Induced emf

• Induced emf is the electromotive force produced in a circuit due to a changing magnetic field.
• It is given by the equation: ε = -N * dΦ/dt where ε is the induced emf, N is the number of turns in the coil, and dΦ/dt is the rate of change of magnetic flux.

Lenz’s Law

• Lenz’s law is a consequence of Faraday’s law that states the direction of the induced current opposes the change in magnetic field that caused it.
• It can be summarized by the following statement: “The induced current always flows in a direction that opposes the change producing it.”

Magnetic Flux

• Magnetic flux (Φ) through a surface is a measure of the total magnetic field passing through that surface.
• It is given by the equation: Φ = B * A * cos(θ) where Φ is the magnetic flux, B is the magnetic field strength, A is the area of the surface, and θ is the angle between the magnetic field and the surface.

Faraday’s Law - Summary

• Faraday’s law can be stated as: “The induced emf in a circuit is equal to the negative rate of change of magnetic flux through the circuit.”
• It can be mathematically represented as: ε = -dΦ/dt

11. Faraday’s Law Of Induction - Induced emf - Example 1 (Magnetic field through a loop)

• Consider a circular loop of radius R placed in a uniform magnetic field with magnitude B.
• If the loop is perpendicular to the magnetic field, the magnetic flux through the loop is given by Φ = B * π * R^2.
• If the magnetic field is changing with time, the induced emf in the loop can be calculated using the equation ε = -dΦ/dt.
• Let’s consider an example: Suppose the magnetic field is decreasing at a rate of 0.1 T/s. Find the induced emf in the loop. Solution:
  • Using the formula, ε = -dΦ/dt = -(d/dt)(B * π * R^2)
  • Since R is a constant, its derivative is zero. Therefore, we only differentiate the magnetic field with respect to time.
  • ε = -(d/dt)(B * π * R^2) = -π * R^2 * dB/dt
  • Given dB/dt = -0.1 T/s, R = 0.5 m
  • Substituting the values, ε = -π * (0.5)^2 * (-0.1) = 0.079 V

12. Faraday’s Law Of Induction - Induced emf - Example 2 (Two coaxial solenoids)

• Consider two coaxial solenoids, one inside the other, with radii r1 and r2 (r2 > r1) and lengths l1 and l2 (l2 > l1).
• Let I1 and I2 be the currents flowing through the inner and outer solenoids, respectively.
• Suppose the current in the inner solenoid is changed at a rate di1/dt.
• The magnetic field due to the inner solenoid at the position of the outer solenoid is given by B1 = μ0 * (N1 * I1) / l1.
• The magnetic field due to the outer solenoid at its own position is given by B2 = μ0 * (N2 * I2) / l2.
• According to Faraday’s law, the induced emf in the outer solenoid is given by ε = -N2 * dΦ2/dt, where N2 is the number of turns in the outer solenoid.

13. Faraday’s Law Of Induction - Induced emf - Example 2 (continued)

• The magnetic flux through the outer solenoid is given by Φ2 = B1 * A2, where A2 is the cross-sectional area of the outer solenoid.
• The cross-sectional area of the outer solenoid can be calculated using A2 = π * (r2^2 - r1^2)
• Differentiating Φ2 with respect to time, we get dΦ2/dt = d/dt(B1 * A2)
• Using the chain rule, dΦ2/dt = dB1/dt * A2 + B1 * dA2/dt
• Since r2 is changing with time, we have dA2/dt = 2π * r2 * dr2/dt
• Substituting these values, we get dΦ2/dt = (μ0 * N1 * I1 * dA2/dt) / l1 + (μ0 * N1 * I1 * A2 * dr2/dt) / l1

14. Faraday’s Law Of Induction - Induced emf - Example 2 (continued)

• The induced emf in the outer solenoid is given by ε = -N2 * dΦ2/dt
• Substituting the value of dΦ2/dt, we get ε = -N2 * [(μ0 * N1 * I1 * dA2/dt) / l1 + (μ0 * N1 * I1 * A2 * dr2/dt) / l1]
• Simplifying further, ε = -(μ0 * N1 * N2 * I1 * dA2/dt) / l1 - (μ0 * N1 * N2 * I1 * A2 * dr2/dt) / l1
• Rearranging, ε = -(μ0 * N1 * N2 * I1 / l1) * [dA2/dt + (A2 * dr2/dt)]

