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

The motional electromotive force induced in a conductor is determined using the relation $e = B v l_{eq}$, where $l_{eq}$ denotes the shortest straight-line distance between the two endpoints of the conductor. As the semicircular ring moves downward, the straight-line separation between its terminal points P and R is equal to the diameter, which is given by $2r$. Multiplying the magnetic field strength, velocity, and this effective length gives the induced potential difference as $2rBv$. Furthermore, analyzing the direction of the magnetic force acting on the charge carriers shows that positive charges accumulate at terminal R, establishing it at a higher potential compared to P.

Transformer operation is governed by power conservation and efficiency, where useful output power equals the input power multiplied by the efficiency percentage. Multiplying the three kilowatt supply power by ninety percent yields an effective output power of two point seven kilowatts. Dividing this output power by the secondary current of six amperes gives a secondary voltage of four hundred fifty volts. Finally, dividing the total input power by the primary supply voltage of two hundred volts determines the primary current of fifteen amperes.

Transformers operate fundamentally on the principle of electromagnetic induction, which mandates a continuously varying magnetic flux to induce an electromotive force in the secondary winding. When connected to a direct current battery, the primary circuit establishes a steady, unchanging current that produces a static magnetic field. Because this magnetic flux remains entirely constant over time, no secondary voltage can be generated, resulting in zero induced potential and no current flow in the secondary circuit regardless of the turns ratio.

The total magnetic flux linked with a current-carrying inductor is directly proportional to the current and is determined by the formula $\Phi_B = L I$. Substituting the given self-inductance of $5$ H and the current of $2.5$ A into this relationship yields the calculation $\Phi_B = 5 \times 2.5$. Multiplying these values results in a magnetic flux of $12.5$ Wb.

The inductive reactance of an AC circuit depends directly on the self-inductance of the coil, which increases significantly when a ferromagnetic material like iron is introduced into its core. Inserting an iron rod amplifies the magnetic field within the windings, thereby raising the overall inductance and the corresponding reactance offered to the alternating current. Consequently, a greater share of the source voltage drops across the inductor itself, leaving a reduced potential difference across the bulb and causing its brightness to diminish.

The induced electromotive force in a rotating loop varies sinusoidally, reversing its polarity every time the plane of the coil passes perpendicular to the magnetic field lines. As the loop completes a full rotation, it achieves this perpendicular orientation twice per cycle, corresponding to the points where the magnetic flux linkage reaches its maximum rate of change with opposing slopes. Each such orientation change flips the direction of the induced current and voltage. Therefore, the direction of the induced electromotive force changes exactly twice during each complete revolution.

The phase relationship in a series AC circuit containing resistance and reactance is governed by the tangent of the phase angle, expressed as $\tan \phi = \frac{X}{R}$. When the inductor is removed, the remaining circuit yields $\tan \left(\frac{\pi}{3}\right) = \frac{X_C}{R}$, which establishes the capacitive reactance as $X_C = \sqrt{3}R$. Similarly, removing the capacitor instead gives an inductive reactance of $X_L = \sqrt{3}R$. Because both reactances are equal in magnitude, the circuit achieves electrical resonance where inductive and capacitive effects cancel out completely, making the impedance equal to the pure resistance and resulting in a unity power factor.

The total charge induced in a circuit is fundamentally determined by the area under the current-time graph, which is represented by the formula $q = \int i \, dt$. By calculating the geometric area under the curve shown \in the figure, the total charge is evaluated to be $0.2$ coulombs. The relationship between this induced charge, the change \in magnetic flux, and the resistance of the circuit is given by the expression $q = \frac{\Delta \phi}{R}$. Rearranging this principle to solve for the flux yields the equation $\Delta \phi = qR$. Multiplying the charge of $0.2$ coulombs by the given coil resistance of $10$ ohms results \in a total change \in magnetic flux of $2$ weber.

The voltage across an inductor is defined by the relation $V = -L \frac{dI}{dt}$, which indicates that $V$ is directly proportional to the slope of the $I-t$ graph. By observing how the rate of change of current shifts across each segment of the plotted curve, we can map out the corresponding voltage levels over time. This sequential tracking of slopes naturally generates the exact waveform depicted in option D.

The phase angle $\phi$ between the applied voltage and current in a series $RL$ circuit is determined by the ratio of the inductive reactance to the resistance, expressed through the formula $\tan\phi = \frac{X_L}{R}$. Substituting the given values where both the resistance and the inductive reactance equal $3\,\Omega$ yields a tangent value of $1$, which corresponds to a phase difference of $\frac{\pi}{4}$ radians.

