Comprehensive Notes on Magnetic Circuits and Electromagnetic Induction

Fundamentals of Magnetic Circuits

• Definition of a Magnetic Circuit:
  • A magnetic circuit is defined as the closed path followed by magnetic flux lines.
  • When an electric current is passed through a coil wound around a ferromagnetic core (such as an iron core), a magnetic path is established within the coil and core.
  • The charge carriers flowing through a conductor constitute an electric current, which sets up a magnetic flux. The closed path followed by this magnetic flux is called a magnetic circuit.
  • In a magnetic circuit, magnetic lines of force leave the North pole, pass through the entire circuit, and return to the starting point.
  • Magnetic circuits consist of materials having high permeability (such as iron, soft steel, etc.), which offer very small opposition to the flow of magnetic flux.

Magnetic Quantities and Definitions

• 1. Magnetic Flux ($\phi$):

  • Definition: The total magnetic lines of force produced by a magnet or a current-carrying coil.
  • Symbol: $\phi$
  • Unit: Weber ($\text{Wb}$)

• 2. Magnetic Flux Density ($B$):

  • Definition: The flux per unit area at right angles to the direction of the flux.
  • Formula:$B = \frac{\phi}{A}$
  • Unit: Weber per square meter ($\text{Wb/m}^2$) or Tesla ($\text{T}$)

• 3. Magnetomotive Force (MMF):

  • Definition: The driving cause that produces magnetic flux in a magnetic circuit.
  • Derivation: Obtained as the product of the current flowing through a coil and the number of turns ($N$) in the coil.
  • Formula:$\text{MMF} = N \times I$
  • Unit: Ampere-turns ($\text{AT}$)

• 4. Magnetic Field Intensity ($H$) or Magnetising Force:

  • Definition: Defined as the MMF per unit length of the magnetic flux path. It is a measure of the ability of a magnetised body to produce magnetic induction in other magnetic substances.
  • Formula:$H = \frac{\text{MMF}}{l} = \frac{NI}{l}$
  • Unit: Ampere-turns per meter ($\text{AT/m}$)

• 5. Reluctance ($S$):

  • Definition: The opposition that the magnetic circuit offers to the flow of magnetic flux. It is defined as the ratio of MMF to magnetic flux.
  • Formula:$S = \frac{\text{MMF}}{\phi} = \frac{NI}{\phi} = \frac{l}{\mu_0 \mu_r A}$
  • Unit: Ampere-turns per Weber ($\text{AT/Wb}$)

• 6. Permeance ($P$):

  • Definition: The reciprocal of reluctance. It represents the readiness with which magnetic flux is developed in a circuit.
  • Formula:$P = \frac{1}{S} = \frac{\phi}{\text{MMF}}$

• 7. Permeability ($\mu$):

  • Definition: The measure of opposition/resistance of a material against the formation of a magnetic field, or simply, the conductivity of a material for magnetic flux.
  • Relationship: The flux density ($B$) is directly proportional to the magnetising force ($H$) that produces it:         $B \propto H \implies B = \mu H = \mu_0 \mu_r H$

• 8. Absolute Permeability of Free Space ($\mu_0$):

  • Definition: The absolute permeability of vacuum or free space.
  • Constant Value:$\mu_0 = 4\pi \times 10^{-7}\,\text{H/m}$

• 9. Relative Permeability ($\mu_r$):

  • Definition: The ratio of the flux density produced in a medium or material to the flux density produced in vacuum by the same magnetising force.
  • Formula:$\mu_r = \frac{\text{Flux density in medium}}{\text{Flux density in vacuum}} = \frac{B}{B_0}$

Analysis of Simple and Composite Magnetic Circuits

• Simple Magnetic Circuit:

