Complete Physics - Electromagnetic Induction and Waves

# Electromagnetic Induction * Magnetic Flux: * Represented by the symbol \phi. * Defined as \phi = B \cdot A = BA\cos\theta, where: * B is the magnetic field strength. * A is the area vector. * \theta is the angle between B and A. * Faraday's Law of Electromagnetic Induction: * States that the induced electromotive force (EMF) in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit. * Expressed as \varepsilon = -N \frac{d\phi}{dt}, where: * \varepsilon is the induced EMF. * N is the number of turns in the coil. * \frac{d\phi}{dt} is the rate of change of magnetic flux. * Motional EMF: * Arises when a conductor moves through a magnetic field. * For a conducting rod of length l moving with velocity v perpendicular to a magnetic field B, the induced EMF is given by \varepsilon = vBl. * If the rod rotates perpendicular to a uniform magnetic field B, the induced EMF between the ends of the rod is |\varepsilon| = \frac{B\omega l^2}{2} = B (2\pi v) \frac{l^2}{2} = B\omega A, where \omega is the angular velocity and A is the area. # Induction and EMF * Faraday's Law: Describes how a changing magnetic flux through a coil induces an electromotive force (EMF). * Induced EMF: The voltage generated in a coil due to a changing magnetic flux. * Induced Current: The current that flows in a closed circuit due to the induced EMF. * Induced Charge: The amount of charge that flows in the circuit due to the induced current; given by dQ = \frac{-Nd\phi}{R}. * Induced Electric Field: A non-conservative electric field is produced by a changing magnetic field, given by E = -\frac{Nd\phi}{dt}. * The induced EMF is present in the circuit as long as the change in magnetic flux is maintained. # Lenz's Law * States that the direction of the induced EMF and induced current are such that they oppose the change that produces them. * "जिससे पैदा हुए, उसी का oppose करेंगे" * "जो घटेगा वही पैदा होगा, जो बढ़ेगा उसका उल्टा पैदा होगा" * If magnetic flux increases into the page ($\otimes), induced current creates magnetic flux out of the page ($\odot), resulting in anticlockwise current. * If magnetic flux decreases into the page ($\otimes), induced current creates magnetic flux into the page ($\otimes), resulting in clockwise current. * If magnetic flux increases out of the page ($\odot), induced current creates magnetic flux into the page ($\otimes), resulting in clockwise current. * If magnetic flux decreases out of the page ($\odot), induced current creates magnetic flux out of the page ($\odot), resulting in anticlockwise current. # Mutual Inductance * The EMF induced in the secondary coil (\varepsilon_s) due to a changing current in the primary coil is given by: \varepsilon_s = -M \frac{dI_p}{dt}, where M is the coefficient of mutual inductance. * Coefficient of Coupling (K): * K = \frac{M}{\sqrt{L_1 L_2}}, where L_1 and L_2 are the self-inductances of the two coils. * For two long co-axial solenoids with N_1 and N_2 turns, each of length l and cross-sectional area A, the mutual inductance is: M = \frac{\mu_0 N_1 N_2 A}{l}. # Self Induction * Occurs when a changing current in a coil induces an EMF in the same coil, opposing the change in current. * N \phi = LI, where: * L is the self-inductance. * \phi is the magnetic flux. * I is the current. * Units of self-inductance: Henry (H), where L = \frac{N\phi}{I} \Rightarrow L = \frac{Weber}{Amp}. * The induced EMF is given by: \varepsilon = -L \frac{di}{dt}. * Dimensions of Inductance: [ML^2T^{-2}A^{-2}]. # Mutual Induction (अन्योन्य प्रेरण) * The changing current in Primary coil 1 induces a changing magnetic flux in coil 2. * N_2 \phi_2 = M I_1 * \varepsilon = -M \frac{di_1}{dt} # Combination of Inductors * Series: * Equivalent inductance: L_{eq} = L_1 + L_2. * Parallel: * Reciprocal of equivalent inductance: \frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2}. * Ideal Coupling: * Coefficient of coupling K = 1. * Mutual inductance: M = \sqrt{L_1 L_2}. # Energy Stored in a Coil * The energy (U) stored in a coil with inductance L carrying a current i is given by: U = \frac{1}{2} L i^2. * Energy Density: u = \frac{U}{V}, where V is the volume. # Self Inductance Formulas * Self