## What is Consideration Energy?

We know that force is the push or pull of an object and performing a day-to-day task we apply force on our bodies. For exerting our body, we need energy. So, here can find a link between the force and the energy. Also, the thing that provides a link between these two quantities is the energy consideration.

Through the concept of energy consideration in motional emf, we can prove that motional emf is correct or valid according to the conservation of energy. Here, we are going to prove the validity through mathematical derivation, that’s why we are discussing the concept of consideration of energy individually.

### Energy Consideration Physics

In this article, we are going to discuss energy consideration of motional emf and illustrative energy consideration example of a loop.

While applying the principles of energy consideration in Motion EMF, we will be focusing on two important concepts viz: Lenz’s law and the law of energy conservation.

One must keep in mind that Lenz’s Law agrees with the law of conservation of energy and to make it understandable, we are taking a conductor placed in the following manner:

Let’s suppose that there is a rectangular frame placed in a magnetic field (B), as shown in the figure below:

If we observe this figure, there is one rod of length ‘L’ labeled as PQ, which can move left-to-right with a velocity ‘v’.

One must note that the rod should be kept in a direction perpendicular to the magnetic field and the reason for keeping it perpendicular to B also has a significance, which we will understand in terms of mathematical expression later.

### Consideration Energy Physics

If we observe the conductor PQ in Fig.1, PQ is a movable arm, and its length is ‘L’, and it is allowed to move with a velocity ‘v’, and the current flowing through this conductor is I. Also, these three quantities viz: L, I, v are perpendicular to each other. Also, the magnetic field marked by ‘x’ is perpendicular to the plane of the sheet on which it is kept and is directed inwards.

We must also consider one more parameter, i.e., the resistance ‘R’. The resistance for PQ is ‘R’ itself, while for QR, RS, and SP, it will be negligible.

So, now the question arises when this conductor PQ starts moving towards the left with a constant velocity, what will be the induced emf generated in this frame? The answer is pretty simple, it will be:

e = BvL….(1)

Now, if wish to find the direction of induced current, we can find out by using Lenz’s law:

From Fig.1, we can see that the rod is moving towards the left and the magnetic field is directed inwards, so the flux is decreasing. As the flux is decreasing, we need to strengthen the magnetic field by adding one more magnetic field pointing downwards. Further, the direction of the current should always be clockwise.

Now, when the current induces in the clockwise direction, it will oppose the decreasing flux.

We must also note that the polarity of EMF at Point P is negative and positive at point Q.

If we are given with induced emf, to ascertain the induced current, we can use Ohm’s law, which is:

I = e/R…….(2)

(Since the resistance is along PQ only)

From eq (1) in (2), we get the current:

I = vBL/R…..(3)

From equation (3), we can say that PQ is a current-carrying conductor with current vBL/R.

The magnetic force on this rod is:

FM = IBL….(4)

= I (B^{→} x L^{→}) = IBL Sin⊖ = IBL Sin 90 degrees = IBL

From eq (3) in (4):

FM = vB^{2}L^{2}/R……(5)

Here, the direction of the magnetic force or (B x L) is outwards by the Right-hand rule. So, when we are pushing the rod inwards, the flux is decreasing, we immediately apply the magnetic force outwards to oppose the changing flux. Now, this force won’t allow the further inward movement of the rod.

If we still wish to move this rod inwards with a constant velocity, we need an additional force viz: FEXT. So, as the rod moves, FM will counterbalance the effect of FEXT, and therefore, FM and FEXT will become equal and opposite.

To make the rod move inwards with velocity v ultimately by FEXT , we will supply the power by an external agent, which is:

PEXT = FEXT X v…..(6)

Since FEXT = FM, now, from eq (6) in eq (5),we get:

PEXT = FM X v = v^{2}B^{2}L^{2}/R….(7)

Now, if we wish to calculate the power dissipated in the loop PQRs, it is:

PDISS = I^{2}R =(vBL/R)^{2} * R =v^{2}B^{2}L^{2}/R….(8)

We can observe that power supplied by an external agent = power dissipated in the loop. It means, whatever amount of work is done by us in moving the rod inwards will always get dissipated in the form of heat.

This principle is correct according to the conservation of energy, the work done on the system = energy generated in the system.

Hence, we proved that the motional emf is valid according to the conservation of energy.

Question 1: A Circular Loop of radius 6.0 m with 40 rotates about its Vertical Diameter with an Angular Speed of 45 rad/s in a Uniform horizontal Magnetic Field of Magnitude 5.0 × 10^{-4}T. If the Coil forms a Closed Loop of resistance 30. Calculate the Value of Current induced in the Coil and the Average Power Dissipation due to the Joule Heating Effect.

Answer: 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 * πr^{2} * ωB = 40 x 45 x 3.14 x 6^{2}x 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

Question 2: What is the basic Cause of Induced EMF?

Answer: The fundamental cause of induced emf is the varying magnetic flux linked with a current-carrying loop.

Question 3: List the Applications Where the Dynamically Induced EMF is used.

Answer: Below is the listed real-life applications of dynamically induced emf:

Back emf in motors

Electric generators

Transformers

Question 4: Who formulated Lenz’s Law?

Answer: Lenz's law was formulated by Emil Lenz. This law describes ‘ the phenomenon of the flux through the circuit.

This law helps determine the direction of the induced EMF and current resulting from EMI.