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

Magnetic monopoles do not exist in nature, which means magnetic field lines always form continuous closed loops rather than originating from isolated sources. Because every field line entering a closed volume must also exit it, the incoming and outgoing fluxes precisely balance each other. Consequently, the total magnetic flux integrated over any arbitrary closed surface is always zero.

The potential energy stored within the magnetic field of an inductor is determined by the formula $U = \frac{1}{2} L I^2$. Substituting the given inductance and current values into this expression results in $U = \frac{1}{2} \times 4 \times 10^{-6} \times 2^2$. Simplifying this calculation gives $8 \times 10^{-6}$ Joules, which corresponds to an energy of $8 \mu\text{J}$.

Active power dissipation in an alternating current circuit depends directly on the power factor, which is the cosine of the phase angle between the voltage and current waveforms. In a purely resistive circuit, the voltage and current remain completely in phase, yielding a maximum power factor of one and allowing continuous energy consumption. Conversely, pure inductors and capacitors introduce a ninety-degree phase difference that reduces the average power dissipation to zero over a complete cycle.

The peak induced electromotive force in a rotating generator coil is given by the formula $E_{\max} = N B A \omega$. Given a rotation speed of one revolution per second, the angular velocity is $\omega = 2\pi$ rad/s. Substituting the values for the number of turns, magnetic field, area, and angular speed into the equation gives $E_{\max} = 100 \times 0.05 \times 1 \times 2\pi$, resulting in a peak voltage of $31.4$ V.

The total opposition to alternating current in a series circuit containing resistance, inductance, and capacitance is governed by the formula $Z = \sqrt{(X_C - X_L)^2 + R^2}$. Based on the circuit parameters, the inductive reactance is 5 ohms, the capacitive reactance is 10 ohms, and the resistance is 10 ohms. Substituting these values into the relation gives the square root of the \sum of the squared reactance difference and the squared resistance, which evaluates to the square root of 125. Simplifying this radical yields $5\sqrt{5}$ ohms.

The behavior of reactive components in alternating current networks is determined by their frequency-dependent resistances, defined by the formulas $X_C = \frac{1}{2 \pi f C}$ and $X_L = 2 \pi f L$. When the frequency becomes extremely high, the capacitive reactance approaches zero, causing the capacitor to behave like a closed short circuit. Conversely, the inductive reactance approaches infinity, making the inductor act like an open circuit that halts all current through its branch. Substituting these limiting behaviors into the network configuration eliminates the open paths and bypasses the capacitors, leaving a simplified framework that results in a total effective impedance of 3 ohms.

An ideal transformer conserves total electrical power, meaning the input power supplied to the primary winding equals the output power consumed by the secondary load. This relationship can be expressed with the equation $V_p I_p = P_{\text{output}}$ where the secondary power is given as 60 W and the primary voltage is 220 V. Substituting these known values into the equation yields $220 × I_p = 60$, resulting in a primary current of 0.27 A.

The natural resonance frequency of a series circuit containing inductance and capacitance is given by the formula $f = \frac{1}{2\pi\sqrt{LC}}$. Substituting the given values for inductance and capacitance yields the expression $f = \frac{1}{2\pi\sqrt{10 \times 10^{-3} \times 1 \times 10^{-6}}}$. Evaluating this mathematical relation produces a frequency of $1.59$ kHz.

Resonance in a series LCR circuit occurs when the inductive reactance equals the capacitive reactance, which establishes the natural frequency of oscillation. The angular resonant frequency is determined using the equation $\omega = \frac{1}{\sqrt{LC}}$ Substituting the given values of 10 H for inductance and 10 microfarads for capacitance gives an angular frequency of 100 rad/s. Converting this and the angular frequency of the source from the voltage equation into standard frequency using $f = \frac{\omega}{2\pi}$ confirms that both frequencies are equal to $\frac{50}{\pi} \text{ Hz}$.

Inductive reactance depends directly on the frequency of the source and is quantified by the relation $X_L = 2\pi f L$. For the alternating current source operating at a frequency of $50\text{ Hz}$ with an inductance of $2\text{ mH}$, substituting these parameters into the formula gives $X_1 = 2\pi \times 50 \times 2 \times 10^{-3}$, which evaluates to $0.628\ \Omega$. When a direct current source replaces the alternating current source, the frequency drops to zero because direct current lacks any oscillation, causing the inductive reactance $X_2$ to vanish and become zero.