Inductance: L = \frac{\mu_0 N^2 A}{l} = \mu_0 n^2 \pi R^2 l where n = \frac{N}{l} # Mutual Inductance Formulas * Mutual Inductance: M = \frac{\mu_0 N_1 N_2 A}{2R_1}. If R_1 >> R_2 then M = \frac{\mu_0 N_1 N_2 A}{l} # Example Problems and Solutions 1. Problem: The magnetic flux linked to a circular coil of radius R is given by \phi = 2t^3 + 3t^2 + 4t + 5 Wb. Find the magnitude of the induced EMF in the coil at t = 5s. * Solution: * \varepsilon = - \frac{d\phi}{dt} = -(6t^2 + 6t + 4) * At t = 5s, \varepsilon = -(6(5)^2 + 6(5) + 4) = -(150 + 30 + 4) = -184 V * Magnitude: |\varepsilon| = 192 V. 2. Problem: A coil with 800 turns and effective area 0.05 m² is kept perpendicular to a magnetic field of 5 \times 10^{-5} T. The coil is rotated by 90° in 0.1 s. Find the induced EMF. * Solution: * \varepsilon = -N \frac{d\phi}{dt} = -N \frac{BA\cos\theta_2 - BA\cos\theta_1}{\Delta t} * \varepsilon = -800 \times \frac{5 \times 10^{-5} \times 0.05 \times (\cos 90° - \cos 0°)}{0.1} = 2 \times 10^{-3} V. 3. Problem: A current of 2.5 A flows through a coil of inductance 5 H. Find the magnetic flux linked with the coil. * Solution: * N\phi = Li * N\phi = 5 \times 2.5 = 12.5 Wb. 4. Problem: A coil of resistance 400 Ω is placed in a magnetic field. The magnetic flux linked with the coil varies with time as \phi = 50t^2 + 4. Find the current in the coil at t = 2 s. * Solution: * i = - \frac{N}{R} \frac{d\phi}{dt} = - \frac{1}{400} \times (100t) * At t = 2 s, i = \frac{200}{400} = 0.5 A. # Alternating Current * Average and RMS Values * Different Types of AC Circuits * Power Dissipated in AC Circuits * Resonance, LC Oscillation * Transformer ## Alternating Current (प्रत्यावर्ती धारा ) * Mean or average value of alternating current or voltage over one complete cycle * I_{av} = \frac{\int_0^T I_0 \sin(\omega t) dt}{\int_0^T dt} = 0 * V_{av} = \frac{\int_0^T V_0 \sin(\omega t) dt}{\int_0^T dt} = 0 * Average value of alternating current for first half cycle * I_{av} = \frac{\int_0^{T/2} I_0 \sin(\omega t) dt}{\int_0^{T/2} dt} = \frac{2I_0}{\pi} \approx 0.637I_0 * Similarly, for alternating voltage, the average value over first half cycle is * V_{av} = \frac{\int_0^{T/2} V_0 \sin(\omega t) dt}{\int_0^{T/2} dt} = \frac{2V_0}{\pi} \approx 0.637V_0 ## Mean value or average value of alternating current over any half cycle * I_{av} = \frac{2I_0}{\pi} = 0.637I_0 ## Root mean square (rms) value of alternating current * I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707I_0 ## Similarly, for alternating voltage * V_{rms} = \frac{V_0}{\sqrt{2}} \approx 0.707V_0 ## Form Factor * Form \, Factor = \frac{I_{rms}}{I_{av}} ## AC circuit ### Average (औसत) ### Power Dissipated *

= V_{rms} I_{rms} \cos\phi V_{rms} I_{rms} \cos(\omega t + \phi) * \cos\phi = Power \, factor ### Pure * Resistive * Capacitive * Inductive ### Combinational * RL Series * LC * LCR ## Inductive Reactance * {X_L} = \omega L = 2 \pi f L ## Capacitive Reactance * {X_C} = \frac{1}{\omega C} = \frac{1}{2 \pi f C} ## Impedance ( प्रतिबाधा ) * Effective Resistance * Ac circuit [ प्रत्यावर्ती परिपथ का प्रभावी प्रतिरोध] ## Quality Factor * Resistive * Inductive * Capacitive * LCR ## Resonance (अनुनाद) ### LCR circuit # * Applied voltage at frequency: Z_{min} i max P_{max} \cos\phi = 1 Resistive * Resonating frequency अनुनादी आवृत्ति \alpha = 0 * Quality factor Q = \frac{\omega L}{R} # Transformer ## Ac current का voltage बदलता ### Efficiency of Transformer * \eta = \frac{output}{input} \times 100 ### Step up * N_p < N_s Voltage बढ़ाना ### Stepdown * N_p > N_s V घटाता ### # Mutual Induction \frac{e_p}{e_s} = \frac{N_p}{N_s} # * Real \eta < 100% * Ideal * \eta = 100% ### Energy loss in Transformee * Cupric loss (तात्रिक हानि) H = I^2 R + * Flux Loss (ThetaH ETTA * Hysteris loss शैथिल्य हानि * Buzzing loss (काम्पनिक हानि) ## 21. A 40 µF capacitor is connected to a 200 V, 50 Hz ac supply. ## The rms value of the current in the circuit is, nearly: ###Given C=40µF V=200 volt f = 50Hz ###Formula Irms = \frac{Vrms}{Z} ### Solution First calculate Z z=xe\Then substitute Vrms, Z inthe above Formula i get * Then 843.14\approx 2.4 ## 22. A small signal voltage V(t) = Vo sin wt is applied across an ideal capacitor C: ###Solution * Phase differece Between Voltage and current 8=90 *average power is

