Physics Tutorial: Induction and Energy Transfers

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In this Physics tutorial, you will learn:

  • The relationship between magnetic and electric field in terms of induction
  • What happens when we do some work to move a coil inside a magnetic field
  • How to change the magnetic flux inside a coil placed inside a magnetic field
  • What is the rate of work done by an external source on a coil placed inside a magnetic field
  • How the emf induced in a moving coil is related to the external magnetic field?
  • The same for the current induced in a moving coil inside a magnetic field
  • What are eddy currents and how they are produced?

Introduction

Do you think there is any energy transfer when an emf is induced in a coil? Where do you base your opinion? (Remember what happens when you use inductors to connect a coil with an electric circuit). What is the direction of energy transfer (if any) in an electric circuit? Do you think there is any relationship between the change in electric flux and energy transfer throughout a circuit? Explain your opinion. In this tutorial, we will discuss about the relationship between induction and energy transfers. This relationship is the key feature in the process of energy transfer when electricity produced in power stations needs to be transferred to the consumers. Induction and Energy Transfer It is a known fact (from the Lentz law) that when a magnet is moving towards or away from a coil, a resistive effect in the form of magnetic force is produced in the coil - an effect which is in the opposite direction to the external force exerted on the coil - in order to move the magnet. This external force does positive work on the system, resulting in an increase in the energy of the system. The current induced in the coil produces a resistance in it, resulting in the delivering of a certain amount of thermal energy. In other words, the energy produced due to the magnet's motion (mechanical energy) is converted into thermal energy of the coil. All this process occurs without any direct contact, but through induction. Therefore, we say "induction results in a transfer of energy between the parts of a system." We have discussed this feature of induction when explaining the methods of energy transfer in Section 13, more precisely in the tutorial 13.4. If the energy lost due to radiation is neglected, we say that faster the magnet is moved, greater the work done by the external force in a certain time and therefore, greater the rate of energy transfer in the loop. This means the power of this energy transfer is greater when the magnet moves faster. The moving direction of magnet is not important; as long as the magnet is moving, it transfers energy to the coil. Physics Tutorials: This image provides visual information for the physics tutorial Induction and Energy Transfers In the figure above, the magnetic flux through the solenoid (coil) is changing because when the magnet gets closer to the coil (direction 1), more magnetic field lines enter the area of coil compared to the case when the magnet moves away from the coil (position 2). In other words, the flux changes because the magnetic field produced by the moving magnet changes. The same effect is obtained when we move a rectangular coil (as the one shown in the figure below) in the left-right and vice-versa direction. Physics Tutorials: This image provides visual information for the physics tutorial Induction and Energy Transfers During this process, the magnetic flux can change in two ways: 1- By pulling the coil at non-uniform velocity, for example pulling it by applying an increasing or decreasing force. In this case, the flux changes due to the change in the number of magnetic field lines, despite the area is the same. 2- By moving the coil in or out of magnetic field. In this case, the area in which the magnetic field lines punch the coil is changing. Thus, when we insert the coil inside the magnetic field the area increases, so the flux increases (remember that magnetic flux in uniform field is Φ = B ∙ A). On the other hand, when we move the coil out of the field the flux decreases as the area punched by magnetic field lines decreases (as shown in the second figure above). The two situations described above, basically represent the same phenomenon - the change in magnetic flux in the coil. However, the setup shown in the last figure (the rectangular coil moving relative to a uniform magnetic field) offers a great advantage regarding the calculation of work done to move the coil out of the magnetic field, as the field lines here are parallel and uniformly distributed, unlike those produced when a bar magnet moves towards or away the solenoid. Therefore, we will consider only the second setup described above when calculating the amount of mechanical work done during this process. Rate of Work (Power) From Dynamics, we know that to pull something at constant velocity, we must apply a constant force, which equals in magnitude the resistive forces. This fact is true in this case as well. We have seen earlier that the pulling force F is opposed by a magnetic force Fm that opposes it. Thus, we have equilibrium only when |F| = |Fm|. In such a case, the coil moves at constant speed. Physics Tutorials: This image provides visual information for the physics tutorial Induction and Energy Transfers The rate of work (i.e. the mechanical power) done by the external force F when moving the coil inside a uniform magnetic field is
P = F ∙ v
Now, let's find an expression for the power P in terms of magnetic field B and other related quantities such as the resistance of coil R and its length l. As stated earlier, when we shift the coil due right, the area punched by the magnetic field lines decreases. As a result, the magnetic flux decreases too. This means a current is induced in the coil; this current produces a magnetic force Fm that opposes the cause of its creation (the pulling force F). 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:
Φ_m = 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
ε_i = (∆Φ_m)/Δt = (B ∙ w ∙ ∆l)/∆t = B ∙ w ∙ v
where v is the moving speed of the coil. Example: 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 a) What is the emf induced in the coil? b) What is the power delivered by the magnetic force if the pulling force is 4N? Solution Clues: l = 20 cm = 0.20 m w = 15 cm = 0.15 m B = 0.2 T Δt = 5 s F = 4 N a) The induced emf in the coil is
ε_i = (∆Φ_m)/Δt = (B ∙ w ∙ ∆l)/∆t = ((0.2 T) ∙ (0.15 m) ∙ (0.20 m-0 m))/(5 s) = 0.0012 V
b) 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
P_m = F_m ∙ v = F ∙ v = (4 N) ∙ (0.04 m/s) = 0.16 W
