Comprehensive Guide to Electromagnetic Induction and Magnetism and Inductive Circuits

Torque and Magnetic Dipole Moments

  • Motors represent the most frequent application of magnetic force acting upon current-carrying wires. When current passes through loops, the magnetic field exerts torque, which causes the rotation of a shaft. This transformation process converts electrical energy into mechanical work.

  • The torque τ\tau acting on a coil consisting of NN loops, where each loop has an area AA and carries a current II, in an external uniform magnetic field BB is determined by the equation: τ=NIABsin(θ)\tau = NIAB \sin(\theta) In this expression, θ\theta represents the angle between the normal to the plane of the loop and the magnetic field direction.

  • In vector notation, torque is expressed as: τ=NIA×B\mathbf{\tau} = N I \mathbf{A} \times \mathbf{B} Here, A\mathbf{A} is a vector oriented perpendicular to the loop's plane with a magnitude equivalent to the area of the loop.

  • The magnetic dipole moment of a coil is defined as the quantity IAIA.

  • The potential energy UU associated with a magnetic field is calculated using: U=μBU = -\mathbf{\mu} \cdot \mathbf{B}

Galvanometers and Sensitivity Factors

  • A galvanometer contains wire loops wound on a soft iron core. This core pivots between the pole faces of a permanent magnet. A current passing through the coil generates torque, while a small spring provides a restoring torque.

  • Current sensitivity in a meter is defined as the deflection per unit current: Sensitivity=ϕI=NABk\text{Sensitivity} = \frac{\phi}{I} = \frac{NAB}{k}

  • To enhance the sensitivity of a current meter, the following adjustments can be made:

    • Increase the magnetic field strength BB.

    • Decrease the spring constant kk (using weaker springs).

    • Increase the number of turns NN in the coil.

    • Increase the area of the coil AA.

  • Voltmeter sensitivity is defined as the deflection per unit potential difference: Sensitivity (Voltmeter)=ϕV\text{Sensitivity (Voltmeter)} = \frac{\phi}{V} Given the relationship V=IRV = IR, this can also be expressed as: ϕV=NABkR\frac{\phi}{V} = \frac{NAB}{kR}

Ammeters, Voltmeters, and Resistance Conversion

  • An ammeter is designed to measure the electric current flowing through it, whereas a voltmeter measures the potential difference between two specific points.

  • A milliammeter can be converted into an ammeter capable of measuring higher currents by connecting a low-resistance resistor, known as a shunt (RshR_{sh}), in parallel. If II is the total current, IcI_c is the current through the moving-coil ammeter, and IshI_{sh} is the current through the shunt: I=Ish+IcI = I_{sh} + I_c The relationship for the shunt is: IcRc=IshRshI_c R_c = I_{sh} R_{sh}

  • To convert a milliammeter into a voltmeter, a high-resistance resistor, known as a multiplier (RsR_s), is connected in series with the milliammeter. If VV is the total potential difference and RcR_c is the resistance of the milliammeter: V=Ic(Rs+Rc)V = I_c (R_s + R_c)

Ampere's Law and the Biot-Savart Law

  • Ampere's Law states that the line integral of the magnetic field vector B\mathbf{B} around any closed path is equal to μ0\mu_0 (permeability of free space) multiplied by the total current II flowing through the circuit: Bdl=μ0I\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I

  • The Biot-Savart Law, formulated by Jean-Baptiste Biot (1774-1862) and Felix Savart (1791-1841), describes the magnetic field produced by current elements. The field dBd\mathbf{B} at a point PP due to a current element IdlI d\mathbf{l} is: dB=μ04πIdl×r^r2d\mathbf{B} = \frac{\mu_0}{4 \pi} \frac{I d\mathbf{l} \times \hat{\mathbf{r}}}{r^2}

  • If θ\theta is the angle between the current element and the line joining it to point PP, the magnitude is: dB=μ04πIdlsin(θ)r2dB = \frac{\mu_0}{4 \pi} \frac{I dl \sin(\theta)}{r^2}

