Magnetism Learning Material
Tutorial IDTitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions

In this Physics tutorial, you will learn:

• What are the quantities (variables) Faraday analysed to make his discoveries in Electromagnetism?
• What is induced current and induced emf?
• What factors do affect the induced emf produced in the presence of magnetic field?
• What does the Faraday's Law of Induction say?
• What is magnetic flux? What is its unit?
• How can we change the magnetic flux in a coil?
• Why the induced emf is also known as motional emf?
• How are the motional emf and electrical energy related to each other?
• The same for the motional emf and electric power

## Introduction

So far, we have seen many examples in which a current carrying wire produces a magnetic field around it. What about the reverse process, i.e. do you think a simple magnet can produce electricity?

How do we call the method of charging objects by bringing another charged object near a neutral one? Do you think this phenomenon is related in any way to the current topic?

This tutorial discusses a phenomenon known as "induced current". This is not a current produced by direct contact between objects; this means it is different from the current we have seen so far.

## What did Faraday Observe during His Experiments?

In tutorial 16.2 "Magnetic Field Produced by Electric Currents", we stated that Faraday did a famous experiment in which he discovered that a moving magnet generates electricity in a conducting wire if moved properly. We can observe better this property by making two experiments as follows.

First experiment. Let's consider a circular loop connected to a very sensitive ammeter as shown in the figure. If we move the magnet towards the loop (direction 1), the ammeter arrow moves and indicates a non-zero current. The same thing occurs when the magnet moves in the opposite direction as well (direction 2). The only case when the ammeter shows zero current is when the magnet is changing direction. It is known than when an object is changing direction, stops for a while (for example, when we throw an object upwards, the velocity at the maximum position is zero).

This experiment shows that:

1. Only a moving magnet is able to produce electric current in the loop; a stationary magnet does not produce any current.
2. Faster the motion of magnet, greater the current produced in the loop.
3. The direction of current changes when the direction of magnet's motion changes. For example, if we move the magnet towards the loop and the current produced is let's say clockwise, the direction of current in the loop will be anticlockwise if we move the same magnet away from the loop.

As we have stated in the tutorial 16.2, the current produced through this method is called induced current, as it is not obtained through direct contact but through induction. The work done to move the charges throughout the loop by means of this method is called induced emf (induced electromotive force).

Second experiment. For this experiment, we need two circular loops to be placed near each other, where one of them is connected to an electric circuit and the other to an ammeter as shown in the figure. When the switch S turns on, the ammeter shows some current flowing through the left circular loop despite the loops are not touching each other. However, after a while the ammeter shows no current despite the switch is ON. When we turn the switch OFF, the current flows again through the left loop as the ammeter shows some current again.

This experiment is a demonstration that an induced current (and an induced emf) is produced in the loop connected to the ammeter (on the left side) due to the magnetic effect of current flowing in the loop on the right connected to the battery. Since this current is not produced through direct contact, it is an induced current. This current is not constant; it changes value and direction continuously every time we change the position of switch.

The two above experiments are demonstrations that a current is produced in the loop only if there is a change of something in proximity of loop. The reason why this change occurs, accompanied with the relevant theoretical explanation for which we will discuss in the next paragraph, was provided by Michael Faraday. That's why it is called the Faraday's Law of Induction.

Based on the above two experiments, Faraday realized that:

1. An emf and current can be induced in the loop by changing the amount of magnetic field around the loop;
2. The amount of magnetic field can be represented through the amount of magnetic field lines passing through the loop.

Combining the above statements, we obtain the Faraday's Law of Induction:

"An induced emf (and current) are induced in a loop only if the number of magnetic field lines passing through the loop is changing."

From this law, we can conclude that the number of actual field lines flowing through the loop is not important; the only thing that matters in this process is the rate of change in the number of magnetic field lines passing through the loop.

Now, let's calculate the amount of magnetic field lines passing through the loop. For this, we have to consider the magnetic flux ΦM, which is analogue to the electric flux ΦE we have discussed in the tutorial 14.6 "Electric Flux and the Gauss Law." The integral expression of the electric flux flowing through a loop is

ΦE = E dA

where E is the vector of electric field and A the area vector of the loop.

