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

The phase relationship in a series alternating current circuit is governed by the net reactance relative to the resistance. When the inductor is removed, the tangent of the phase angle is given by $\tan\phi = \frac{X_{C}}{R}$, and removing the capacitor instead gives $\tan\phi = \frac{X_{L}}{R}$. Since both configurations produce the same phase difference of $\frac{\pi}{3}$, the inductive and capacitive reactances must be equal in magnitude, meaning the circuit operates at resonance where impedance equals resistance and the power factor is unity.

The rms current in a purely capacitive alternating current circuit is determined by the product of the angular frequency, capacitance, and RMS voltage, expressed as $I_{\text{rms}} = \omega C V_{\text{rms}}$. The angular frequency is evaluated as two times pi times the source frequency, while the capacitance and voltage are substituted directly into the relation. Performing this multiplication with 40 microfarads, 200 volts, and a frequency of 50 Hz results in an rms current of 2.5 A.

Eddy currents are loops of electrical current induced within bulk conductors by changing magnetic fields, which create opposing magnetic forces or localized heating. This electromagnetic phenomenon is deliberately applied in magnetic braking systems, induction furnaces, and electromagnetic dampers to control motion or generate high temperatures efficiently. In contrast, an electric heater operates purely on the principle of Joule heating, where electrical energy is converted directly into thermal energy through the resistance of a heating element without relying on induced circulating currents.

The magnitude of the induced electromotive force in a coil is given by Faraday's law of electromagnetic induction as $e = N \frac{\Delta \phi}{\Delta t}$ where $N$ is the number of turns and $\Delta \phi$ is the change \in magnetic flux. The initial magnetic flux linked with the coil is calculated using the relation $\phi_1 = NBA \cos 0^\circ$, substituting the given values of turns, area, and magnetic field to obtain $2 × 10^{-3}$ weber. Rotating the coil by $90^\circ$ reduces the final flux to zero because the magnetic field becomes parallel to the plane of the coil. Dividing this total flux change by the time interval of $0.1\text{ s}$ gives the final induced emf value of $0.02\text{ V}$.

The opposition offered by a circuit to alternating current is determined by its impedance, while direct current encounters only the ohmic resistance of the circuit. Dividing the source voltage by the respective currents using the relations $Z = \frac{V}{I_{\text{ac}}}$ and $R = \frac{V}{I_{\text{dc}}}$ yields an impedance of 60 ohms and a resistance of 30 ohms. Because the total impedance exceeds the pure resistance, the circuit must contain a frequency-dependent reactive component. An inductor introduces reactance in an AC circuit while acting as a simple conductor to steady DC, which identifies the circuit as a series LR arrangement.

The electromotive force induced between the center and the rim of a rotating conducting wheel in a uniform magnetic field is derived by integrating the motional emf across its radius, resulting in the formula $E = \frac{1}{2} B \omega r^2$ where $B$ is the magnetic field strength, $\omega$ is the angular velocity, and $r$ is the radius of the wheel. Substituting the given values into this expression gives $E = \frac{1}{2} × 0.1 × 10 × (0.5)^2$. Evaluating this product yields the induced emf value of 0.125 volts.

The magnetic potential energy stored in an inductor carrying a current is expressed by the relation $U = \frac{1}{2} L I^2$ where $L$ is the inductance and $I$ is the current. Substituting the given energy and current values into this formula yields the equation $25 \times 10^{-3} = \frac{1}{2} L (60 \times 10^{-3})^2$ which rearranges to solve for the inductance as $L = \frac{50}{3.6}$, ultimately resulting in 13.89 henrys.

The average power loss in a series RLC circuit is determined by the root-mean-square current and the resistance, given by the formula $P_{\text{av}} = \left( \frac{V_{\text{rms}}}{Z} \right)^2 R$ where $V_{\text{rms}}$ is the root-mean-square voltage, $R$ is the resistance, and $Z$ is the total impedance of the circuit. The impedance is calculated by combining the resistance, inductive reactance, and capacitive reactance through the relation $Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$ which evaluates to approximately 56 ohms using the given angular frequency and component values. Substituting the root-mean-square voltage of 10 divided by the square root of 2, along with the impedance and resistance values into the power equation, yields an average power loss of 0.79 watts.

At the exact moment the switch is closed, inductors block any sudden change in current and temporarily behave as open circuits. Consequently, no current flows through the branches containing the inductors at this initial instant, leaving only the resistive branch connected across the source. Using Ohm's law, the current drawn from the battery is given by the formula $i = \frac{\varepsilon}{R}$. Substituting the provided values yields $i = \frac{18}{9.0}$, which results in a battery current of 2 A.

The total charge induced in a circuit is determined by the relation $q = \frac{\Delta \phi}{R}$, where $\Delta \phi$ is the total change in magnetic flux linkage and $R$ is the resistance. The magnetic flux linked with the inner coil due to the magnetic field of the long solenoid is given by $\phi = N B A$, with $B = \mu_0 n i$ representing the magnetic field inside the solenoid and $A = \pi r^2$ being the cross-sectional area of the coil. Substituting the expression for flux change into the charge equation yields $q = \frac{N \mu_0 n \pi r^2 \Delta i}{R}$. Inserting the given numerical values for the solenoid turns per meter, coil turns, radii, current change, and resistance into this formula gives the total charge value of 32 microcoulombs.