15. Faraday’s Law Of Induction - Induced emf - Example 2 (continued)

• Suppose the current in the inner solenoid is changing at a constant rate di1/dt = 100 A/s.
• The area of the outer solenoid is A2 = π * (0.02^2 - 0.01^2) m^2.
• The radius of the outer solenoid is changing at a constant rate dr2/dt = 0.001 m/s.
• Let N1 = 1000, N2 = 2000, l1 = 0.1 m.
• Substituting these values, we get ε = -(4π * 10^-7 * 1000 * 2000 * 100 / 0.1) * [(π * (0.02^2 - 0.01^2) * 0.001) + (π * (0.02^2 - 0.01^2) * 0.001)]
• Simplifying this expression will give the value of induced emf in the outer solenoid.

16. Faraday’s Law Of Induction - Lenz’s Law - Concept

• Lenz’s law states that the induced current in a circuit will always flow in such a direction as to oppose the change in magnetic field that produced it.
• This law is based on the principle of conservation of energy.
• Lenz’s law ensures that when a magnetic field is changing, the work done by the induced current is always negative (i.e., it is dissipated as heat or stored as potential energy).
• The direction of the induced current can be determined using the right-hand rule or by applying Lenz’s law.

17. Faraday’s Law Of Induction - Lenz’s Law - Right-Hand Rule

• The right-hand rule can be used to determine the direction of the induced current in a wire loop.
• Point your thumb in the direction of the changing magnetic field.
• Curl your fingers in the direction of the loop.
• The direction in which your fingers curl represents the direction of the induced current.

18. Faraday’s Law Of Induction - Lenz’s Law - Example

• Suppose a bar magnet is approaching a conducting loop from the left side.
• According to Lenz’s law, the induced current will flow in a direction that creates a magnetic field opposing the approach of the bar magnet.
• Using the right-hand rule, you can determine the direction of the induced current and magnetic field.
• This principle is used in electromagnetic braking systems, where the application of magnetic fields opposes the motion of objects.

19. Faraday’s Law Of Induction - Magnetic Flux - Example

• Consider a rectangular loop of wire.
• The loop is placed in a uniform magnetic field B, and the angle between the magnetic field and the normal to the surface of the loop is θ.
• The area of the loop is A.
• The magnetic flux through the loop is given by Φ = B * A * cos(θ).
• Let’s consider an example: Suppose a rectangular loop with an area of 0.1 m^2 is placed in a magnetic field of 0.5 T at an angle of 30 degrees with the normal to the loop. Find the magnetic flux through the loop.

20. Faraday’s Law Of Induction - Magnetic Flux - Example (continued)

• Substituting the given values, we have Φ = 0.5 * 0.1 * cos(30)
• Simplifying, Φ = 0.05 * 0.866
• Calculating this expression will give the value of magnetic flux through the loop.

21. Faraday’s Law Of Induction - Induced emf - Example 2 (Two coaxial solenoids)

• Consider two coaxial solenoids, one inside the other, with radii r1 and r2 (r2 > r1) and lengths l1 and l2 (l2 > l1).
• Let I1 and I2 be the currents flowing through the inner and outer solenoids, respectively.
• Suppose the current in the inner solenoid is changed at a rate di1/dt.
• The magnetic field due to the inner solenoid at the position of the outer solenoid is given by B1 = μ0 * (N1 * I1) / l1.
• The magnetic field due to the outer solenoid at its own position is given by B2 = μ0 * (N2 * I2) / l2.
• According to Faraday’s law, the induced emf in the outer solenoid is given by ε = -N2 * dΦ2/dt, where N2 is the number of turns in the outer solenoid.