Similarly, the magnetic flux ΦM flowing through a similar loop is

ΦM = B dA

The result of this integral is the dot product of the magnetic field vector B and the area vector A, which is perpendicular to the surface pierced by the magnetic field lines and is numerically equal to the area enclosed by the loop. The magnetic field B between the poles produced by the bar magnet shown above is uniform, i.e. the magnetic field lines are parallel to each other (vertical in direction). The loop is deflected by an angle θ to the horizontal direction; this means the area vector A forms the same angle θ to the magnetic field lines. Therefore, the value of the integral if the magnetic field is uniform, is

ΦM = B ∙ A

and the magnitude of magnetic flux ΦM, is

ΦM = B ∙ A ∙ cos⁡θ

In addition, if the loop is placed perpendicular to the direction of magnetic field lines, i.e. the area vector is parallel to the magnetic field lines, we obtain

ΦM = B ∙ A ∙ cos⁡00
= B ∙ A ∙ 1
= B ∙ A

The unit of magnetic flux (Tesla × square metre) is known as Webber, [Wb]. Thus,

1 Wb = 1 T ∙ m2

The concept of magnetic flux helps us state the Faraday' Law of induction in a more comprehensive and practical way:

"The magnitude of the electromotive force induced in a loop is equal to the rate of change of the magnetic flux in this loop."

We have seen in tutorial 16.2 that the induced current in a coil containing N loops when we move a magnet towards to or away from it, is in the opposite direction to the moving magnet. The same thing is true for the induced emf produced in the coil as well. Thus, based on the Faraday's Law of Induction, we can say that the induced emf tends to oppose the flux change, i.e.

εi = -∆ΦM/Δt

for not-so-small intervals of time and

εi = -M/dt

for infinitely small intervals of time. Obviously, the last formula gives a more precise result for the emf induced in a loop or coil due to a magnetic field. Therefore, it is known as the mathematical expression of the Faraday's Law of induction.

### Example 1

A circular loop of radius equal to 3 cm is placed between the poles of a horseshoe magnet at 300 to the horizontal direction, as shown in the figure. What is the induced emf produced in the loop if we rotate the coil from the actual position to the horizontal position in 2 seconds? The magnetic field between the poles of magnet is 20 mT.

### Solution 1

First, we must find the magnetic flux flowing through the loop in both positions. In the actual position, the angle between magnetic field lines (down-up) and the area vector (normal to the loop) is still 300 (from geometry it is known that two angles that have perpendicular sides to each other are equal).: We have the following clues:

r = 3 cm = 3 × 10-2 m
B = 20 mT = 2 × 10-2 T
θ1 = 30°
θ2 = 0°
Δt = 2 s

Thus, for the angle θ1 = 30°

ΦM(1) = B ∙ A ∙ cos⁡θ1
= B ∙ π ∙ r2 ∙ cos⁡θ1
= (2 × 10-2 T) ∙ (3.14) ∙ (3 × 10-2 m)2 ∙ cos⁡300
= (2 × 10-2 T) ∙ (3.14) ∙ (3 × 10-2 m)2 ∙ (0.866)
= 4.89 × 10-5 Wb

When rotating the loop at θ2 = 0° we obtain

ΦM(2) = B ∙ A ∙ cos⁡θ2
= B ∙ π ∙ r2 ∙ cos⁡θ2
= (2 × 10-2 T) ∙ (3.14) ∙ (3 × 10-2 m)2 ∙ cos⁡00
= (2 × 10-2 T) ∙ (3.14) ∙ (3 × 10-2 m)2 ∙ 1
= 5.65 × 10-5 Wb

Therefore, the induced electromotive force εi produced in the loop when rotating it inside the uniform magnetic field, is

εi = -∆ΦM/Δt
= -M(2)M(1) )/Δt
= -(5.65 × 10-5 Wb-4.89 × 10-5 Wb)/2s
= -0.38 V

Hence, the induced emf during this process is 0.38V (the negative sign is because induced emf is in the opposite direction to the cause of its generation (moving direction of loop).

Note that here we have used the symbol Δ instead of d to represent the time interval as it is too large to consider as infinitely small.

If the coil has N-turns, we obtain for the induced emf:

εi = -N ∙ ∆ΦM/Δt

The magnetic flux through a coil can be changed in four ways:

1. By changing the value of magnetic field in the space containing the coil,
2. By changing the area of coil,
3. By changing the angle of coil to the magnetic field lines, and
4. By changing the number of turns (loops) in the coil

## Induced Electromotive Force as Motional Emf

Let's extend the discussion about the induced emf considering it from another viewpoint. As a special case of electromagnetic induction is when a magnetic force acts on a straight current carrying wire placed inside a uniform magnetic field is moving due right as shown in the figure below. The direction of the resulting magnetic force is vertically down, based on the Fleming's Left Hand Rule explained in the previous tutorials. The scalar equation of this magnetic force is

FM = Q ∙ v ∙ B

Since the wire is moving due right, the magnetic force is balanced by another force acting in the upward direction. This force is the electric force Fe produced on the moving charges. We can write the equation that shows mathematically this balance of forces as

