Electromagnetic Induction And Alternating Currents – NEET UG PhysicsPractice Questions & PYQs

Alternating current tends to concentrate near the outer surface of a conductor due to the skin effect, meaning a collection of thinner strands offers a larger total surface area than a single solid wire of equivalent cross-section. This makes bundled strands much more efficient for carrying alternating current by reducing effective resistance. In contrast, direct current utilizes the entire cross-sectional area uniformly without any skin effect limitations, allowing either single or multiple-strand configurations to be deployed with equal effectiveness.

The magnitude of the induced electromotive force in an inductor is governed by the relation $|\varepsilon| = L \frac{di}{dt}$. Given an inductance of $40 \ \mathrm{mH}$, a current change from 1 A to 11 A, and a time duration of 4 ms, the rate of change of current can be determined directly. Substituting these values into the formula yields $|\varepsilon| = (40 \times 10^{-3}) \times (10 / (4 \times 10^{-3}))$, which evaluates to 100 volts.

Inductors store energy within the magnetic field established by the flow of electric current through their windings. When current passes through the conductive turns, it generates a magnetic flux that confines and holds this potential energy. Unlike capacitors which rely on electric fields between plates, inductors depend entirely on magnetic induction phenomena.

The motional electromotive force induced across a straight conductor moving perpendicularly through a uniform magnetic field is given by the equation $\varepsilon = B l v$. Substituting the given values of magnetic field intensity, conductor length, and speed into this relationship gives $\varepsilon = 0.9 \times 0.4 \times 7$. Multiplying these values together results in an induced electromotive force of $2.52$ volts.

The conservation of energy in an ideal transformer requires that the primary power equals the secondary power, expressed as $V_P I_P = V_S I_S$. First, find the secondary voltage by scaling the primary voltage of 230 V by the turns ratio of 25:1, resulting in 5750 V. Multiplying this secondary voltage by the secondary current of 2 A gives the total output power, which when divided by the primary voltage yields a primary current of 50 A.

Motional electromotive force is generated when a conductor moves across a magnetic field, cutting through magnetic flux lines according to the relation $E = B l v$. As the loop exits the field region with a velocity of $2 \times 10^{-2}\ \text{m/s}$, it is the shorter side of length $3 \times 10^{-2}\ \text{m}$ that cuts perpendicularly through the magnetic field lines of $0.3\ \text{T}$. Substituting these values into the formula yields the product $0.3 \times 2 \times 10^{-2} \times 3 \times 10^{-2}$, which multiplies out to $1.8 \times 10^{-4}\ \text{V}$.

In an alternating current circuit containing an inductance coil, the voltage $V$ leads the current $I$ by a phase difference of ninety degrees. Because of this phase separation, the voltage waveform reaches its corresponding points a quarter-cycle earlier than the current waveform. Consequently, the voltage drops over time following the exact profile shown in graph A.

The self-inductance of a coil is determined by the magnetic flux linkage per unit current, which scales quadratically with the number of turns. When current passes through a winding, it generates a proportional magnetic field that cuts across every single turn, creating a cumulative flux linkage. Mathematically, substituting the magnetic field equation $B = \frac{\mu_0 N i}{l}$ into the flux formula yields $L = \frac{\mu_0 N^2 A}{l}$, demonstrating that self-inductance varies directly with $N^2$.

The voltage transformation in an ideal transformer is directly proportional to the ratio of the number of turns in the secondary coil to the primary coil, while the frequency of the alternating supply remains completely unchanged. Setting up the transformer turns ratio equation $\frac{N_s}{N_p} = \frac{E_s}{E_p}$ allows us to relate the given primary voltage of 20 V and winding counts of 500 and 5000. Multiplying the primary voltage by the turns ratio of 10 gives a secondary voltage of 200 V while preserving the original frequency of 50 Hz.

The amplitude of the induced alternating current is determined by dividing the peak electromotive force by the resistance, where the peak emf is given by the formula $E_0 = N B A \omega$. The coil has a radius of 30 cm, a resistance of $\pi^2$ ohms, and rotates at 200 rpm in a magnetic field of $10^{-2}$ tesla. Substituting these parameters and the calculated area into the current relation $I_0 = \frac{N B A \omega}{R}$ results in a peak current of $6 \times 10^{-3}$ amperes, which corresponds to $6 \text{ mA}$.