  • Consider a circular solenoid or toroidal coil having a mean magnetic path length of $l$\,meters, cross-sectional area of $A$\,$\text{m}^2$, mean core radius $R_m$, and a coil of $N$\,turns carrying $I$\,amperes.
  • Magnetic Field Intensity:         $H = \frac{NI}{l} = \frac{NI}{2\pi R_m}$
  • Magnetic Flux Density:         $B = \mu_0 \mu_r H = \mu_0 \mu_r \frac{NI}{2\pi R_m}$
  • Total Magnetic Flux:         $\phi = B \times A = \mu_0 \mu_r \frac{NI}{2\pi R_m} A = \frac{NI}{\frac{l}{\mu_0 \mu_r A}}$
  • Reluctance ($S$) of the Circuit:         $S = \frac{l}{\mu_0 \mu_r A}$
  • The reluctance $S$ in a magnetic circuit is directly analogous to resistance ($R$) in an electric circuit.

• Composite Magnetic Circuit (Series Circuit):

  • Consists of multiple different magnetic materials having different relative permeabilities ($\mu_{r1}, \mu_{r2}, \mu_{r3}$) along with an air gap ($\mu_{rg} = 1$).
  • Flux lines find a single closed path through all the different magnetic materials and the air gap. Hence, a single flux ($\phi$) exists throughout, making it a series magnetic circuit.
  • Each section offers reluctance depending on its dimensions ($l_1, l_2, l_3, l_g$), cross-sectional areas ($A_1, A_2, A_3, A_g$), and relative permeabilities.
  • Total Reluctance ($S$\,or\,$S_T$):         $S = S_1 + S_2 + S_3 + S_g$$S = \frac{l_1}{\mu_0 \mu_{r1} A_1} + \frac{l_2}{\mu_0 \mu_{r2} A_2} + \frac{l_3}{\mu_0 \mu_{r3} A_3} + \frac{l_g}{\mu_0 A_g}$
  • Total MMF:         $\text{Total MMF} = \phi \times S = \phi (S_1 + S_2 + S_3 + S_g)$$\text{Total MMF} = H_1 l_1 + H_2 l_2 + H_3 l_3 + H_g l_g$

• Parallel Magnetic Circuit:

  • A magnetic circuit in which the flux has more than one path is known as a parallel magnetic circuit.
  • The total flux ($\phi$) exists in a common central section containing the exciting coil. It then divides into two or more branch paths ($\phi_1, \phi_2$), follows the different paths, and recombines at the other end of the common section:         $\phi = \phi_1 + \phi_2$
  • Let $S_1$ be the reluctance of branch path $ABCDA$, $S_2$ be the reluctance of branch path $AEFDA$, and $S_c$ be the reluctance of common central path $AD$
  • The branch paths $ABCDA$ and $AEFDA$ are in parallel.
  • In a parallel magnetic circuit, the MMF required for the parallel paths is equal to the MMF required for any one of the parallel paths:         $\text{Total MMF} = \text{MMF for Central Path } AD + \text{MMF for Path } ABCDA \text{ (or } AEFDA \text{)}$$\text{Total MMF} = \phi S_c + \phi_1 S_1 = \phi S_c + \phi_2 S_2$

Analogy Between Magnetic and Electric Circuits

• Comparison Table:
  • Path: Closed path for magnetic flux vs. Closed path for electric current.
  • Flow Characteristic: Magnetic flux does not actually "flow" (it is set up) vs. Electric current actually flows in a conductor.
  • Driving Cause: Magnetomotive Force (MMF, unit: Ampere-turns) vs. Electromotive Force (EMF, unit: Volts).
  • Quantity: Flux $\phi$ (unit: Weber, $\text{Wb}$) vs. Current $I$ (unit: Ampere, $\text{A}$).
  • Formula for Quantity: $\phi = \frac{\text{MMF}}{\text{Reluctance}}$ vs. $I = \frac{\text{EMF}}{\text{Resistance}}$.
  • Opposition: Reluctance $S = \frac{l}{\mu_0 \mu_r A}$ opposes flux vs. Resistance $R = \frac{\rho l}{A}$ opposes current.
  • Property: Permeance $P = \frac{1}{\text{Reluctance}}$ vs. Conductance $G = \frac{1}{\text{Resistance}}$.
  • Density: Flux density $B = \frac{\phi}{A}$ vs. Current density $J = \frac{I}{A}$.
  • Intensity: Magnetic field intensity $H$ vs. Electric field intensity $E$
  • Directional Flow: Flux lines flow from North pole to South pole vs. Current flows from Positive to Negative terminal.

