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What are Eddy Currents?
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Energy Consideration

Last Updated : 28 Mar, 2025
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Force is the influence that causes an object to move or change its motion. In our daily activities, we apply force to our bodies. To generate this force, our body requires energy. Therefore, there is a connection between force and energy, and this connection is explained through the concept of energy.

By looking at the idea of energy in motion-induced Emf, we can show that it follows the law of conservation of energy. To prove this, we will use a mathematical approach. This is why we are discussing the energy concept first, to understand how it works in this context.

What is Consideration Energy?

Energy consideration refers to the analysis of how energy is transferred, transformed, or conserved within a system. It involves evaluating the different forms of energy (kinetic, potential, thermal, etc.) and understanding how these energy changes occur in physical processes while ensuring the law of conservation of energy is upheld.

Energy Consideration In Physics

⇒ Here, we will talk about how energy works in motion-induced EMF and give an example of energy in a loop to help explain it.

⇒ When looking at how energy works in motion-induced EMF, we'll focus on two key ideas: Lenz's law and the law of conservation of energy.

⇒ In physics, Energy consideration refers to the process of taking into account the energy involved in a particular situation or process. This can include analyzing the energy being transferred, stored, or used in some way.

⇒ In general, understanding and analyzing energy considerations is important in a wide range of fields, including physics, engineering, and technology.

For example,

  • In the field of thermodynamics, energy considerations are important in understanding how heat is transferred between different systems and how work can be performed using thermal energy.
  • In the field of electrical engineering, energy considerations are important in the design and operation of devices such as transformers and generators, which rely on electromagnetic induction to transfer energy.

Energy Consideration: A Quantitative Study

We will be concentrating on Lenz's law and the law of energy conservation as we apply the ideas of energy consideration in Motion EMF.

Lenz's Law and the law of conservation of energy are compatible, so bear that in mind. To illustrate this, let's use the example of a conductor set up as follows:

  • As indicated in the image below, suppose that a rectangular frame (A) is positioned in a magnetic field (B), If we look at this image, we can see one rod with a length of "l" and the label "CD," which has a left-to-right velocity of "v."
  • It is important to remember that the rod should always be kept perpendicular to the magnetic field, and there is a rationale for doing so as well, which is mentioned below in terms of a mathematical formula.
Energy Consideration
Energy Consideration
  • Consider a rectangular conductor. We can infer from the illustration that the rectangular conductor's sides are AB, CD, BC, and DA. Now, three of the sides of this rectangular conductor are fixed, but one of them, side AB, is free.
  • Let 'r' represent the conductor's adjustable resistance. As a result, in comparison to this movable resistance, the resistance of the remaining three sides of the rectangular conductor—sides CD, DA, and BC—is very low.

If we alter the flux in a magnetic field that is always present, an emf is produced.  

i.e  E = dΦ/dt

We can state that current is present if there is an induced emf E and a moveable resistance r in the conductor,

I =  Blv/R. 

As long as there is a magnetic field, there will also be a force F acting since F = ILB. This force, which is determined by force, is directed outward in the direction opposite to the rod's velocity.

 F = B²l²v/R

Power = force × velocity = B²l²v²/R

The work that is being done in this instance is mechanical, and the mechanical energy is lost as Joule heat.

It is stated as PJ = I²R = B²l²v²/R. 

The mechanical energy then changes into electrical energy and then heat energy. As a result of Faraday's law,

we know that |E| =ΔΦB/Δt

Thus, we have,

|E| = IR = (ΔQ/Δt)R

As a result,

ΔQ=  ΔΦB/R.

OR

The definition of electromagnetic induction, states that an induced current is produced in a conductor when there is a changing magnetic field around it.

Mathematically, this can be represented as,

ΔΦ = ε

where ΔΦ is the change in the magnetic flux through the conductor (Φ is measured in webers), and ε is the induced electromotive force (measured in volts).

Next, we can use Ohm's law, which states that the current through a conductor is equal to the voltage across it divided by the resistance of the conductor,

I = V/R

Substituting the expression for ε from the equation above, we get,

I = ΔΦ/R

Finally, we can use the definition of electrical charge, which states that the electrical charge Q is equal to the current multiplied by the time for which it flows,

Q = I × t

Substituting the expression for I from the equation above, we get,

ΔQ = ΔΦB/R

This equation states that the change in electrical charge (ΔQ) is equal to the change in magnetic flux (ΔΦ) divided by the resistance (R) of the conductor. This equation is a useful way to analyze the energy being transferred through the process of electromagnetic induction.

Also, Read

  • Lenz’s Law and Conservation of Energy
  • Law of Conservation of Energy

Solved Examples - Energy Consideration

Example 1: In a uniform horizontal magnetic field with a magnitude of 5.0 10-4 T, a circular loop with a radius of 6.0 m and a circumference of 40 revolves with an angular speed of 45 rad/s about its vertical diameter. If a closed loop of resistance 30 occurs in the coil. Calculate the average power dissipation caused by the Joule heating effect as well as the value of current induced in the coil.