The root-mean-square value of any periodic potential waveform is determined by taking the square root of the average of the squared voltage over a complete time period $T$, represented by the formula $V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} V^2 dt}$. Based on the provided graph, the potential is equal to $V_0$ throughout the first half of the cycle from $0$ to $\frac{T}{2}$ and drops to zero for the remaining half from $\frac{T}{2}$ to $T$. Splitting the integral over these two time intervals allows the zero-voltage portion to drop out, leaving us to integrate only from $0$ to $\frac{T}{2}$. Solving this definite integral gives $V_{rms} = \sqrt{\frac{V_0^2}{T} \left( \frac{T}{2} \right)}$, which neatly reduces to $\frac{V_0}{\sqrt{2}}$ after canceling the period $T$.

Inductive reactance is directly proportional to frequency, meaning that doubling the frequency from $50\,\text{Hz}$ to $100\,\text{Hz}$ proportionally doubles the reactance of the coil to $40\,\Omega$. Incorporating this updated reactance alongside the constant resistance of $30\,\Omega$ into the impedance equation $Z = \sqrt{R^2 + X_L^2}$ gives a total impedance of $50\,\Omega$, which allows the circuit current to be calculated as $4.0\,\text{A}$ when driven by a $200\,\text{V}$ source.

Faraday's law of electromagnetic induction relates the induced emf in a coil to the rate of change of current through the equation $e = -L \frac{di}{dt}$. Because the current-time graph is composed of straight-line segments, the slope $\frac{di}{dt}$ is constant along each interval, producing horizontal steps of induced emf that alternate between positive and negative values as the slope transitions.

Energy conservation in an ideal LC circuit requires that the initial electrostatic energy stored in the capacitor transforms into a combination of electrostatic energy at the reduced potential and magnetic energy stored in the inductor. Equating these energy states as $\frac{1}{2}CV_1^2 = \frac{1}{2}CV_2^2 + \frac{1}{2}LI^2$ and isolating the current term yields the final expression $\left(\frac{C(V_1^2-V_2^2)}{L}\right)^{1/2}$.

When a series LCR circuit satisfies the condition $V_L = V_C$, the inductive and capacitive potential drops completely neutralize each other, establishing electrical resonance. At this resonance state, the total impedance of the circuit is purely resistive, causing the entire source voltage to appear across the resistor.

Consequently, the voltmeter $V_3$ connected across the resistor reads a potential difference of 100 V. Dividing this voltage by the circuit resistance yields the total circuit current, resulting in the ammeter A reading 2.0 A.

The efficiency of a transformer is defined as the ratio of output power to input power, expressed by the formula $\eta = \frac{V_s I_s}{V_p I_p}$. Rearranging this expression to solve for the primary current gives $I_p = \frac{V_s I_s}{\eta V_p}$. Substituting the given values into this rearranged formula yields $I_p = \frac{440 \times 2}{0.80 \times 220}$, which evaluates to $5.0 \text{ ampere}$.

The magnitude of canyon-like induced electromotive forces in a changing area loop exposed to a uniform magnetic field is governed by Faraday's law in the form $|\varepsilon| = \frac{d}{dt}(B\pi r^2)$. Substituting the magnetic field strength of $0.025\,\text{T}$, the instantaneous radius of $2 \times 10^{-2}\,\text{m}$, and the shrinkage rate of $10^{-3}\,\text{m/s}$ into the expanded derivative expression $2\pi r B \frac{dr}{dt}$ yields an induced emf of $\pi\,\mu\text{V}$.

Faraday's law states that the magnitude of the induced electromotive force is equal to the rate of change of magnetic flux, given by $\varepsilon = \frac{d\phi}{dt}$ where magnetic flux through the circular loop is $\phi = B \pi r^2$. Differentiating this flux expression with respect to time yields $\varepsilon = B \pi (2r) \frac{dr}{dt}$. Substituting the given numerical values of magnetic field, radius, and rate of shrinking into this derivative gives $\varepsilon = 0.04 \times \pi \times (2 \times 2 \times 10^{-2}) \times (2 \times 10^{-3})$, which simplifies to $3.2\pi \mu\text{V}$.

The magnitude of the induced electromotive force in a moving loop depends directly on the rate at which magnetic flux changes as it exits the magnetic field region. While certain loops with straight, uniform boundaries maintain a steady rate of area extraction under constant velocity, loops with curved geometries or varying widths experience a continuously changing rate of flux reduction. Consequently, circular and elliptical loops, along with specific non-uniform orientations, produce an induced emf that varies continuously over time.

The average power dissipated in an alternating current series circuit is given by the expression $P = E_{\text{rms}} I_{\text{rms}} \cos\phi$, incorporating the root-mean-square values and the power factor. Expressing the root-mean-square current as $I_{\text{rms}} = \frac{E_{\text{rms}}}{Z}$ and the power factor as $\cos\phi = \frac{R}{Z}$ in terms of circuit impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$ allows power to be rewritten as $P = \frac{E_{\text{rms}}^2 R}{Z^2}$. Replacing the root-mean-square emf with $\varepsilon$ and substituting the expanded denominator yields the final power formula $\frac{\varepsilon^2 \mathrm{R}}{\mathrm{R}^2 + (\mathrm{L}\omega - 1/\mathrm{C}\omega)^2}$.