Physics Lesson 16.7.4 - Motional Emf and Electrical Energy

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Welcome to our Physics lesson on Motional Emf and Electrical Energy, this is the fourth lesson of our suite of physics lessons covering the topic of Faraday's Law of Induction, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Motional Emf and Electrical Energy

When a motional emf causes a current, a second magnetic force produced because of the current I produced in the bar. (Remember, the first magnetic force is due to the interaction of the moving bar and magnetic field B; it lies is in the direction of bar). The new magnetic force F^'_m produced in this case, opposes the applied force F_app that is used to move the bar. The magnitude of this new magnetic force is

F^'_m = I ∙ B ∙ L ∙ sin⁡θ

where I is the current induced in the circuit, L is the length of the bar and θ is the angle formed by the bar and the direction of the magnetic field (here sin θ = 1 because θ = 90°). Look at the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Faraday's Law of Induction

If the external applied force F_app is numerically equal to the new magnetic force discussed above, the bar moves at constant speed v. The external force F_app does some work W_app against the magnetic force during the bar's motion. This work is equal to the electrical energy produced in the circuit.

From the definition of electromotive force, we have for the induced emf in the circuit:

ε_i = W/Q

(Here W is the work done to move the charges throughout the bar, not the work done to move the bar itself; they are completely different things.)

Therefore, we obtain:

ε_i = F_M ∙ L/Q

where F_M is the original magnetic force discussed earlier (here F_M is directed downwards). Thus, since

F_M = Q ∙ v ∙ B

we obtain the known equation for the induced emf after substitutions,

ε_i = Q ∙ v ∙ B ∙ L/Q
= B ∙ v ∙ L

The power P delivered by the applied force F_app is

P = F_app ∙ v
= F^'_m ∙ v
= I ∙ B ∙ L ∙ v
= (ε_i/R) ∙ B ∙ L ∙ v
= (B ∙ v ∙ L/R) ∙ B ∙ L ∙ v
= B² ∙ v² ∙ L²/R

Thus, we can write

P = ε_i²/R

The last formula is interpreted as follows:

"The power delivered by the applied force in the above setup is equal to the rate at which the electrical energy is dissipated in the resistor."

The direction of current is in accordance with the law of conservation of energy. If the current was flowing is in the opposite direction, the direction of the new magnetic force would reverse; this means the bar would experience some acceleration as both the applied force F_app and the new magnetic force F'm are in the same direction. Even if no external force was present, the bar would accelerate because of the new magnetic force. This means the bar moves without doing any work on it and therefore, it produces electrical energy out of nothing. This is impossible as it represents a violation of the law of energy conservation (energy can neither produced from nothing neither it can disappear; it only converts from one form into another).

Example 3

A 400 g conducting bar, which is 20 cm long, is free to slide between two copper tracks. A 4 Ω resistor is connected between the upper ends of the tracks. The system is placed inside a constant magnetic field of magnitude equal to 5T. The magnetic field lines are perpendicular to the plane of the system. The bar is sliding downwards at constant velocity due to the effect of gravity.

Calculate:

The current in the resistor
The moving velocity of the bar
The power in the resistor

Solution 3

Clues:

m = 400 g = 0.4 kg
L = 20 cm = 0.2 m
R = 4 Ω
B = 5 T
θ = 90°
I = ?
v = ?
P = ?

Based on the Fleming's Left Hand Rule, we find out that the direction of magnetic force produced on the bar is upwards. Since the bar is moving at constant speed, this magnetic force is balanced by the gravitational force acting on the bar. Numerically, we have
F_M = F_g
I ∙ B ∙ L = m ∙ g
I = m ∙ g/B ∙ L
= (0.4 kg) ∙ (9.81 m/s² )/(5 T) ∙ (0.2 m)
= 3.924 A
To find the moving speed of bar, first we must calculate the emf induced in the circuit. We have
ε_i = I ∙ R
= (3.924 A) ∙ (4 Ω)
= 15.696 V
Hence, giving that
ε_i = B ∙ v ∙ L
we obtain for the speed v after rearranging the above equation
v = ε_i/B ∙ L
= 15.696 V/(5 T) ∙ (0.2 m)
= 15.696 m/s
From the definition of electric power, we have for the power delivered in the resistor
P = ε_i ∙ I
= 15.696 V ∙ 3.924 A
= 61.591 W

You have reached the end of Physics lesson 16.7.4 Motional Emf and Electrical Energy. There are 4 lessons in this physics tutorial covering Faraday's Law of Induction, you can access all the lessons from this tutorial below.

More Faraday's Law of Induction Lessons and Learning Resources

Magnetism Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
16.7	Faraday's Law of Induction
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
16.7.1	What did Faraday Observe during His Experiments?
16.7.2	Faraday's Law of Induction
16.7.3	Induced Electromotive Force as Motional Emf
16.7.4	Motional Emf and Electrical Energy

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