Example 1: Changing Magnetic Field

Problem: A circular loop of radius 0.05 m is placed in a uniform magnetic field. The field increases from 0.2 T to 0.8 T in 0.5 seconds. What is the induced emf?

Solution:

1. Calculate area: \(A = \pi r^2 = \pi(0.05)^2 = 0.00785 \text{ m}^2\)
2. Calculate field change: \(\Delta B = 0.8 - 0.2 = 0.6 \text{ T}\)
3. Calculate rate of change: \(\frac{dB}{dt} = \frac{0.6}{0.5} = 1.2 \text{ T/s}\)
4. Calculate emf: \(\mathcal{E} = -A\frac{dB}{dt} = -(0.00785)(1.2) = -0.00942 \text{ V}\)

Answer: The induced emf is -9.42 mV.

Example 2: Moving Conductor

Problem: A metal rod of length 0.2 m moves at 5 m/s perpendicular to a 0.4 T magnetic field. What is the induced emf?

Solution:

1. Use motional emf formula: \(\mathcal{E} = BLv\)
2. Substitute values: \(\mathcal{E} = (0.4)(0.2)(5)\)
3. Calculate: \(\mathcal{E} = 0.4 \text{ V}\)

Answer: The induced emf is 0.4 V.

Example 3: Rotating Loop

Problem: A rectangular loop of area 0.01 m² rotates at 120 rad/s in a 0.3 T magnetic field. What is the maximum induced emf?

Solution:

1. Use rotating loop formula: \(\mathcal{E} = BA\omega\sin(\omega t)\)
2. Maximum occurs when \(\sin(\omega t) = \pm 1\)
3. Maximum emf: \(|\mathcal{E}| = BA\omega = (0.3)(0.01)(120)\)
4. Calculate: \(|\mathcal{E}| = 0.36 \text{ V}\)

Answer: The maximum induced emf is 0.36 V.

Example 4: Magnetic Force on Moving Rod

Problem: A metal rod of length 0.2 m moves at 5 m/s perpendicular to a 0.4 T magnetic field. The rod has resistance 0.1 Ω. What is the magnetic force on the rod?

Solution:

1. Calculate induced emf: \(\mathcal{E} = BLv = (0.4)(0.2)(5) = 0.4 \text{ V}\)
2. Calculate induced current: \(I = \frac{\mathcal{E}}{R} = \frac{0.4}{0.1} = 4 \text{ A}\)
3. Calculate magnetic force: \(F = ILB = (4)(0.2)(0.4) = 0.32 \text{ N}\)
4. The force opposes the motion (Lenz's Law)

Answer: The magnetic force is 0.32 N, opposing the motion.

Derivation 1: Motional EMF Formula

Derive: \(\mathcal{E} = BLv\)

Method 1: Using Faraday's Law

1. Consider a conductor moving through a magnetic field
2. Flux change: \(\Delta\Phi_B = B \cdot \Delta A = B \cdot L \cdot \Delta x\)
3. Rate of change: \(\frac{d\Phi_B}{dt} = B \cdot L \cdot \frac{dx}{dt} = BLv\)
4. By Faraday's Law: \(\mathcal{E} = -\frac{d\Phi_B}{dt} = BLv\)

Method 2: Using Lorentz Force

1. Electrons in the conductor experience Lorentz force: \(F = qvB\)
2. This creates a separation of charge along the conductor
3. The electric field created is: \(E = \frac{F}{q} = vB\)
4. The potential difference is: \(\mathcal{E} = EL = vBL\)

Result: \(\mathcal{E} = BLv\) (NOT on equation sheet)

Derivation 2: Rotating Loop Maximum EMF

Derive: Maximum \(\mathcal{E} = BA\omega\)

Step-by-Step:

1. Flux through rotating loop: \(\Phi_B = BA\cos(\omega t)\)
2. Rate of change: \(\frac{d\Phi_B}{dt} = -BA\omega\sin(\omega t)\)
3. By Faraday's Law: \(\mathcal{E} = -\frac{d\Phi_B}{dt} = BA\omega\sin(\omega t)\)
4. Maximum occurs when \(\sin(\omega t) = \pm 1\)
5. Maximum EMF: \(|\mathcal{E}_{max}| = BA\omega\)

Result: Maximum \(\mathcal{E} = BA\omega\) (NOT on equation sheet)

Derivation 3: Magnetic Force on Moving Rod

Derive: \(F = \frac{B^2L^2v}{R}\)

Step-by-Step:

1. Motional EMF: \(\mathcal{E} = BLv\)
2. Induced current: \(I = \frac{\mathcal{E}}{R} = \frac{BLv}{R}\)
3. Magnetic force on current-carrying wire: \(F = ILB\)
4. Substitute: \(F = \frac{BLv}{R} \cdot LB = \frac{B^2L^2v}{R}\)
5. This force opposes the motion (Lenz's Law)

Result: \(F = \frac{B^2L^2v}{R}\) (NOT on equation sheet)

Example: Motional EMF Derivation in Action

Problem: A 0.4 m rod moves at 5 m/s perpendicular to a 0.3 T magnetic field. What is the induced emf?

• \( \mathcal{E} = vBL = (5)(0.3)(0.4) = 0.6 \) V

Answer: The induced emf is 0.6 V.

Derivation: Motional EMF Formula

Derive: \(\mathcal{E} = vBL\)

Step-by-Step Derivation:

1. Magnetic force on charge: \(F = qvB\)
2. This force creates electric field: \(E = \frac{F}{q} = vB\)
3. Electric field creates potential difference: \(\mathcal{E} = EL\)
4. Substitute: \(\mathcal{E} = vBL\)

Result: \(\mathcal{E} = vBL\) (NOT on equation sheet)

Derivation: EMF for Sliding Rod (Faraday's Law)

1. Magnetic flux: \(\Phi_B = B \cdot A = B \cdot (Lx)\), where \(x\) is position
2. \(\frac{d\Phi_B}{dt} = B L \frac{dx}{dt} = B L v\)
3. Faraday's Law: \(\mathcal{E} = -\frac{d\Phi_B}{dt} = -BLv\) (sign from Lenz's Law)
4. Magnitude: \(\mathcal{E} = vBL\)

Example 1: Basic Motional EMF

Problem: A 0.5 m long conductor moves at 10 m/s perpendicular to a 0.2 T magnetic field. What is the induced EMF?

Solution:

1. Use motional EMF formula: \(\mathcal{E} = vBL\)
2. Substitute values: \(\mathcal{E} = (10)(0.2)(0.5)\)
3. Calculate: \(\mathcal{E} = 1.0 \text{ V}\)

Answer: The induced EMF is 1.0 V.

Example 2: Power Generation

Problem: A conductor with resistance 2 Ω moves through a 0.1 T field at 5 m/s. The conductor is 0.3 m long. What current flows?

Solution:

1. Calculate EMF: \(\mathcal{E} = vBL = (5)(0.1)(0.3) = 0.15 \text{ V}\)
2. Use Ohm's Law: \(I = \frac{\mathcal{E}}{R} = \frac{0.15}{2} = 0.075 \text{ A}\)

Answer: The current is 0.075 A.

Example 3: Direction of Current

Problem: A conductor moves upward through a magnetic field pointing into the page. What is the direction of induced current?

Solution:

1. Use right-hand rule: thumb up (velocity), fingers into page (B-field)
2. Palm faces right, so positive charges move right
3. Conventional current flows right
4. Electrons flow left (opposite direction)

Answer: Conventional current flows to the right.