Magnetic Leakage and Fringing

• Magnetic Leakage (Leakage Flux):

  • The part of the magnetic flux that does not follow the intended/desired path in a magnetic circuit is known as leakage flux ($\phi_l$).
  • In practical magnetic circuits with an iron core and air gap, a large part of the flux passes through the iron core and the air gap (termed Useful Flux, $\phi_u$), while some flux leaks through the air surrounding the core.
  • Total Magnetic Flux produced by coil:         $\phi_T = \phi_u + \phi_l$
  • Leakage Coefficient (Leakage Factor, $\lambda$):$\lambda = \frac{\text{Total Magnetic Flux}}{\text{Useful Magnetic Flux}} = \frac{\phi_T}{\phi_u}$
  • The value of leakage coefficient $\lambda$ for electrical machines usually ranges from $1.15$ to $1.25$

• Magnetic Fringing:

  • When magnetic lines of force pass through an air gap between magnetic poles, they tend to bulge outward.
  • This bulging occurs because magnetic field lines repel each other when passing through air.
  • This effect is known as magnetic fringing, and it effectively increases the cross-sectional area of the air gap.

Electromagnetic Induction, Laws, and Hand Rules

• Faraday's Laws of Electromagnetic Induction:

  • First Law: Whenever a conductor cuts magnetic flux, or whenever the magnetic flux linking a circuit changes, an EMF is induced in the conductor/circuit.
  • Second Law: The magnitude of the induced EMF is directly proportional to the rate of change of flux linkages.
  • Mathematical Expression:$e = N \frac{d\phi}{dt}$         where $e$ is induced EMF, $N$ is number of turns, and $\frac{d\phi}{dt}$ is rate of change of flux.

• Lenz's Law:

  • The direction of an induced EMF is such that it opposes the very cause producing it.
  • Mathematical Expression:$e = -N \frac{d\phi}{dt}$

• Fleming's Left-Hand Rule:

  • Purpose: Determines the direction of force acting on a current-carrying conductor placed in a magnetic field.
  • Rule: Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular to one another ($90^\circ$).
    • Forefinger: Direction of Magnetic Field.
    • Middle Finger: Direction of Current.
    • Thumb: Direction of Motion / Force acting on the conductor.
  • Application: Used in DC Motors.

• Fleming's Right-Hand Rule:

  • Purpose: Determines the direction of induced EMF/current in a conductor moving in a magnetic field.
  • Rule: Stretch the thumb, forefinger, and middle finger of the right hand mutually perpendicular to one another ($90^\circ$).
    • Forefinger: Direction of Magnetic Field.
    • Thumb: Direction of Motion of the conductor.
    • Middle Finger: Direction of Induced EMF / Current.
  • Application: Used in DC Generators.

Types of Induced EMF

• 1. Dynamically Induced EMF:

  • Occurs when the conductor moves within a stationary magnetic field such that the flux linkage changes in magnitude.
  • Example: DC Generator or AC Generator.
  • Derivation:
    • Consider a conductor of length $l$\,meters moving with velocity $v$\,$\text{m/s}$ at right angles to a uniform stationary field of flux density $B$\,$\text{Wb/m}^2$
    • Distance moved in $dt$\,seconds: $dx = v \cdot dt$
    • Area swept by conductor: $dA = l \cdot dx = l \cdot v \cdot dt$
    • Magnetic flux cut: $d\phi = B \cdot dA = B \cdot l \cdot v \cdot dt$
    • Induced EMF ($N=1$):             $e = \frac{d\phi}{dt} = B l v$
    • If the conductor moves at an angle $\theta$ to the magnetic field:             $e = B l v \sin(\theta)$