Power in an alternating current circuit relates root-mean-square voltage and current through the expression $P = V_{\text{rms}} I_{\text{rms}}$. Rearranging this formula yields the root-mean-square current $I_{\text{rms}} = \frac{P}{V_{\text{rms}}}$, which calculates to $\frac{100}{200} = 0.5\text{ A}$. The peak current $I_0$ is then determined by multiplying the root-mean-square current by $\sqrt{2}$, resulting in $0.5 \times \sqrt{2}$, which equals $0.707\text{ A}$.

In an alternating current circuit containing a pure capacitor, the current leads the voltage across the capacitor by a phase angle of $\frac{\pi}{2}$ radians because the reactive nature of capacitance causes the current waveform to peak ahead of the voltage waveform. Therefore, Statement I correctly identifies this fundamental phase relationship. However, Statement II incorrectly claims the phase difference is $\pi$, making that statement false.

Magnetic flux through a surface measures the total amount of magnetic field passing perpendicularly through a given area. It is calculated using the dot product of the magnetic field vector and the area vector, written as $\phi_B = B A \cos \theta$ Since the plane of the loop is perpendicular to the field, the angle between the magnetic field and the area vector is zero. Multiplying the magnetic field of 0.5 T by the 1 square meter area of the loop yields a magnetic flux of 0.5 Wb.

Faraday's law of electromagnetic induction combined with Ohm's law dictates that a rotating coil in a magnetic field generates an alternating electromotive force, producing a maximum current proportional to the peak induced voltage divided by the circuit resistance. The peak induced current is calculated using the formula $i_{\max} = \frac{NBA\omega}{R}$ where the area of the circular coil is given by $A = \pi r^2$. Substituting the given values for the number of turns, magnetic field, radius, angular speed, and resistance into this expression results in a maximum induced current of 1 A.

The root mean square value of an alternating current represents its effective direct current equivalent, defined as the steady current that would produce the same heating effect in a given resistance. Mathematically, this relationship is expressed by the formula $E_{\text{rms}} = \frac{E_0}{\sqrt{2}}$ which shows that the peak voltage $E_0$ is equal to $\sqrt{2}$ times the RMS value of the AC source.

Faraday's law of electromagnetic induction states that the magnitude of the induced electromotive force is equal to the time rate of change of magnetic flux, expressed as $|e| = \frac{d\phi}{dt}$. Differentiating the given flux function $\phi = 2t^3 + 4t^2 + 2t + 5$ with respect to time $t$ provides the instantaneous emf equation $e(t) = 6t^2 + 8t + 2$. Substituting the specified time value of $5\text{ s}$ into this derivative gives $6(5)^2 + 8(5) + 2$, which evaluates to $150 + 40 + 2$, resulting in a final induced emf of 192 V.

An ideal transformer assumes no power losses, meaning the electric power supplied to the primary circuit equals the power consumed by the secondary load, expressed as $V_p I_p = P_{\text{out}}$. Substituting the primary voltage of $220\text{ V}$ and the total output power of $44\text{ W}$ into this conservation equation gives $220 \times I_p = 44$. Solving this expression for the primary current $I_p$ results in $\frac{44}{220}$, which simplifies to $0.2\text{ A}$.

The impedance of an alternating current series circuit is defined by the formula $Z = \frac{V_{\text{rms}}}{I_{\text{rms}}}$ where the total root-mean-square voltage combines the individual component potentials through phasor addition as $V_{\text{rms}} = \sqrt{V_{R}^{2} + (V_{L} - V_{C})^{2}}.$ Substituting the given component voltages yields a total source voltage of 50 volts. Dividing this total voltage by the root-mean-square current of 10 amperes, derived from the given peak amplitude, results in an impedance of 5 ohms.

Mutual inductance is defined as the magnetic flux passing through a secondary loop per unit current in the primary loop. Assuming a current flows through the larger outer loop of radius $R_{1}$, the magnetic field generated at its center is expressed as $B = \frac{\mu_{0}I}{2R_{1}}$. Since the inner loop of radius $R_{2}$ is much smaller, the magnetic field is uniform across its cross-sectional area of $\pi R_{2}^{2}$, showing that mutual inductance scales directly with the square of the inner radius and inversely with the outer radius.

The resonant angular frequency of a series circuit is determined by the formula $\omega_{0} = \frac{1}{\sqrt{LC}}$, which calculates to 50 rad/s using the given parameters. The half-power frequencies are found at offsets of $\frac{R}{2L}$ above and below this resonant frequency. Evaluating this offset yields 4 rad/s, which when subtracted from and added to 50 rad/s gives 46 rad/s and 54 rad/s.