22. Faraday’s Law Of Induction - Induced emf - Example 2 (continued)

• The magnetic flux through the outer solenoid is given by Φ2 = B1 * A2, where A2 is the cross-sectional area of the outer solenoid.
• The cross-sectional area of the outer solenoid can be calculated using A2 = π * (r2^2 - r1^2)
• Differentiating Φ2 with respect to time, we get dΦ2/dt = d/dt(B1 * A2)
• Using the chain rule, dΦ2/dt = dB1/dt * A2 + B1 * dA2/dt
• Since r2 is changing with time, we have dA2/dt = 2π * r2 * dr2/dt
• Substituting these values, we get dΦ2/dt = (μ0 * N1 * I1 * dA2/dt) / l1 + (μ0 * N1 * I1 * A2 * dr2/dt) / l1

23. Faraday’s Law Of Induction - Induced emf - Example 2 (continued)

• The induced emf in the outer solenoid is given by ε = -N2 * dΦ2/dt
• Substituting the value of dΦ2/dt, we get ε = -N2 * [(μ0 * N1 * I1 * dA2/dt) / l1 + (μ0 * N1 * I1 * A2 * dr2/dt) / l1]
• Simplifying further, ε = -(μ0 * N1 * N2 * I1 * dA2/dt) / l1 - (μ0 * N1 * N2 * I1 * A2 * dr2/dt) / l1
• Rearranging, ε = -(μ0 * N1 * N2 * I1 / l1) * [dA2/dt + (A2 * dr2/dt)]

24. Faraday’s Law Of Induction - Induced emf - Example 2 (continued)

• Suppose the current in the inner solenoid is changing at a constant rate di1/dt = 100 A/s.
• The area of the outer solenoid is A2 = π * (0.02^2 - 0.01^2) m^2.
• The radius of the outer solenoid is changing at a constant rate dr2/dt = 0.001 m/s.
• Let N1 = 1000, N2 = 2000, l1 = 0.1 m.
• Substituting these values, we get ε = -(4π * 10^-7 * 1000 * 2000 * 100 / 0.1) * [(π * (0.02^2 - 0.01^2) * 0.001) + (π * (0.02^2 - 0.01^2) * 0.001)]
• Simplifying this expression will give the value of induced emf in the outer solenoid.

25. Faraday’s Law Of Induction - Lenz’s Law - Concept

• Lenz’s law states that the induced current in a circuit will always flow in such a direction as to oppose the change in magnetic field that produced it.
• This law is based on the principle of conservation of energy.
• Lenz’s law ensures that when a magnetic field is changing, the work done by the induced current is always negative (i.e., it is dissipated as heat or stored as potential energy).
• The direction of the induced current can be determined using the right-hand rule or by applying Lenz’s law.

26. Faraday’s Law Of Induction - Lenz’s Law - Right-Hand Rule

• The right-hand rule can be used to determine the direction of the induced current in a wire loop.
• Point your thumb in the direction of the changing magnetic field.
• Curl your fingers in the direction of the loop.
• The direction in which your fingers curl represents the direction of the induced current.

27. Faraday’s Law Of Induction - Lenz’s Law - Example

• Suppose a bar magnet is approaching a conducting loop from the left side.
• According to Lenz’s law, the induced current will flow in a direction that creates a magnetic field opposing the approach of the bar magnet.
• Using the right-hand rule, you can determine the direction of the induced current and magnetic field.
• This principle is used in electromagnetic braking systems, where the application of magnetic fields opposes the motion of objects.

28. Faraday’s Law Of Induction - Magnetic Flux - Example

• Consider a rectangular loop of wire.
• The loop is placed in a uniform magnetic field B, and the angle between the magnetic field and the normal to the surface of the loop is θ.
• The area of the loop is A.
• The magnetic flux through the loop is given by Φ = B * A * cos(θ).
• Let’s consider an example: Suppose a rectangular loop with an area of 0.1 m^2 is placed in a magnetic field of 0.5 T at an angle of 30 degrees with the normal to the loop. Find the magnetic flux through the loop.

29. Faraday’s Law Of Induction - Magnetic Flux - Example (continued)

• Substituting the given values, we have Φ = 0.5 * 0.1 * cos(30)
• Simplifying, Φ = 0.05 * 0.866
• Calculating this expression will give the value of magnetic flux through the loop.

30. Summary

• Faraday’s law of induction states that a change in the magnetic field induces an electromotive force (emf) in a closed loop wire.
• The induced emf in a circuit is directly proportional to the rate of change of magnetic flux through the loop.
• Lenz’s law states that the induced current always flows in a direction that opposes the change in magnetic field that caused it.