• 2. Statically Induced EMF:

  • Occurs when the conductor remains stationary while the magnetic field moves or varies with time.
  • Example: Transformer.
  • a. Self-Induced EMF:
    • EMF induced in a coil due to a change in its own magnetic flux linking with itself.
    • Formula:             $e = -N \frac{d\phi}{dt} = -L \frac{dI}{dt}$
    • where Self-Inductance $L$ is defined as:             $L = \frac{N\phi}{I} = \frac{N^2 \mu_0 \mu_r A}{l}$
  • b. Mutually Induced EMF:
    • EMF induced in a secondary coil due to a changing magnetic flux produced by a neighboring primary coil.
    • Let Coil 1 ($N_1$ turns) carry current $I_1$, producing mutual flux $\phi_{12}$ linking Coil 2 ($N_2$ turns).
    • Mutual Inductance $M$:             $M = \frac{N_2 \phi_{12}}{I_1}$
    • Induced EMF in Coil 2:             $e_2 = -N_2 \frac{d\phi_{12}}{dt} = -M \frac{dI_1}{dt}$
    • Similarly, EMF induced in Coil 1 due to current change in Coil 2:             $e_1 = -M \frac{dI_2}{dt}$

Worked Numerical Problems

• Problem 1 (Magnetic Field Intensity):

  • Statement: A solenoid is wound with a coil of $200$\,turns carrying a current of $1.5$\,A. Find the value of magnetic intensity when the length of the coil is $80$\,cm.
  • Given: $N = 200$, $I = 1.5\,\text{A}$, $l = 80\,\text{cm} = 80 \times 10^{-2}\,\text{m}$.
  • Calculation:$H = \frac{NI}{l} = \frac{200 \times 1.5}{80 \times 10^{-2}} = 375\,\text{AT/m}$

• Problem 2 (Magnetic Flux Density):

  • Statement: A magnet with a square cross-section of side $5$\,cm produces a magnetic flux of $0.3$\,mWb. Find the magnetic flux density.
  • Given: $\phi = 0.3\,\text{mWb} = 0.3 \times 10^{-3}\,\text{Wb}$, Area $A = 5\,\text{cm} \times 5\,\text{cm} = 25 \times 10^{-4}\,\text{m}^2$
  • Calculation:$B = \frac{\phi}{A} = \frac{0.3 \times 10^{-3}}{25 \times 10^{-4}} = 0.12\,\text{Wb/m}^2$

• Problem 3 (Relative Permeability):

  • Statement: A current of $2$\,A passes through a coil of $350$\,turns wound on an iron ring of mean diameter $12$\,cm. The flux density established is $1.4$\,$\text{Wb/m}^2$. Find the relative permeability of iron.
  • Given: $I = 2\,\text{A}$, $N = 350$, Mean diameter $D = 12\,\text{cm} \implies R_m = 6\,\text{cm} = 6 \times 10^{-2}\,\text{m}$, $B = 1.4\,\text{Wb/m}^2$, $\mu_0 = 4\pi \times 10^{-7}\,\text{H/m}$.
  • Calculation:$l = 2\pi R_m = 2\pi \times 6 \times 10^{-2} = 12\pi \times 10^{-2}\,\text{m}$$B = \mu_0 \mu_r \frac{NI}{l} \implies \mu_r = \frac{B \cdot l}{\mu_0 N I}$$\mu_r = \frac{1.4 \times 12\pi \times 10^{-2}}{4\pi \times 10^{-7} \times 350 \times 2} = 600$

• Problem 4 (Required Current for Mild Steel Ring):