When an alternating voltage is applied across an ideal capacitor, the current leads the voltage by a phase angle of ninety degrees. Throughout each complete cycle, the capacitor alternately stores electrical energy in its electric field during the charging phase and returns that exact same energy to the source during the discharging phase. As a result, the net work done and energy consumed by an ideal capacitor over a full cycle is precisely zero.

The power factor in an alternating current circuit is defined as the ratio of resistance to impedance, which is expressed in terms of component voltages as $\cos\phi = \frac{V_R}{\sqrt{V_R^2 + (V_L - V_C)^2}}$. Substituting the given potential differences yields $\cos\phi = \frac{80}{\sqrt{80^2 + (100 - 40)^2}}$. Evaluating the denominator terms results in the square root of sixty-four hundred plus thirty-six hundred, which simplifies to one hundred. Dividing the resistive voltage by this total impedance voltage gives a power factor of zero point eight.

The sharpness of tuning and selectivity of a series circuit are determined by its quality factor, which is expressed by the formula $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$, where $R$ is resistance, $L$ is inductance, and $C$ is capacitance. For better tuning and communication performance, a higher quality factor is required, necessitating a combination of low resistance, high inductance, and low capacitance. Evaluating this relation across the given choices reveals that the combination with 15 ohms of resistance, 3.5 henrys of inductance, and 30 microfarads of capacitance yields the maximum quality factor.

Power loss in an alternating current series circuit is determined by the relation $P = I_{\text{rms}}^2 R$, where $I_{\text{rms}}$ is the root-mean-square current and $R$ is the resistance. The impedance of the circuit is calculated using the formula $Z = \sqrt{R^2 + (X_C - X_L)^2}$, which requires finding the inductive reactance $X_L = \omega L$ and the capacitive reactance $X_C = \frac{1}{\omega C}$. Using the angular frequency from the voltage equation along with the given inductance and capacitance values yields an inductive reactance of 6.8 ohms and a capacitive reactance of 58.8 ohms. Substituting these reactances and the resistance into the impedance equation gives the total impedance, which can then be combined with the root-mean-square voltage derived from the peak voltage to compute the final power dissipation of 0.51 watts.

The self-inductance of a solenoid is defined by the flux linkage relation $N \phi = L i$, where $N$ is the total number of turns, $\phi$ is the magnetic flux linked with each individual turn, $L$ is the self-inductance, and $i$ is the current flowing through the circuit. Multiplying the given flux per turn by the total number of turns yields the total magnetic flux linkage. Dividing this total flux linkage by the current of 4 amperes directly gives the self-inductance value of 1 henry.

Faraday's law of electromagnetic induction dictates that the generated emf is equal to the negative rate of change of magnetic flux enclosed by a loop, expressed as $e = -\frac{d\Phi}{dt}$. Magnetic flux is determined by the product of the magnetic field and the cross-sectional area where that field genuinely exists.

Because loop 1 encompasses the entire region of radius $r$ where the magnetic field is restricted, its magnetic flux is given by $B\pi r^2$, which differentiates with respect to time to produce an emf of $-\frac{dB}{dt}\pi r^2$. Conversely, loop 2 sits completely outside the restricted magnetic field zone, meaning no magnetic flux threads through its surface at any point, resulting in an induced emf of zero.

Power dissipated in a purely resistive circuit is related to the source voltage and resistance through the relation $P = \frac{V^2}{R}$, which rearranges to $V^2 = PR$. When an inductance is added in series, the new power is given by $P' = \frac{V^2}{Z} \cos\phi$, where the power factor is the ratio of resistance to impedance. Substituting the voltage relation and the power factor into this expression produces $P' = \frac{PR}{Z} \times \frac{R}{Z}$. Simplifying this algebraic product yields the final power expression of $P \left(\frac{R}{Z}\right)^2$.

Introducing a dielectric material like mica increases the capacitance of a capacitor, which subsequently decreases its capacitive reactance according to the relationship $X_C = \frac{1}{\omega C}$. Because the voltage drop across the capacitor depends directly on its reactance for a given circuit current, a lower reactance results in a smaller potential difference. Consequently, the air-filled capacitor maintains a lower capacitance and higher reactance, producing a larger voltage drop than the mica-filled counterpart.

Electromagnetic induction dictates that a changing magnetic flux through a circuit generates an induced current to oppose that very change. As the electron travels along the straight path $XY$ toward the loop $abcd$, the magnetic flux passing through the loop increases in a specific direction. Once the electron passes the loop and moves away, that same magnetic flux decreases, prompting the induced current to automatically reverse its direction.

When a conducting frame moves through a non-uniform magnetic field, the net induced electromotive force is determined by the motional EMF generated across its two active sides parallel to the long current-carrying wire. Because the magnetic field produced by a straight wire varies inversely with the perpendicular distance $r$, the two sides of the square frame parallel to the wire experience different field magnitudes.

If the center of the frame is at a distance $x$ from the wire, the inner side lies at a distance of $x - a/2$ and the outer side lies at $x + a/2$. The net induced EMF is proportional to the difference in magnetic fields between these two positions, which translates to a variation of $\frac{1}{x - a/2} - \frac{1}{x + a/2}$. Rewriting this expression gives $\frac{1}{2x-a} - \frac{1}{2x+a}$, and combining the fractions yields a final proportionality to $\frac{1}{(2x-a)(2x+a)}$.