= Vrms ,irms , cos0 * capacitor does not consume any energy from the voltage source. 23. . The instantaneous values of alternating current and voltages in a circuit are given as ###solution average power consumed p = Vrms \times irms \times cosp \= \frac{%0}{\sqrt{2}} \times \frac{I0}{\sqrt{2}} \times cos \times/3 and finally P \frac{V0 10}{4} ##4 In an a.c circuit an alternating voltage le = 200 √2 sin 100 t volts is connected to a capacitor of capacity 1 μF. The r.m.s. value of the current in the circuit is 1 = Vims. &Tife so, after putting the values given we get ##25. In an a.c circuit the e.m.f. (e) and the current (i) at any instant are given respectively by **The average power in the circuit over one cycle of a.c. is** ##solution Average pse Vrms lim wo therefore the answer is \frac{Eolo}{2}cos o ##26. In an A.C. circuit with voltage Vand current I the power dissipated is **Solution** $Pseudocode$: AVERAGE POWER = Vrms Ims wo therefore the answer is VI ##Q 27. In an a.c. circuit, the r.m.s. value of current, irms is related to the peak current, in by the relation ###Given $Pseudocode$: Relation peak and Rms current ###Formula RMS value equals i\/2 or $= {\tfrac{1}{2}}i$ ##Q28. The net impedance of circuit (as shown in figure) will be ###Given 220 V, 50 Hz\1562 f=50Hz ###Formula * Inductive reactance = 2 * 50 * 10-3 / Capacitive reactance = 1/ (2 * f * C) = 10 Net impedance = Z = \sqrt{(5 - 104 +(101252 +10\approx =5/5/13 \{\ approximately }11.18 ##Q 29. An ac source is connected to a capacitor C. Due to decrease in its operating frequency **Solution** xc9 \{\ approximately } 1 /(2\Pelt f*C) so decreasing f will cause capacitive reactance to increase and hence will lead to displacement current decreases. ##Q_30_. In a series LCR circuit, the inductance L is 10 mH, capacitance C is µF and resistance R is 100 Q. The frequency at which resonance occurs is ### Given data And the formula for frequency. * **Solution** f = \text{5kHz} * Hence the answer should be 5 kHz. ##Q31. Given below are two statements. **Solution** * Statement 1: is correct. * Statement II: is also correct. Because of basic definition of capacitor's. When AC is applied Then voltage and current differ PI/2. Thus Both statements I and II are correct ##Q32. An inductor of inductance L, a capacitor of capacitance Cand a resistor of resistance 'R' are connected in series to an ac source of potential difference 'V' volts as shown in figure. Potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is 10√2 A. The impedance of the circuit is ###Solution The Formula For Impedance We know is \sqrt{R^2 + (XL . XC 2}$$ Now as per the question. Impedance Must be *5\sqrt{2} ##Q 33, An inductor 20 mH, a capacitor 100 uF and a resistor 502 are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is ###Solution The power loss in A.C. circuit = 0.79 W ## Q34 An inductor 20 mH, a capacitor 50 F and a resistor 402 are connected in series across a source of emf V = 10sin 340 t. The power loss in A.C. circuit is: ###solutionThe power loss in A.C. circuit : 0. 51 W ##Q35 5 - A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z', the power drawn will be: We know, Power = Vrms\* Irms \* cos(O) New power drawn = P(R\Z) ##Q_36_ - the question, the reading of voltmeter V1 and V2 are 300volts each. The reading of the voltmeter V3 and ammeter A are respectively. We can resolve this problem using kirchhoffs laws ##Q37 solution For Q_37 is We know, P=E^2/R ##Q38 A transformer having efficiency of 90percentage is working 200 v and 3kW power supply. If the current in the secondary coil is 6A, The voltage the across the secondary coil and the current in the primary coil respectively are *Solution*: = [ 300V, 15A* ##Q39*1 A 220 volts input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. if the efficiency of the transformer is 80percentage , the current drawn by the primary windings of the transformer, is *Solution*:-> [ 5.0 amphere* ##Q40 A-step-down transtormer, connected to main supply ot 220v is made to operate at 11 V, 44 W lamp ignoring-power losses in the transformer ,what is the current in the primary Circuit: *Solution*:- [ 0.2A