Solution:

Given 
R = 30 , B = 5.0 x 10-4 T, r = 6 m,  = 45 rad/s, and N = 40

We know that I = e/R and 

e = NωAB 
   = N × πr2 × ωB 
   = 40 x 45 x 3.14 x 62x  5.0 × 10-4
    = 1.01736 V 

So, I = 1.01736/10 
        = 0.101736 A

and, Power loss = eI/2 
                          =1.01736 x 0.101736 / 2
                          = 0.052 W.

Example 2: A 0.3 T uniform magnetic field directed normally to an 8 cm by 2 cm rectangular wire loop with a minor cut is traveling out of the region. The loop’s velocity is normal to its longer side, which is 8 cm long. Calculate the emf created across the cut and the duration of the induced voltage.

Solution: 

Given  

A = 8 cm × 2 cm = 16 cm², 
Number of turns in the loop N = 1, 
ω = v/r, where v is the velocity of the loop and the distance travelled by the loop 
r = 8 cm. 

Therefore, the angular velocity of the loop is given by ω = v/r = v/8 cm. The emf generated by the loop is given by 
Emf (E) = NABω 
            = 1 × 0.3 T × 16 cm² × ω.

Let’s say the velocity of the loop is v = 10 cm/s. Then the angular velocity of the loop is given by ω = v/r = 10 cm/s / 8 cm = 1.25 s⁻¹. 

Emf (E) = NABω 
            = 1 × 0.3 T × 16 cm² × 1.25 s⁻¹ 
            = 4 volts. 

The duration of the induced voltage is equal to the time it takes for the loop to travel a distance of 8 cm, which is 

t = r/v 
  = 8 / 10 
  = 0.8 seconds.

If the loop’s velocity is normal to its shorter side, which is 2 cm long, the distance travelled by the loop is

 r = 2 cm and the angular velocity of the loop is given by 

ω = v/r 
    = v/2 .

Duration of the induced voltage is equal to the time it takes for the loop to travel a distance of 2 cm, which is

 t = r/v 
   = 2 / 10 
   = 0.2 seconds.

Example 3: A wire with a resistance of 20 ohms and a length of 2 meters is placed in a uniform magnetic field of strength 1 tesla. The wire is rotated through an angle of 90 degrees in 0.5 seconds. Calculate the heat generated by the wire during this time.

Solution:

Change in flux through the wire is given by ΔΦ = BAcosθ, 

where B is the magnetic field strength, A is the cross-sectional area of the wire, and θ is the angle through which the wire is rotated. 

Since the wire is rotated through an angle of 90 degrees, cosθ = 0 and ΔΦ = 0. 

Therefore, the heat generated by the wire is given by

 ΔQ = ΔΦB/R 
       = 0/20 
       = 0 joules.

Example 4: A heater with a resistance of 10 ohms is connected to a 120-volt power supply. Calculate the power dissipated by the heater.

Solution:

Current flowing through the heater is given by Ohm’s law as

I= V/R 
  = 120 
  = 12 amperes. 

Power dissipated by the heater is then given by 

P = I²R

P = 12² × 10 
   = 1440 watts.

Conclusion

"Energy consideration" refers to analyzing a situation by accounting for the energy involved, ensuring that the total energy remains constant according to the law of conservation of energy. This principle is often applied in physics to explain phenomena like electromagnetic induction, where forces and energy transformations are at play. A key aspect of this is Lenz's law, which determines the direction of the induced current in a way that opposes changes in magnetic flux, helping maintain energy balance.


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    Spherical mirrors are generally constructed from glass. A spherical surface is a part cut from a hollow sphere. This curved surface of the glass has a silver coating on one side and a polished surface on the other, where the reflection of light takes place. The term "convex mirror" refers to a mirro
    11 min read
    Refraction of Light
    Refraction is an important term used in the Ray Optics branch of Physics. Refraction of light is defined as the change in direction or the bending of a wave passing from one medium to another due to the change in speed of the wave. Some natural phenomena occurring in nature where refraction of light
    11 min read
    Total Internal Reflection
    In Physics, total internal reflection is the complete reflection of a light ray within the medium (air, water glass, etc). For example, the total internal reflection of rays of light takes place in a Diamond. Since Dimond has multiple reflecting surfaces through which the Total internal reflection t
    8 min read
    Image formation by Spherical Lenses
    You might have used a microscope in the science lab for magnifying the micro-size object. It basically magnifies tiny objects and we can see the enlarged image of that object. Telescopes are used by scientists to the planets and stars which are far- far away from the earth. You might see the spectac
    8 min read
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