  • Statement: A mild steel ring of mean circumference $60$\,cm and cross-sectional area $4$\,$\text{cm}^2$ is wound with $300$\,turns. Calculate the current required to produce a flux of $500$\,$\mu\text{Wb}$, given $\mu_r = 400$
  • Given: Mean length $l = 60\,\text{cm} = 60 \times 10^{-2}\,\text{m}$, Area $A = 4\,\text{cm}^2 = 4 \times 10^{-4}\,\text{m}^2$, $N = 300$, $\phi = 500 \times 10^{-6}\,\text{Wb}$, $\mu_r = 400$
  • Calculation:$\phi = \frac{NI \mu_0 \mu_r A}{l} \implies I = \frac{\phi \cdot l}{N \mu_0 \mu_r A}$$I = \frac{500 \times 10^{-6} \times 60 \times 10^{-2}}{300 \times 4\pi \times 10^{-7} \times 400 \times 4 \times 10^{-4}} = 7.4604\,\text{A}$

• Problem 5 (Toroidal Air-Cored Coil Analysis):

  • Statement: A toroidal air-cored coil ($\mu_r = 1$) with $2000$\,turns has a mean radius of $25$\,cm and a turn diameter of $6$\,cm. Current $I = 10$\,A. Find (a) MMF, (b) Flux, (c) Flux density.
  • Given: $N = 2000$, $I = 10\,\text{A}$, $R_m = 25\,\text{cm} = 25 \times 10^{-2}\,\text{m}$, turn radius $r = 3\,\text{cm} = 3 \times 10^{-2}\,\text{m}$.
  • Calculation:
    • (a) MMF:             $\text{MMF} = NI = 2000 \times 10 = 20000\,\text{AT}$
    • (b) Reluctance and Flux:             $l = 2\pi R_m = 2\pi \times 25 \times 10^{-2}\,\text{m}$$A = \pi r^2 = \pi (3 \times 10^{-2})^2\,\text{m}^2$$S = \frac{l}{\mu_0 \mu_r A} = \frac{2\pi \times 25 \times 10^{-2}}{\pi (3 \times 10^{-2})^2 \times 4\pi \times 10^{-7} \times 1} = 4.46 \times 10^8\,\text{AT/Wb}$$\phi = \frac{\text{MMF}}{S} = \frac{20000}{4.46 \times 10^8} = 4.48 \times 10^{-5}\,\text{Wb}$
    • (c) Flux Density:             $B = \frac{\phi}{A} = \frac{4.48 \times 10^{-5}}{\pi (3 \times 10^{-2})^2} = 1.6 \times 10^{-2}\,\text{Wb/m}^2$

• Problem 6 (Air Gap Calculations):

  • Statement: Flux between two electromagnetic poles is $5 \times 10^{-2}$\,Wb across air gap area $0.2$\,$\text{m}^2$ and length $1.2$\,cm. Find (a) flux density, (b) field intensity, (c) reluctance, and (d) MMF drop.
  • Given: $\phi = 5 \times 10^{-2}\,\text{Wb}$, $A = 0.2\,\text{m}^2$, $l_g = 1.2 \times 10^{-2}\,\text{m}$, $\mu_r = 1$
  • Calculation:
    • (a) $B = \frac{\phi}{A} = \frac{5 \times 10^{-2}}{0.2} = 0.25\,\text{Wb/m}^2$
    • (b) $H = \frac{B}{\mu_0 \mu_r} = \frac{0.25}{4\pi \times 10^{-7} \times 1} = 1.9894 \times 10^5\,\text{A/m}$
    • (c) $S_g = \frac{l_g}{\mu_0 A} = \frac{1.2 \times 10^{-2}}{0.2 \times 4\pi \times 10^{-7}} = 47746.4\,\text{AT/Wb}$
    • (d) $\text{MMF} = \phi \times S_g = 5 \times 10^{-2} \times 47746.4 = 2387\,\text{AT}$

• Problem 7 (Iron Ring with Radial Air Gap):

