Inductance

  • Definition: Inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it.
  • Characteristics:
    • It affects how current and magnetic fields interact within circuits.
    • It can be understood in two main contexts: self-inductance and mutual inductance.

Key Concepts of Inductance

Sequence of Events

  1. An electric current sets up a magnetic field $B$.
  2. Any change in current will also change $B$.
  3. Changes in $B$ will also change the magnetic flux.
  4. A change in flux induces electromotive force (emf), as stated by Faraday’s law of induction.
  5. The induced emf opposes the change in magnetic flux.
  6. According to Lenz’s law, the induced flux tries to oppose the change in current.

Mutual Inductance

  • Definition: Mutual inductance ($M$) occurs when a changing current in one circuit generates a changing magnetic field that induces an emf in another circuit.
  • Significance: It allows energy transfer between circuits without direct electrical connection.

Self-Inductance

  • Definition: Self-inductance ($L$) is the induction of emf within a circuit due to a change in its own current.
  • Mechanism: A changing current induces a changing magnetic field which can affect the current that originally produced it.
  • Implication: Self-inductance serves as a crucial property for inductors, aiding in energy storage and conversion.

Understanding Inductance

  • Conceptualization: Inductance quantifies how effectively an emf is induced in a device due to changes in its own current.
  • Explanation of Effects:
    • An increase in current ($I$) generates an emf that leads to current flowing in the opposite direction.
    • Conversely, a decrease in $I$ generates an emf that encourages current flow in the same direction as the original current.

Inductors

  • Definition of an Inductor: A circuit element specifically designed to provide self-inductance.
  • Construction: Typically consists of circular loops or coils of conductive wire insulated to prevent short circuits.
  • Ferrite Core Inductor:
    • Enhances the magnetic field similar to how a dielectric material enhances electric fields in capacitors.
  • Classification: Presented as low-value inductors.

Magnetic Field and Flux Relationship

  • The strength of the magnetic field $B$ due to a current loop is proportional to the current $I$:
    B=extμ0I2RB = \frac{ ext{μ}_{0}I}{2R}
  • The magnetic flux ($Φ{B}$) linked to the coil depends on $B$: Φ</em>BextproportionaltoIΦ</em>{B} ext{ proportional to } I
  • Rewriting the relationship gives:
    ΦB=LIΦ_{B} = LI
  • Where $L$, the self-inductance, is dependent on the geometry of the inductor.

Energy Stored in Inductor

  • Concept: Inductors store magnetic energy, akin to how capacitors store electric energy.
  • Energy Calculation:
    U=12LI2U = \frac{1}{2} LI^2
  • This formula determines the energy stored using the inductor's self-inductance $L$ and the current $I$ through it.

Induced EMF Expressions

Induced EMF in Inductor

  • The induced emf ($ ext{ℰ}$) due to a change in current is mathematically expressed as:
    extE=−LextΔIextΔtext{ℰ} = -L \frac{ ext{Δ}I}{ ext{Δ}t}
  • Units:
    • Two unit definitions yielding henry (H):
    1. Weber/amp (from $Φ_{B} = LI$)
    2. Volt-s/amp = ohm-s.
  • Conversion:
    • 1 H is a large unit; common units include milli-henries (mH) and micro-henries (µH).

Multiple Turns in Inductor

  • For a loop with $N$ turns:
    NΦB=LINΦ_{B} = LI
  • Rearranging from Faraday's Law gives:
    extE=−NextΔΦ<em>BextΔt=−extΔ(NΦ</em>B)extΔt=−LextΔIextΔtext{ℰ} = -N \frac{ ext{Δ}Φ<em>{B}}{ ext{Δ}t} = - \frac{ ext{Δ}(NΦ</em>{B})}{ ext{Δ}t} = -L \frac{ ext{Δ}I}{ ext{Δ}t}
  • Rearranged for self-inductance:
    L=extEextΔtextΔIL = \frac{ ext{ℰ} ext{Δ}t}{ ext{Δ}I}