Magnetic Field Configurations

  • Long Straight Wire: At a perpendicular distance aa from a wire carrying current II: B=μ0I2πaB = \frac{\mu_0 I}{2 \pi a}

  • Center of a Circular Loop: For a loop of radius rr carrying current II: B=μ0I2rB = \frac{\mu_0 I}{2 r}

  • Axis of a Circular Loop: At a distance xx from the center of a loop with radius aa: B=μ0Ia22(a2+x2)3/2B = \frac{\mu_0 I a^2}{2 (a^2 + x^2)^{3/2}} At the center (x=0x = 0), this returns to B=μ0I2aB = \frac{\mu_0 I}{2 a}.

  • Solenoid: At the center of a solenoid with NN turns and length LL, carrying current II: B=μ0nIB = \mu_0 n I Where n=N/Ln = N/L is the linear turn density. At either end of the solenoid, the field is: Bend=12μ0nIB_{end} = \frac{1}{2} \mu_0 n I

  • Right-Hand Source Rule: To determine the direction of BB, grasp the wire with the right hand so the thumb points in the direction of the current II; the curled fingers indicate the sense of the magnetic field.

Forces Between Parallel Conductors

  • When two parallel wires carry currents I1I_1 and I2I_2 at a distance dd apart, the field from one exerts a force on the other. The force per unit length is defined as: FB=μ0I1I22πd\frac{F_B}{\ell} = \frac{\mu_0 I_1 I_2}{2 \pi d}

  • Interaction rules:

    • Conductors carrying currents in the same direction attract each other.

    • Conductors carrying currents in opposite directions repel each other.

  • Definition of the Ampere: When the force per unit length between two long, parallel wires separated by 1m1\,m carrying identical currents is 2×107N/m2 \times 10^{-7}\,N/m, the current in each wire is defined as 1A1\,A.

Faraday’s Law and Lenz’s Law

  • Faraday's Law of Induction: A changing magnetic field induces an electromotive force (EMF). The induced EMF in a wire loop is proportional to the rate of change of magnetic flux through the loop: ϵ=NΔΦBΔt\epsilon = -N \frac{\Delta \Phi_B}{\Delta t}

  • Magnetic Flux: Defined as the total number of field lines passing through a loop: ΦB=BAcos(θ)\Phi_B = BA \cos(\theta) Unit: Weber (WbWb). 1Wb=1Tm21\,Wb = 1\,T \cdot m^2.

  • Lenz's Law: The minus sign in Faraday's law indicates that the induced current generates a magnetic field that tends to oppose the change in the original magnetic field.

  • Flux changes can occur through:

    • Changing the magnetic field strength BB.

    • Changing the area of the loop AA.

    • Changing the angle θ\theta between the loop and the field.

Motional EMF and Electric Fields

  • EMF in a Moving Conductor: For a conductor of length ll moving with velocity vv perpendicular to a magnetic field BB: ϵ=Blv\epsilon = Blv The induced current tends to slow the moving bar, requiring an external force to maintain motion. This principle is utilized in the measurement of blood velocity.

  • Electric Field Generalization: A changing magnetic flux induces an electric field regardless of whether conductors are present.

Electric Generators

  • A generator transforms mechanical energy into electrical energy using an axle rotated by external forces like steam or falling water.

  • AC Generator: Uses slip rings to maintain contact. The induced EMF is: ϵ=NBAωsin(ωt)=ϵ0sin(ωt)\epsilon = NBA\omega \sin(\omega t) = \epsilon_0 \sin(\omega t) Where ϵ0=NBAω\epsilon_0 = NBA\omega is the peak EMF.

  • DC Generator: Uses a split-ring commutator. The commutator reverses the current every half revolution to ensure the torque maintains rotation in the same direction.

Back EMF and Eddy Currents in Motors

  • In a motor, rotation within a magnetic field induces an EMF called back EMF (ϵ\epsilon), which opposes the supply voltage VV. The armature current is: I=VϵRaI = \frac{V - \epsilon}{R_a} Where RaR_a is the armature resistance.