The magnitude of average induced electromotive force is given by Faraday's law as $\varepsilon = \frac{N \Delta \Phi}{\Delta t}$, where the change in magnetic flux corresponds to the initial flux because the final flux drops to zero upon removal. Rearranging the expression to solve for time gives $t = \frac{N B A}{\varepsilon}$, and substituting the given values yields a duration of 0.1 s.

Faraday's law of electromagnetic induction states that the induced electromotive force is proportional to the rate of change of magnetic flux, and combining this with Ohm's law gives the current expression $i = \frac{N A}{R} \frac{dB}{dt}$. Substituting the given values into this relationship results in $i = \frac{20 \times (25 \times 10^{-4}) \times 1000}{100}$, which evaluates directly to 0.5 A.

Faraday's law of electromagnetic induction dictates that the induced electromotive force in a coil is proportional to the time derivative of the current flowing through the mutually coupled primary coil. Differentiating the time-varying current equation $I = I_0 \sin \omega t$ with respect to time yields the rate of change of current as $I_0 \omega \cos \omega t$. Multiplying this derivative by the mutual inductance gives the induced electromotive force expression $\varepsilon = M I_0 \omega \cos \omega t$, which reaches its maximum value of $5\pi$ when $\cos \omega t = 1$ under the given parameters.

The magnetic potential energy stored in an inductor is calculated using the relation $E = \frac{1}{2} L i^2$, where $L$ is inductance and $i$ is current. Converting the inductance to standard units and substituting it alongside the current into the formula yields $E = 0.5 \times (0.1) \times (1)^2$, resulting in a stored energy of 0.05 J.

Eddy currents are generated within a conductor whenever it experiences a changing magnetic flux. According to electromagnetic induction principles, variations in the surrounding magnetic field alter the flux linked with the metal surface. This temporal change induces localized circular electromotive forces that drive swirling currents throughout the bulk of the material.

Faraday's law of electromagnetic induction establishes that a changing magnetic flux generates an electromotive force, which subsequently drives a current through any conducting circuit. Multiplying the constant magnetic field by the time-varying area yields the magnetic flux expression $\phi = BA_0(1 + \sin t)$. Taking the time derivative of this flux determines the induced electromotive force as $-BA_0 \cos t$, and dividing by the resistance provides the induced current. Squaring this current and multiplying by the resistance results in the dissipated power formula $P = \frac{B^2 A_0^2 \cos^2 t}{R}$. Because the power depends directly on $\cos^2 t$, the resulting graph displays a strictly non-negative, periodic curve with maximum peaks occurring at multiples of $\pi$ and dropping to zero at odd multiples of $\pi/2$.

The induced electromotive force and current generated according to Faraday's law require mechanical work to move a conductor or change a magnetic field, ensuring that electrical energy is created at the expense of external work. This fundamental alignment dictates that electromagnetic induction obeys the principle of conservation of energy by transforming mechanical or source energy into electrical form without violation.

Electrical resonance in a series circuit occurs when the inductive reactance equals the capacitive reactance, causing the two reactive effects to completely cancel each other out. With zero net reactance, the impedance of the circuit reaches its minimum possible value and consists solely of the pure resistance. Consequently, the applied voltage and the resulting alternating current are completely in phase with one another, making the phase difference equal to zero.

The current amplitude in a series LCR circuit driven by an alternating voltage source is found by dividing the peak voltage by the total impedance of the circuit. Extracting the angular frequency from the voltage equation allows for the calculation of inductive reactance as $X_L = \omega L$ and capacitive reactance as $X_C = \frac{1}{\omega C}$. Since the inductive and capacitive reactances are equal, the circuit achieves resonance, meaning the impedance simplifies purely to the resistance value of $100\ \Omega$. Using the peak voltage of $220\ \text{V}$, the peak current evaluates via $i_0 = \frac{V_0}{Z}$ to give $2.2\ \text{A}$.

Resonance in a series RLC circuit occurs at a specific frequency where the inductive and capacitive reactances perfectly cancel each other out, governed by the formula $f_0 = \frac{1}{2\pi\sqrt{LC}}$. Substituting the inductance and capacitance values into the radical expression gives a time product whose square root evaluates to $1 \times 10^{-5}\ \text{seconds}$. Completing the division results in a frequency of approximately $15915\ \text{Hz}$, which converts directly to $15.9\ \text{kHz}$.