced in the coil due to its motion relative to the magnetic field therefore is
    i = B ∙ w ∙ v/R
    = (5.0 T) ∙ (0.25 m) ∙ (0.03 m/s)/(2.0 Ω)
    = 0.01875 A
  2. Since the coil is moving at constant speed, the pulling force F is equal to the resistive magnetic force Fm acting in the coil due to its motion inside the magnetic field. Thus,
    F = FM = i ∙ B ∙ w
    = (0.01875 A) ∙ (5.0 T) ∙ (0.25 m)
    = 0.0234 N
  3. The induced emf can be calculated in many ways. One of them is by applying the Ohm's Law. Thus,
    εi = i ∙ R
    = (0.01875 A) ∙ (2.0 Ω)
    = 0.0375 V
  4. The rate of word done at the coil represents the useful power of the external source. Thus, we have
    P = F ∙ v
    = (0.0234 N) ∙ (0.03 m/s)
    = 0.0007 W
  5. The thermal energy can be calculated in many ways. For example, since energy = power × time, we can write
    E = P ∙ ∆t
    = 0.0007 W ∙ 10 s
    = 0.007 J
    Another method for calculating the thermal energy would be
    E = (B2 ∙ w2 ∙ v2/R) ∙ ∆t
    = (5.0 T)2 ∙ (0.25 m)2 ∙ (0.03 m/s)2/(2.0 Ω) ∙ (10 s)
    = 0.007 J
    As you see, the result is the same in both methods used.

You have reached the end of Physics lesson 16.10.5 Power and Energy. 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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