  • Statement: An iron ring (mean diameter $15$\,cm, area $1.7$\,$\text{cm}^2$) has a radial air gap of $0.5$\,mm. Wound with $1500$\,turns, a current of $1$\,A produces a flux of $0.1$\,mWb. Calculate the relative permeability of iron (assuming no leakage).
  • Given: $\phi = 0.1 \times 10^{-3}\,\text{Wb} = 10^{-4}\,\text{Wb}$, $A = 1.7 \times 10^{-4}\,\text{m}^2$, $D = 15\,\text{cm} \implies R_m = 7.5\,\text{cm}$, $l_g = 0.5 \times 10^{-3}\,\text{m}$, $N = 1500$, $I = 1\,\text{A}$.
  • Calculation:
    • Total length $l = 2\pi R_m = 15\pi \times 10^{-2}\,\text{m}$.
    • Iron path length $l_{\text{iron}} = 15\pi \times 10^{-2} - 0.5 \times 10^{-3} = 0.47074\,\text{m}$.
    • $B = \frac{\phi}{A} = \frac{10^{-4}}{1.7 \times 10^{-4}} = 0.5882\,\text{T}$.
    • For air gap: $H_g = \frac{B}{\mu_0} = \frac{0.5882}{4\pi \times 10^{-7}} = 4.681 \times 10^5\,\text{A/m}$.
    • $\text{AT}_{\text{air gap}} = H_g l_g = 4.681 \times 10^5 \times 0.5 \times 10^{-3} = 234.1\,\text{AT}$.
    • Total AT available: $NI = 1500 \times 1 = 1500\,\text{AT}$.
    • AT for iron path: $\text{AT}_{\text{iron}} = 1500 - 234.1 = 1265.9\,\text{AT}$.
    • $H_{\text{iron}} = \frac{\text{AT}_{\text{iron}}}{l_{\text{iron}}} = \frac{1265.9}{0.47074} = 2689.17\,\text{A/m}$.
    • $\mu_r = \frac{B}{\mu_0 H_{\text{iron}}} = \frac{0.5882}{4\pi \times 10^{-7} \times 2689.17} = 174$

• Problem 8 (Ring Connected to Voltage Supply):

  • Statement: An iron ring ($A = 400$\,$\text{mm}^2$, mean diameter $20$\,cm) is wound with $500$\,turns. Given $\mu_r = 250$, coil resistance $R = 480$\,$\Omega$, and supply voltage $V = 240$\,V, find the total flux.
  • Given: $A = 400 \times 10^{-6}\,\text{m}^2$, $D = 20\,\text{cm} \implies R_m = 10\,\text{cm}$, $N = 500$, $\mu_r = 250$, $R = 480\,\Omega$, $V = 240\,\text{V}$.
  • Calculation:
    • $I = \frac{V}{R} = \frac{240}{480} = 0.5\,\text{A}$.
    • $l = 2\pi R_m = 2\pi \times 10 \times 10^{-2} = 0.62831\,\text{m}$.
    • $H = \frac{NI}{l} = \frac{500 \times 0.5}{0.62831} = 397.887\,\text{AT/m}$.
    • $B = \mu_0 \mu_r H = 4\pi \times 10^{-7} \times 250 \times 397.887 = 0.125\,\text{Wb/m}^2$.
    • $\phi = B \times A = 0.125 \times 400 \times 10^{-6} = 50 \times 10^{-6}\,\text{Wb} = 50\,\mu\text{Wb}$.

• Problem 9 (Ring Constructed from Metal Bar):

  • Statement: An iron ring prepared from a $10$\,mm bar has an average diameter of $8.3$\,cm and $598$\,turns. Current $I = 0.975$\,A, $\mu_r = 875$. Find the flux.
  • Given: Bar radius $r = 5\,\text{mm} = 5 \times 10^{-3}\,\text{m}$, ring mean radius $R_m = 4.15\,\text{cm} = 4.15 \times 10^{-2}\,\text{m}$, $N = 598$, $I = 0.975\,\text{A}$, $\mu_r = 875$
  • Calculation:
    • $A = \pi r^2 = \pi (5 \times 10^{-3})^2 = 7.85 \times 10^{-5}\,\text{m}^2$.
    • $l = 2\pi R_m = 2\pi \times 4.15 \times 10^{-2} = 0.2607\,\text{m}$.
    • $\phi = \frac{NI \mu_0 \mu_r A}{l} = \frac{598 \times 0.975 \times 4\pi \times 10^{-7} \times 875 \times 7.85 \times 10^{-5}}{0.2607} = 0.193\,\text{mWb}$.