  • Mechanical power developed in the motor is calculated as: Pmech=ϵIP_{mech} = \epsilon I

  • In a generator, the induced current produces a counter torque, requiring increased external torque to maintain rotation.

Transformers and Power Transmission

  • A transformer consists of primary and secondary coils linked by an iron core. Transformers require AC to function.

  • EMF Ratio: EsEp=NsNp\frac{E_s}{E_p} = \frac{N_s}{N_p}

  • Current Ratio (assuming no losses): IpIs=NsNp\frac{I_p}{I_s} = \frac{N_s}{N_p}

  • Types:

    • Step-up: N_s > N_p, resulting in E_s > E_p.

    • Step-down: N_s < N_p, resulting in E_s < E_p.

Inductance: Self and Mutual

  • Self Inductance (LL): A varying current in a coil induces an EMF in itself: ϵ=LdIdt\epsilon = -L \frac{dI}{dt} Unit: Henry (HH). 1H=1Vs/A=1Ωs1\,H = 1\,V \cdot s/A = 1\,\Omega \cdot s.

  • Mutual Inductance (MM): A changing current in one coil induces an EMF in a secondary coil: ϵ2=MdI1dt\epsilon_2 = -M \frac{dI_1}{dt} and ϵ1=MdI2dt\epsilon_1 = -M \frac{dI_2}{dt}

Energy Stored in Magnetic Fields

  • Energy can be stored in the magnetic field of an inductor: U=12LI2U = \frac{1}{2} L I^2

  • For a solenoid, energy density (energy per unit volume) is: u=B22μ0u = \frac{B^2}{2 \mu_0}

LR Circuits: Growth and Decay

  • Time Constant (\tau): Defined as: τ=LR\tau = \frac{L}{R}

  • Current Growth: Upon closing a switch, current increases according to: i=VR(1et/τ)i = \frac{V}{R} (1 - e^{-t/\tau}) At t=τt = \tau, the current reaches approximately 63% of its final value (I0=V/RI_0 = V/R).

  • Current Decay: If the battery is removed, current decreases following: i=I0et/τi = I_0 e^{-t/\tau} At t=τt = \tau, the current decreases to about 37% of its original value.

Worked Examples

  • Example 15.28 (Mechanical Power):

    • Armature resistance Ra=0.08ΩR_a = 0.08\,\Omega; Supply V=120VV = 120\,V; Current I=50AI = 50\,A.

    • Power supplied: V×I=120×50=6000WV \times I = 120 \times 50 = 6000\,W.

    • Heat loss in resistance: I2Ra=502×0.08=200WI^2 R_a = 50^2 \times 0.08 = 200\,W.

    • Mechanical power = 6000200=5800W6000 - 200 = 5800\,W.

  • Example 15.29 (Transformer Turns):

    • Step-down from 2.2kV2.2\,kV to 110V110\,V. Secondary turns Ns=25N_s = 25.

    • EpEs=NpNs2200110=Np25\frac{E_p}{E_s} = \frac{N_p}{N_s} \rightarrow \frac{2200}{110} = \frac{N_p}{25}.

    • Primary turns Np=500N_p = 500.

  • Example 15.49 (LR Circuit Current):

    • R=20ΩR = 20\,\Omega, L=0.30HL = 0.30\,H, V=90VV = 90\,V. Find current at t=0.05st = 0.05\,s.

    • Time constant τ=L/R=0.30/20=0.015s\tau = L/R = 0.30/20 = 0.015\,s.

    • i=VR(1et/τ)=9020(1e0.05/0.015)i = \frac{V}{R} (1 - e^{-t/\tau}) = \frac{90}{20} (1 - e^{-0.05/0.015}).

    • i=4.5(1e3.33)=4.5(10.0357)4.34Ai = 4.5 (1 - e^{-3.33}) = 4.5 (1 - 0.0357) \approx 4.34\,A.