FM = Fe
Q ∙ v ∙ B = Q ∙ E
E = v ∙ B

Since the electric field produced by the moving charges is constant, we express the relationship between the electric field E and the potential difference ΔV across the ends of conducting wire as

∆V = E ∙ L

Combining the last two equation, we obtain

∆V = B ∙ L ∙ v

This potential difference is equal to the emf ε of the wire as the resistance of wire is negligible. Obviously, this is an induced emf as it is caused due to the motion of wire and is not constant; it depends on the following factors:

1. Magnitude of magnetic field - a stronger field causes a higher induced emf than a weaker magnetic field;
2. Length of conductor - a longer conductor causes a higher induced emf than a shorter one;
3. Velocity of conductor - a greater velocity causes a higher induced emf in the wire than a lower velocity.
Another factor that affects the induced emf produced in the wire is the angle between the moving direction and the wire itself. So far, we have considered only the case in which the moving direction is perpendicular to the wire's length. In this case, the induced emf is maximal. However, when we push the wire in another direction, the potential difference decreases until it becomes zero when the wire is pushed in the direction of its length. Hence, we write the fourth factor affecting the amount of induced emf as
4. The angle formed by the wire and its moving direction. Hence, the formula of emf induced in the conducting wire when it moves inside a magnetic field is

εi = B ∙ v ∙ L ∙ sin⁡θ

If the conductor that is sliding due right is part of a closed conducting path as shown in the figure below, then, electrons will move in the clockwise direction throughout the circuit due to the existence of emf produced by the source. (Remember the direction of conventional current is opposite to the direction of electron's flow).

The value of current I flowing through the circuit in this case, is

I = εi/R = B ∙ v ∙ L/R

where R is the resistance in the circuit.

### Example 2

At what speed should we move a 50 cm bar as shown in the figure below to produce a current of 0.4 A if the bar touches the two opposite sides of a circuit having a resistance of 20 Ω? The plane of circuit (including the bar) is normal to a 2 T magnetic field the direction of which is into the page. ### Solution 2

Clues:

L = 50 cm = 0.50 m
I = 0.4 A
R = 20 Ω
B = 2 T
v = ?

From the equation

I = ε/R = B ∙ v ∙ L/R

we obtain for the speed v after rearranging the terms

v = I ∙ R/B ∙ L
= (0.4 A) ∙ (20 Ω)/(2 T) ∙ (0.50 m)
= 8 m/s

## 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 Fapp 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. If the external applied force Fapp is numerically equal to the new magnetic force discussed above, the bar moves at constant speed v. The external force Fapp does some work Wapp 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:

εi = 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:

εi = FM ∙ L/Q

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

FM = Q ∙ v ∙ B

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

εi = Q ∙ v ∙ B ∙ L/Q
= B ∙ v ∙ L

The power P delivered by the applied force Fapp is

P = Fapp ∙ v
= F'm ∙ v
= I ∙ B ∙ L ∙ v
= (εi/R) ∙ B ∙ L ∙ v
= (B ∙ v ∙ L/R) ∙ B ∙ L ∙ v
= B2 ∙ v2 ∙ L2/R

Thus, we can write

P = εi2/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 Fapp 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:

1. The current in the resistor
2. The moving velocity of the bar
3. 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 = ?

1. 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
FM = Fg
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
2. To find the moving speed of bar, first we must calculate the emf induced in the circuit. We have
εi = I ∙ R
= (3.924 A) ∙ (4 Ω)
= 15.696 V
Hence, giving that
εi = B ∙ v ∙ L
we obtain for the speed v after rearranging the above equation
v = εi/B ∙ L
= 15.696 V/(5 T) ∙ (0.2 m)
= 15.696 m/s
3. From the definition of electric power, we have for the power delivered in the resistor
P = εi ∙ I
= 15.696 V ∙ 3.924 A
= 61.591 W

## Whats next?

Enjoy the "Faraday's Law of Induction" physics tutorial? People who liked the "Faraday's Law of Induction" tutorial found the following resources useful:

1. Physics tutorial Feedback. Helps other - Leave a rating for this tutorial (see below)
2. Magnetism Revision Notes: Faraday's Law of Induction. Print the notes so you can revise the key points covered in the physics tutorial for Faraday's Law of Induction
3. Magnetism Practice Questions: Faraday's Law of Induction. Test and improve your knowledge of Faraday's Law of Induction with example questins and answers
4. Check your calculations for Magnetism questions with our excellent Magnetism calculators which contain full equations and calculations clearly displayed line by line. See the Magnetism Calculators by iCalculator™ below.
5. Continuing learning magnetism - read our next physics tutorial: Lentz Law