Induced Voltages and Inductance (Chapter 20)

20.1 Induced emf and Magnetic Flux

  • Faraday’s Experiment Components:
      - Primary Circuit: Consists of a primary coil connected directly to a battery.
      - Secondary Circuit: Consists of a secondary coil connected to an ammeter. There is no battery in the secondary circuit.

  • Experimental Results:
      - When the switch in the primary circuit is closed, the ammeter in the secondary circuit reads a current and then returns to zero.
      - When the switch in the primary circuit is opened again, the ammeter reads a current in the opposite direction and then returns to zero.
      - While there is a steady current in the primary circuit, the ammeter reads zero.

  • Conclusions from Faraday’s Experiment:
      - An electrical current is produced by a changing magnetic field.
      - A steady magnetic field does not produce a current unless the coil is moving.
      - Current in the secondary circuit is instantaneous, occurring only while the magnetic field through the secondary coil is changing.
      - The secondary circuit acts as if a source of emf were connected to it for a short time.
      - This phenomenon is described as an induced emf produced by a changing magnetic field.

  • Definition of Magnetic Flux:
      - The emf is induced by a change in magnetic flux rather than simply a change in the magnetic field.
      - Magnetic flux (ΦB\Phi_B) is a measure of the number of magnetic field lines crossing a given area.
      - It is proportional to the strength of the magnetic field passing through the plane of a loop and the area of the loop.
      - The number of field lines per unit area increases as field strength increases.

  • Mathematical Definition of Magnetic Flux:
      - For a loop of wire with area AA in a uniform magnetic field BB:
      - ΦB=BA=BAcos(θ)\Phi_B = B_{\perp} A = BA \cos(\theta)
      - θ\theta is defined as the angle between the magnetic field and the normal (perpendicular) to the plane of the loop.

  • Units:
      - SI unit: weber (WbWb).
      - Conversion: 1Wb=1Tm21\,Wb = 1\,T\,m^2.

  • Orientation and Flux Magnitude:
      - Maximum Flux: Occurs when field lines are perpendicular to the plane of the loop (θ=0\theta = 0^{\circ}, because cos(0)=1\cos(0^{\circ}) = 1).
        - ΦB,max=BA\Phi_{B,max} = BA
      - Zero Flux: Occurs when field lines are parallel to the plane of the loop (θ=90\theta = 90^{\circ}, because cos(90)=0\cos(90^{\circ}) = 0).
        - ΦB=0\Phi_B = 0
      - Negative Flux: Magnetic flux can be negative if θ=180\theta = 180^{\circ}.

Example 20-1: Magnetic Flux Calculations

  • Scenario: A conducting circular loop with radius r=0.250mr = 0.250\,m is in the xy-plane. A uniform magnetic field B=0.360TB = 0.360\,T points in the positive z-direction (same as the normal to the plane).

  • (a) Calculate the magnetic flux through the loop:
      - Area A=πr2=π(0.250m)2A = \pi r^2 = \pi (0.250\,m)^2
      - ΦB=BAcos(0)=(0.360T)(π)(0.250m)2(1)\Phi_B = BA \cos(0^{\circ}) = (0.360\,T)(\pi)(0.250\,m)^2(1)

  • (b) Loop rotated clockwise around x-axis, normal points at 45.045.0^{\circ} to z-axis:
      - θ=45.0\theta = 45.0^{\circ}
      - ΦB=BAcos(45.0)=(0.360T)(π)(0.250m)2cos(45.0)\Phi_B = BA \cos(45.0^{\circ}) = (0.360\,T)(\pi)(0.250\,m)^2 \cos(45.0^{\circ})

  • (c) Change in flux (ΔΦB\Delta \Phi_B):
      - ΔΦB=ΦB,finalΦB,initial\Delta \Phi_B = \Phi_{B,final} - \Phi_{B,initial}

20.2 Faraday’s Law of Induction and Lenz’s Law

  • Electromagnetic Induction Experiment:
      - Moving a magnet toward a loop: Ammeter reads current in one direction.
      - Holding magnet stationary: Ammeter reads zero current.
      - Moving magnet away from loop: Ammeter reads current in opposite direction.
      - Moving the loop toward or away from a stationary magnet produces the same result.
      - Conclusion: Current is set up (induced current) as long as there is relative motion between the magnet and loop.