• Problem 10 (EMF Induced by Changing Flux Density):

  • Statement: A $100$-turn coil has an average diameter of $10$\,mm. Find induced EMF when flux density $B$ changes from $5$\,$\text{mWb/m}^2$ to $0.1$\,$\text{Wb/m}^2$ in $35$\,ns.
  • Given: $N = 100$, coil radius $r = 5\,\text{mm} = 5 \times 10^{-3}\,\text{m}$, $dt = 35 \times 10^{-9}\,\text{s}$, $B_1 = 5 \times 10^{-3}\,\text{Wb/m}^2$, $B_2 = 0.1\,\text{Wb/m}^2$
  • Calculation:
    • $A = \pi r^2 = \pi (5 \times 10^{-3})^2 = 78.5 \times 10^{-6}\,\text{m}^2$.
    • $\phi_1 = B_1 A = 5 \times 10^{-3} \times 78.5 \times 10^{-6} = 0.3925 \times 10^{-6}\,\text{Wb}$.
    • $\phi_2 = B_2 A = 0.1 \times 78.5 \times 10^{-6} = 7.85 \times 10^{-6}\,\text{Wb}$.
    • $d\phi = \phi_2 - \phi_1 = 7.4575 \times 10^{-6}\,\text{Wb}$.
    • $e = N \frac{d\phi}{dt} = 100 \times \frac{7.4575 \times 10^{-6}}{35 \times 10^{-9}} = 21307\,\text{V}$.

• Problem 11 (Inductance and Current Reversal):

  • Statement: Calculate inductance of $400$-turn coil on iron-cored toroidal ring (mean diameter $30$\,cm, circular cross-section diameter $5$\,cm). Also find induced EMF when current of $4$\,A is reversed in $30$\,ms.
  • Given: $N = 400$, $R_m = 15\,\text{cm}$, turn radius $r = 2.5\,\text{cm}$, $dI = 4 - (-4) = 8\,\text{A}$, $dt = 30 \times 10^{-3}\,\text{s}$.
  • Calculation:
    • $l = \pi \times 30 \times 10^{-2} = 0.942\,\text{m}$.
    • $A = \pi (2.5 \times 10^{-2})^2 = 1.963 \times 10^{-4}\,\text{m}^2$.
    • $L = \frac{N^2 \mu_0 \mu_r A}{l} = \frac{400^2 \times 4\pi \times 10^{-7} \times 1 \times 1.963 \times 10^{-4}}{0.942} = 4.1877 \times 10^{-4}\,\text{H}$.
    • $e = L \frac{dI}{dt} = 4.1877 \times 10^{-4} \times \frac{8}{30 \times 10^{-3}} = 0.1117\,\text{V}$.

• Problem 12 (Self-Inductance Rate Calculations):

  • Statement: A coil has self-inductance $L = 30$\,mH. Calculate induced EMF when current: (i) increases at rate of $300$\,A/s, (ii) rises from $0$ to $10$\,A in $0.06$\,s.
  • Given: $L = 30 \times 10^{-3}\,\text{H}$.
  • Calculation:
    • (i) $e = L \frac{dI}{dt} = 30 \times 10^{-3} \times 300 = 9\,\text{V}$.
    • (ii) $dI = 10\,\text{A}, dt = 0.06\,\text{s} \implies e = 30 \times 10^{-3} \times \frac{10}{0.06} = 5\,\text{V}$.

