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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

Φ_E = ∫▒E*⃗* dA*⃗*

where EΦ_M = ∫▒B*⃗* dA*⃗*

The result of this integral is the dot product of the magnetic field vector BΦ_M = B*⃗* ∙ A*⃗*

and the magnitude of magnetic flux ΦM, is Φ_M = B ∙ A ∙ cosθ

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 Φ_M = B ∙ A ∙ cos〖0^0 〗 = B ∙ A ∙ 1 = B ∙ A

The unit of magnetic flux (Tesla × square metre) is known as Webber, [Wb]. Thus, 1 Wb = 1 T ∙ m^{2}

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. ε_i = -(∆Φ_M)/Δt

for not-so-small intervals of time and ε_i = -(dΦ_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: 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 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 = 300 θ2 = 00 Δt = 2 s Thus, for the angle θ1 = 300 Φ_M(1) = B ∙ A ∙ cos〖θ_{1} 〗 = B ∙ π ∙ r^{2} ∙ cos〖θ_{1} 〗 = (2 × 10^{-2} T) ∙ (3.14) ∙ (3 × 10^{-2} m)^{2} ∙ cos〖30^0 〗 = (2 × 10^{-2} T) ∙ (3.14) ∙ (3 × 10^{-2} m)^{2} ∙ (0.866) = 4.89 × 10^{-5} Wb

When rotating the loop at θ2 = 00 we obtain Φ_M(2) = B ∙ A ∙ cos〖θ_{2} 〗 = B ∙ π ∙ r^{2} ∙ cos〖θ_{2} 〗 = (2 × 10^{-2} T) ∙ (3.14) ∙ (3 × 10^{-2} m)^{2} ∙ cos〖0^0 〗 = (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 ε_i = -(∆Φ_M)/Δt = -(〖(Φ〗_M(2) -Φ_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: ε_i = -N ∙ (∆Φ_M)/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 F_M = 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 F_M = F_e 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 ε_i = B ∙ v ∙ L ∙ sinθ

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 = ε_i/R = (B ∙ v ∙ L)/R

where R is the resistance in the circuit. Example: 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 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 ∙ sinθ

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 = 900). 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: 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 FM is the original magnetic force discussed earlier (here FM 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 Fapp is P = F_{a}pp ∙ v = F_m^' ∙ v = I ∙ B ∙ L ∙ v = (ε_i/R) ∙ B ∙ L ∙ v = ((B ∙ v ∙ L)/R) ∙ B ∙ L ∙ v = (B^{2} ∙ v^{2} ∙ L^{2})/R

Thus, we can write P = (ε_i^{2})/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: 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: a) The current in the resistor b) The moving velocity of the bar c) The power in the resistor Solution Clues: m = 400 g = 0.4 kg L = 20 cm = 0.2 m R = 4 Ω B = 5 T θ = 900 I = ? v = ? P = ? a) 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/s^{2} ))/((5 T) ∙ (0.2 m) ) = 3.924 A

b) 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

c) 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

Summary The experiments performed by Faraday to study the interaction between magnetism and electricity indicate 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. 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). In other words, a current is produced in the loop only if there is a change of something in proximity of loop. Based on his 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." The magnetic flux ΦM flowing through a loop is Φ_M = ∫▒B*⃗* dA*⃗*

The result of this integral is the dot product of the magnetic field vector BΦ_M = B ∙ A ∙ cosθ

where A is the area vector of the loop and θ is the angle formed by the magnetic field lines and the area vector. The unit of magnetic flux (Tesla × square metre) is known as Webber, [Wb]. Thus, 1 Wb = 1 T ∙ m^{2}

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." Based on the Faraday's Law of Induction, we can say that the induced emf tends to oppose the flux change, i.e. ε_i = -(∆Φ_M)/Δt

for not-so-small intervals of time and ε_i = -(dΦ_M)/dt

for infinitely small intervals of time. 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. If the coil has N-turns, we obtain for the induced emf: ε_i = -N ∙ (∆Φ_M)/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 The induced emf is also known as motional emf. The induced emf 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. 4- The angle formed by the wire and its moving direction. Combining all these factors, we obtain the formula of emf induced in the conducting wire when it moves inside a magnetic field: ε_i = B ∙ v ∙ L ∙ sinθ

The value of current I flowing through the circuit in this case, is I = ε_i/R = (B ∙ v ∙ L)/R

where R is the resistance in the circuit. When a motional emf causes a current, a second magnetic force produced because of the current I produced in the 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 ∙ sinθ

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 = 900). 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. The power P delivered by the applied force Fapp is P = F_{a}pp ∙ v = (ε_i^{2})/R

The equation above means that: "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. **1)** The magnetic flux through a coil of 200 turns is changing at a rate of 0.4 Wb/s What is the total emf induced in the coil?

- 80 V
- -80 V
- 500 V
- -500 V

**Correct Answer: B**

**2)** A 60 cm long conducting bar is placed inside a 0.05 T magnetic field in the vertical direction shown in the figure. The magnetic field lines are directed into the page. What is the induced emf across the wire if it is moved at 2 m/s due right?

- 0.06 V
- 0.6 V
- 6 V
- 60 V

**Correct Answer: A**

**3)** A 1.2 m long metal bar is free to slide vertically between two vertical copper tracks. A constant magnetic field of magnitude 5 T is directed perpendicular to the axes of copper tracks, as shown in the figure. The bar moves under the effect of gravity at 2 m/s constant speed. What is the induced current in the 40 Ω resistor?

- 12 A
- 1.2 A
- 0.3 A
- 0.25 A

**Correct Answer: C**

We hope you found this Physics tutorial "Faraday's Law of Induction" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of Magnetism with our Physics tutorial on Lentz Law .

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