  • Faraday’s Law of Induction:
      - An emf is induced in a circuit when the magnetic flux through the circuit changes with time.
      - The instantaneous emf induced equals the negative rate of change of magnetic flux with respect to time.
      - Formula for a circuit with NN loops:
        - ϵ=NΔΦBΔt\epsilon = -N \frac{\Delta \Phi_B}{\Delta t}
      - SI unit: volt (VV).
      - Change in flux (ΔΦB\Delta \Phi_B) can be produced by varying BB, AA, or θ\theta.

  • Lenz’s Law:
      - Determines the polarity/direction of the induced current.
      - The induced current travels in a direction such that it creates a magnetic field whose flux opposes the change in the original flux through the circuit.
      - Two magnetic fields are involved:
        - 1. The external changing magnetic field.
        - 2. The magnetic field produced by the induced current (BindB_{ind}).
      - Applying Lenz's Law:
        - If external flux is increasing: BindB_{ind} points in the opposite direction to the external field.
        - If external flux is decreasing: BindB_{ind} points in the same direction as the external field.
        - If flux is unchanged: Induced current and BindB_{ind} are zero.

  • Right-Hand Rule for Current: Once the direction of BindB_{ind} is known, use the right-hand rule to find the direction of the induced current.

Examples of Lenz's Law Application

  • Magnet toward coil: Induces magnetic field that repels the magnet.

  • Magnet away from coil: Induces magnetic field that attracts the magnet.

  • Scenario: decreasing outward flux: Induced current will be counterclockwise to create outward field lines.

  • Scenario: decreasing inward flux: Induced current will be clockwise to create inward field lines.

  • Scenario: increasing inward flux: Induced current will be counterclockwise to oppose the increase.

  • Scenario: increasing flux to the left: Induced current will be counterclockwise.

Example 20-2: Induced Current in a Square Coil

  • Setup: Coil with N=25N = 25 turns, square cross-section 1.80cm1.80\,cm on a side (A=(0.0180m)2A = (0.0180\,m)^2). Total resistance R=0.350ΩR = 0.350\,\Omega. Magnetic field is perpendicular to the coil.

  • (a) Find induced emf if field changes from 0.00T0.00\,T to 0.500T0.500\,T in 0.800s0.800\,s:
      - ΔΦB=A×(BfBi)=(0.0180m)2×(0.500T0.00T)\Delta \Phi_B = A \times (B_f - B_i) = (0.0180\,m)^2 \times (0.500\,T - 0.00\,T)
      - ϵ=NΔΦBΔt\epsilon = -N \frac{\Delta \Phi_B}{\Delta t}

  • (b) Find magnitude of induced current:
      - I=ϵRI = \frac{|\epsilon|}{R}

  • (c) Find direction of induced current:
      - Since BB is increasing, the induced field must oppose it.

Applications of Faraday’s Law

  • Ground Fault Interrupters (GFI): Safety devices protecting against electrical shock by detecting changes in magnetic flux caused by current imbalances.

  • Electric Guitar Pickups: A vibrating string induces a voltage in a pickup coil.

  • Apnea Monitors: Used for infants to alert caregivers if breathing (which effects flux in a coil) stops.

20.3 Motional emf

  • Definition: The potential difference produced in a straight conductor of length ll moving with velocity vv through a uniform magnetic field BB (where vBv \perp B).

  • Mechanism:
      - Electrons in the conductor experience a magnetic force Fm=qvBF_m = qvB directed toward one end (e.g., downward).
      - Charge separation occurs: electrons accumulate at one end (negative), while the other becomes positive.
      - This separation creates an internal electric field (EE).
      - Equilibrium is reached when electric force qEqE balances magnetic force qvBqvB.

  • Mathematical Relationships:
      - The potential difference is ΔV=El=Blv\Delta V = El = Blv.
      - If the moving bar is part of a closed circuit with resistance RR:
        - ϵ=Blv\epsilon = Blv
        - Induced current I=ϵR=BlvRI = \frac{\epsilon}{R} = \frac{Blv}{R}

  • Conservation of Energy:
      - Moving the bar through the field requires an applied force to overcome the magnetic force on the induced current.
      - Power delivered to the resistor: P=I2RP = I^2 R.
      - Work done by the applied force equals the energy dissipated in the resistor.

Example 20-4: Sliding Bar Motional emf

  • Data: l=0.500ml = 0.500\,m, v=2.00m/sv = 2.00\,m/s, B=0.250TB = 0.250\,T, R=0.500ΩR = 0.500\,\Omega.

  • (a) Induced voltage: ϵ=Blv=(0.250T)(0.500m)(2.00m/s)=0.250V\epsilon = Blv = (0.250\,T)(0.500\,m)(2.00\,m/s) = 0.250\,V.