Induced Current We can still apply the Ohm's Law
i = ε_i/R
to calculate the current induced in the loop if the resistance of coil is known. Combining this law with the equation
ε_i = B ∙ w ∙ v
provided in the previous paragraph, we obtain for the induced current i:
i = (B ∙ w ∙ v)/R
Giving that the magnetic force produced in a current carrying wire is
F_m = i ∙ B ∙ L
where L is the length of wire (do not confuse it with the inductance explained in the previous tutorial), we obtain for the specific setup discussed in this tutorial where the length of the metal bar L here is represented by the width w of the rectangular coil:
F_m = i ∙ B ∙ w
This is true because there is no magnetic force in the direction of the two long sides of the coil but only in the direction of the lateral ones. The only forces acting on the coil are the pulling force (due right) and the magnetic force (due left) which are balanced as the coil moves at constant velocity. Therefore, we obtain for the magnetic force acting on the loop:
F_m = ((B ∙ w ∙ v)/R) ∙ B ∙ w = (B2 ∙ w2 ∙ v)/R
From the last equation, we conclude that if the magnetic force Fm is constant, the moving speed v of the coil is constant as well. This is because the other parameters such as the magnetic field B, the width of loop w and the resistance R are all constants. 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 = F_m ∙ v = ((B2 ∙ w2 ∙ v)/R) ∙ v = (B2 ∙ w2 ∙ v2)/R = (ε_i2)/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: 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: a) The current induced in the coil b) The average pulling force applied in the coil c) The emf induced in the coil d) The rate at which the external source is doing work on the coil e) The total thermal energy transferred to the coil Solution 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 = ? a) 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
b) 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 = F_m = i ∙ B ∙ w = (0.01875 A) ∙ (5.0 T) ∙ (0.25 m) = 0.0234 N
c) The induced emf can be calculated in many ways. One of them is by applying the Ohm's Law. Thus,
ε_i = i ∙ R = (0.01875 A) ∙ (2.0 Ω) = 0.0375 V
d) 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
e) 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. Eddy Currents The rectangular coil used so far is made of a rectangular frame and an empty internal space. Let's replace it with a solid rectangular plate, as shown in the figure below. Physics Tutorials: This image provides visual information for the physics tutorial Induction and Energy Transfers Again, the moving force F (here due right) encounters an opposition by the magnetic force Fm in the opposite direction. However, the induced current does not flow in a single direction and only in the lateral frame as in the case of rectangular frame discussed earlier, rather, electrons swirl around in circular paths as shown in the figure. The currents produced due to this phenomenon are known as eddy currents (eddy = whirlpool). Again, in this case, the current induced in the plate generates an induced emf and therefore, it produces some energy dissipated in the form of heat. Summary Induction results in a transfer of energy between the parts of a system. If the energy lost due to radiation is neglected, we say that faster the magnet is moved, greater the work done by the external force in a certain time and therefore, greater the rate of energy transfer in the loop. This means the power of this energy transfer is greater when the magnet moves faster. The moving direction of magnet is not important; as long as the magnet is moving, it transfers energy to the coil. The magnetic flux produced when we move a coil inside a uniform magnetic field can change in two ways: 1- By pulling the coil at non-uniform velocity, for example pulling it by applying an increasing or decreasing force. In this case, the flux changes due to the change in the number of magnetic field lines, despite the area is the same. 2- By moving the coil in or out of magnetic field. In this case, the area in which the magnetic field lines punch the coil is changing. Thus, when we insert the coil inside the magnetic field the area increases, so the flux increases (remember that magnetic flux in uniform field is Φ = B ∙ A). On the other hand, when we move the coil out of the field the flux decreases as the area punched by magnetic field lines decreases. The magnitude of the induced emf in the coil is
ε_i = B ∙ w ∙ v
where w is the width of the rectangular coil and v is its moving velocity. Giving that the magnetic force produced in a current carrying wire is
F_m = i ∙ B ∙ w
we obtain for the magnetic force induced in the coil
F_m = (B2 ∙ w2 ∙ v)/R
The last equation means that if the magnetic force Fm is constant, the moving speed v of the coil is constant as well. This is because the other parameters such as the magnetic field B, the width of loop w and the resistance R are all constants. The rate of work (i.e. the mechanical power) done by the external force F when moving the coil inside a uniform magnetic field is
P = F ∙ v = (B2 ∙ w2 ∙ v2)/R = (ε_i2)/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
The above two equations give the same value. 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. The circular currents produced when we replaced the rectangular frame with a solid rectangular plate are known as eddy currents (eddy = whirlpool).

Induction and Energy Transfers Revision Questions

1) What is the magnitude of the uniform magnetic field shown in the figure if the emf induced in the 10cm × 6cm rectangular coil when pulling it completely out of the field is 0.2 V. The entire process lasts for 0.4s.

Physics Tutorials: This image provides visual information for the physics tutorial Induction and Energy Transfers
  1. 0.83 T
  2. 3.3 T
  3. 13.3 T
  4. 33.3 T

Correct Answer: C

2) A 36 cm2 square-shaped coil (l = w) is inserted at constant speed (0.2 m/s) inside a 4 T uniform magnetic field. What is the resistance produced in the coil due to the magnetic field if the current induced in the coil is 0.8 A?

Physics Tutorials: This image provides visual information for the physics tutorial Induction and Energy Transfers
  1. 0.1 Ω
  2. 1.0 Ω
  3. 0.048 Ω
  4. 0.06 Ω

Correct Answer: D

3) A 50 cm × 40 cm rectangular loop is pulled at constant speed out a 2.0 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 5 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 0.8 Ω, calculate the thermal energy transferred to the coil during this process.

  1. 0.04 J
  2. 0.032 J
  3. 0.064 J
  4. 0.5 J

Correct Answer: A

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