Problem Examples

Example Problem 1

  • Problem: Find the induced emf in an inductor (where $L = 1 H$) if a current of 10 A drops to zero in 1.0 ms.
  • Answer: 10000extV10000 ext{ V}

Example Problem 2

  • Problem: An induced emf of 2.0 V is measured across a coil with 50 closely wound turns as the current increases from 0.0 to 5.0 A in 0.10 s. (a) What is the self-inductance of the coil? (b) Assuming current is at 5.0 A, what is the flux through each turn?
    • Solutions:
    • For (a):
      L=EdIdt=2.0extV5.0extA/0.10exts=4.0×10−2extHL = \frac{E}{\frac{dI}{dt}} = \frac{2.0 ext{ V}}{5.0 ext{ A}/0.10 ext{ s}} = 4.0 × 10^{-2} ext{ H}
    • For (b):
      Φm=LIN=(4.0×10−2extH)(5.0extA)50extturns=4.0×10−3extWbΦ_{m} = \frac{LI}{N} = \frac{(4.0 × 10^{-2} ext{ H}) (5.0 ext{ A})}{50 ext{ turns}} = 4.0 × 10^{-3} ext{ Wb}

Example Problem 3

  • Problem: A changing current produces an emf of 10 V across a 0.25-H inductor. What is the rate of change of current?
  • Answer: 40extA/s40 ext{ A/s}

Methodology for Determining Self-Inductance

  1. Assume a current $I$ flows through the inductor.
  2. Determine the magnetic field $B$ produced by the current (using Ampère's law if symmetry permits).
  3. Obtain the magnetic flux, $Φ_{m}$.
  4. With the flux known, calculate self-inductance from the equation:
    L=NΦmIL = \frac{NΦ_{m}}{I}

Inductance of a Cylindrical Solenoid

  • Considerations:
  • Properties of a long cylindrical solenoid with length $l$, cross-sectional area $A$, and $N$ turns.
  • Assumptions: The length of the solenoid is significantly larger than its diameter to ignore end effects.
Magnetic Field in a Solenoid
  • For a current $I$:
    B=extμ<em>0NlIB = ext{μ}<em>0 \frac{N}{l} IΦ</em>m=BA=extμ0NlIAΦ</em>{m} = BA = ext{μ}_0 \frac{N}{l} I A
  • Expression for self-inductance of the long solenoid:
    L<em>solenoid=extμ</em>0N2AlL<em>{solenoid} = ext{μ}</em>0 \frac{N^2 A}{l}

Problem Examples for Solenoids

Example Problem 4

(a) Calculate the self-inductance of a tightly wound solenoid (diameter 0.10 cm, cross-sectional area 0.90 cm², length 40 cm).

  • Average solution yields: 4.5×105extH4.5 × 10^5 ext{ H}
    (b) If the current decreases uniformly from 10 to 0 A in 0.10 s, the induced emf across the solenoid is: 4.5×10−3extV4.5 × 10^{-3} ext{ V}

Example Problem 5

(a) Find the magnetic flux through one turn of a solenoid with a self-inductance of 8.0×10−5extH8.0 × 10^{-5} ext{ H} when a current of 3.0 A flows.

  • Answer: 2.4×10−7extWb2.4 × 10^{-7} ext{ Wb}
    (b) The cross-sectional area of the solenoid is calculated to be: 6.4×10−5extm26.4 × 10^{-5} ext{ m}^2

Energy Storage in a Magnetic Field

  • Comparison: Energy stored in a capacitor is in the electric field while an inductor stores energy in its magnetic field.
  • The magnetic energy density formula is given by:
  • u<em>m=B22extμ</em>0u<em>{m} = \frac{B^2}{2 ext{μ}</em>0}
  • Total energy $U$ stored in a solenoid can be derived from the energy density over the solenoid's volume:
    U=12LI2U = \frac{1}{2} LI^2

Example Problem 6

  • Situation: Current of 0.20 A flows through a coil; energy stored in the magnetic field is 6.0 × 10^−3 J.
  • Query: What is the self-inductance of the coil?
  • Solution steps omitted for brevity, requiring appropriate calculations to arrive at an answer.