• Problem 13 (Mutual Induction Between Coils):

  • Statement: Two identical coils A and B ($N_1 = N_2 = 1000$) lie in parallel planes such that $60\%$ of flux in coil A links coil B. $10$\,A in coil A produces $\phi_1 = 10^{-4}$\,Wb. Current changes from $+12$\,A to $-12$\,A in $0.02$\,s. Find EMF induced in coil B.
  • Given: $N_1 = N_2 = 1000$, $I_1 = 10\,\text{A}$, $\phi_1 = 10^{-4}\,\text{Wb}$, $\phi_{12} = 0.6 \times 10^{-4}\,\text{Wb}$, $\frac{dI_1}{dt} = \frac{12 - (-12)}{0.02} = 1200\,\text{A/s}$.
  • Calculation:
    • $M = \frac{N_2 \phi_{12}}{I_1} = \frac{1000 \times 0.6 \times 10^{-4}}{10} = 6 \times 10^{-3}\,\text{H}$.
    • $e_{m2} = M \frac{dI_1}{dt} = 6 \times 10^{-3} \times 1200 = 7.2\,\text{V}$.

• Problem 14 (Composite Air Gap Electromagnetic Circuit):

  • Statement: Calculate MMF required to establish flux density $B = 1.3$\,$\text{Wb/m}^2$ in an air gap electromagnet with iron path length $2$\,m and air gap length $2$\,mm. Relative permeability of iron $\mu_r = 1500$
  • Given: $B = 1.3\,\text{Wb/m}^2$, $l_{\text{iron}} = 2\,\text{m}$, $l_{\text{air}} = 2 \times 10^{-3}\,\text{m}$, $\mu_{r,\text{iron}} = 1500$, $\mu_{r,\text{air}} = 1$
  • Calculation:
    • $\text{Total MMF} = B \left( \frac{l_{\text{iron}}}{\mu_0 \mu_{r,\text{iron}}} + \frac{l_{\text{air}}}{\mu_0 \mu_{r,\text{air}}} \right)$
    • $\text{Total MMF} = 1.3 \times \left( \frac{2}{4\pi \times 10^{-7} \times 1500} + \frac{2 \times 10^{-3}}{4\pi \times 10^{-7} \times 1} \right) = 3448.3571\,\text{AT}$

• Problem 15 (Excitation Current with Saw Cut and Leakage Factor):

  • Statement: An iron ring ($A = 5$\,$\text{cm}^2$, mean circumference $200$\,cm) has a saw cut of $2$\,mm and $N = 400$\,turns. Find exciting current to produce $0.5$\,mWb in air gap given $\mu_{r,\text{iron}} = 600$ and leakage factor $\lambda = 1.2$
  • Given: $A = 5 \times 10^{-4}\,\text{m}^2$, $l_{\text{iron}} = 200 \times 10^{-2}\,\text{m} = 2\,\text{m}$, $l_{\text{air}} = 2 \times 10^{-3}\,\text{m}$, $N = 400$, $\phi_{\text{air}} = 0.5 \times 10^{-3}\,\text{Wb}$, $\lambda = 1.2$
  • Calculation:
    • $\phi_{\text{iron}} = \phi_{\text{air}} \times \lambda = 0.5 \times 10^{-3} \times 1.2 = 0.6 \times 10^{-3}\,\text{Wb}$.
    • $NI = \frac{\phi_{\text{iron}} l_{\text{iron}}}{\mu_0 \mu_{r,\text{iron}} A} + \frac{\phi_{\text{air}} l_{\text{air}}}{\mu_0 \mu_{r,\text{air}} A}$
    • $400 \times I = \frac{0.6 \times 10^{-3} \times 2}{4\pi \times 10^{-7} \times 600 \times 5 \times 10^{-4}} + \frac{0.5 \times 10^{-3} \times 2 \times 10^{-3}}{4\pi \times 10^{-7} \times 1 \times 5 \times 10^{-4}}$
    • $400 \times I = 3183.098 + 1591.549 = 4774.647$
    • $I = \frac{4774.647}{400} = 11.937\,\text{A}$