  • (b) Current and Power:
      - I=ϵR=0.250V0.500Ω=0.500AI = \frac{\epsilon}{R} = \frac{0.250\,V}{0.500\,\Omega} = 0.500\,A.
      - P=I2R=(0.500A)2(0.500Ω)=0.125WP = I^2 R = (0.500\,A)^2(0.500\,\Omega) = 0.125\,W.

  • (c) Magnetic force on bar:
      - Fm=BIl=(0.250T)(0.500A)(0.500m)=0.0625NF_m = BIl = (0.250\,T)(0.500\,A)(0.500\,m) = 0.0625\,N.

  • (d) Applied force via power:
      - P=Fappv    Fapp=Pv=0.125W2.00m/s=6.25×102NP = F_{app} v \implies F_{app} = \frac{P}{v} = \frac{0.125\,W}{2.00\,m/s} = 6.25 \times 10^{-2}\,N.

20.4 Generators and Motors

  • AC (Alternating Current) Generators:
      - Convert mechanical energy to electrical energy.
      - Construction: A wire loop rotated in a magnetic field. Heat sources (coal) or falling water (hydroelectric) turn turbines to rotate the loop.
      - Components: Slip rings (rotate with the loop) and stationary brushes (connect to external circuit).
      - Operation: As the loop rotates, flux changes.
        - ϵ=ϵmax\epsilon = \epsilon_{max} when the loop plane is parallel to BB.
        - ϵ=0\epsilon = 0 when the loop plane is perpendicular to BB.
      - Formula: ϵ=NBAωsin(ωt)\epsilon = NBA \omega \sin(\omega t), where ω\omega is constant angular speed.

  • DC (Direct Current) Generators:
      - Essentially the same components as AC generators.
      - Major Difference: Uses a split ring or commutator instead of slip rings.
      - Output: Pulsating DC current (always same polarity). To produce steady DC, multiple loops and commutators are used to superimpose multiple outputs.

  • Motors:
      - Convert electrical energy into mechanical energy (generator run in reverse).
      - Back emf: An induced emf that tends to reduce the applied current as the motor coil rotates.
      - Initially, when a motor is turned on, there is no back emf, leading to a very large starting current.
      - As rotation speed increases, back emf opposes the applied voltage, reducing the current.

20.5 Self-Inductance

  • Definition: Self-induction occurs when the changing flux through a circuit arises from the circuit itself (due to its own changing current).

  • Mechanism: As current increases, the magnetic flux it produces also increases. This induces an emf that opposes the change, resulting in a gradual rather than instantaneous current increase.

  • Formula for Self-Induced emf:
      - ϵ=LΔIΔt\epsilon = -L \frac{\Delta I}{\Delta t}
      - LL is the proportionality constant called self-inductance.
      - SI unit: henry (HH). 1H=1Vs/A1\,H = 1\,V\,s/A.

  • Inductance Formula for a Solenoid:
      - L=NΔΦBΔIL = \frac{N \Delta \Phi_B}{\Delta I}
      - L=μ0N2AlL = \frac{\mu_0 N^2 A}{l}, where NN is turns, AA is area, and ll is length.

  • Example 20-7: Solenoid Inductance:
      - Given: N=300N = 300, l=25.0cml = 25.0\,cm, A=4.00×104m2A = 4.00 \times 10^{-4}\,m^2.
      - ϵ\epsilon calculation if II decreases at 50.0A/s50.0\,A/s (ΔIΔt=50.0A/s\frac{\Delta I}{\Delta t} = -50.0\,A/s).

20.6 RL Circuits

  • Inductors: Circuit elements with high inductance, represented by a coil symbol.

  • Inductanc e as Opposition: While resistance RR opposes current magnitude, inductance LL opposes the rate of change of current.

  • Current Behavior: Current does not reach the maximum value ϵ/R\epsilon / R instantaneously.

  • Time Constant ((\tau)):
      - τ=LR\tau = \frac{L}{R}
      - The time required for the current to reach 63.2%63.2\% of its final value (ϵ/R\epsilon / R).

  • Current Equation:
      - I=ϵR(1et/τ)I = \frac{\epsilon}{R} (1 - e^{-t / \tau})

  • Steady State: When current reaches its maximum, the rate of change is zero, and back emf is zero.

20.7 Energy Stored in a Magnetic Field

  • Concept: A battery must do work to establish a current in an inductor because the induced emf opposes the setup.

  • Magnetic Energy: The work done is stored as energy in the inductor’s magnetic field.

  • Formula:
      - PEL=12LI2PE_L = \frac{1}{2} L I^2