Electromagnetic Induction – JEE Main PhysicsPractice Questions & PYQs

Faraday's law of induction indicates that an induced electromotive force arises only when the magnetic flux passing through a conducting loop changes over time, as described by the relationship $\epsilon = -\frac{d\phi}{dt}$. As the square loop moves at a constant speed of 2 cm/s, it covers a total distance of 20 cm after 10 s of motion. Given that the loop side length is 15 cm and the magnetic field region spans 50 cm, the entire loop resides fully inside the uniform magnetic field at this instant. Because the magnetic flux enclosed by the loop remains constant while it is completely submerged in the uniform field, the rate of change of flux is zero, meaning the induced EMF is zero.

Maxwell's equations act as the definitive set of laws describing how electric charges and currents create electric and magnetic fields. Gauss's law for electrostatics equates the electric flux through a surface to the total enclosed charge as $\oint \vec{E} \cdot \vec{da} = \frac{1}{\epsilon_0} \int \rho dV$, whereas Gauss's law for magnetostatics dictates the zero net magnetic flux through a closed surface, represented by $\oint \vec{B} \cdot \vec{da} = 0$. The interaction between time-varying magnetic fields and electric fields is governed by Faraday's law, $\oint \vec{E} \cdot \vec{dl} = -\frac{d}{dt} \int \vec{B} \cdot \vec{da}$, and the generation of magnetic fields by electric current is described by Ampere's law, $\oint \vec{B} \cdot \vec{dl} = \mu_0 I$. Aligning these fundamental principles requires linking Gauss's law of magnetostatics with the second equation, Faraday's law with the third, Ampere's law with the fourth, and Gauss's law of electrostatics with the first.

Faraday's law of electromagnetic induction states that the induced electromotive force in a circuit is equal to the negative rate of change of magnetic flux, represented by the formula
$\varepsilon = -\frac{d\phi}{dt}$
By differentiating the flux expression $\phi = 5t^2 - 36t + 1$ with respect to time, the rate of change is found to be $10t - 36$. Evaluating this expression at $t = 2$ s results in a value of $-16$, and applying the negative sign from Faraday's law yields an induced EMF magnitude of $16$ V. Using Ohm's law, which relates current, voltage, and resistance through
$I = \frac{\varepsilon}{R}$
dividing the induced EMF of $16$ V by the resistance of $8 \Omega$ gives an induced current of $2$ A.

Faraday's law of induction states that the magnitude of an induced electromotive force in a circuit is determined by the rate of change of magnetic flux through it. The square loop, with a side of 10 cm, has an area of 0.01 square meters, and since it is oriented in the east-west plane, its area vector is directed north-south. The magnetic field of 0.20 T is oriented north-east, meaning it splits into two equal components, each with a magnitude of $\frac{0.20}{\sqrt{2}}$ T, along the east and north directions. When calculating the magnetic flux, the dot product of the magnetic field and the area vector ensures that only the north component contributes to the flux, giving $\Phi = 0.01 \times \frac{0.20}{\sqrt{2}}$. Given that the magnetic field decreases to zero over a time interval of 1 s, the magnitude of the induced emf is found by dividing the initial flux by the time, resulting in an emf of $\sqrt{2} \times 10^{-3}$ V. Matching this result to the expression $\sqrt{x} \times 10^{-3}$ V confirms that x is 2.

The average induced electromotive force in a loop is determined by Faraday's Law of Induction, which relates the voltage generated to the negative rate of change of magnetic flux through the area of the loop, defined by:
$\varepsilon = - \frac{\Delta \phi}{\Delta t}$
The surface area of the rectangular loop is $5\text{ m}^2$, calculated by multiplying the length of $2.5\text{ m}$ and the width of $2\text{ m}$. With a magnetic field of $4\text{ T}$ and an angle of $60^\circ$, the initial magnetic flux through the loop is $4 \times 5 \times \cos 60^\circ$, which equals $10\text{ Wb}$ since $\cos 60^\circ$ is $0.5$. Because the loop is removed from the field, the final flux becomes $0\text{ Wb}$ after $10\text{ s}$. Inserting these values into the induction formula yields an average emf of $- \frac{0 - 10}{10}$, resulting in $+1\text{ V}$.

The generation of an induced electromotive force depends entirely on a time-varying magnetic flux, which is mathematically expressed as $\Phi = \vec{B} \cdot \vec{A}$. Within a uniform magnetic field, moving the coil at any speed maintains constant values for the magnetic field, area, and orientation, resulting in no change in flux. Conversely, rotating the coil alters the angle between the coil's surface and the magnetic field lines, while changing the physical area of the loop directly modifies the amount of flux passing through it, both of which successfully induce an emf.

The induced electromotive force in the loop is fundamentally determined by Faraday's law, which states that an emf is generated by a time-varying magnetic flux. Inside a solenoid, the magnetic field is uniform and given by $B = \mu_0 n I$, where $n$ represents the turns per unit length and $I$ is the current. Because the current varies sinusoidally as $I = I_0 \sin(\omega t)$, the magnetic field oscillates, inducing an emf in the loop with an amplitude proportional to the product of the peak magnetic field, the loop area, and the angular frequency.

First, convert the dimensions and properties to SI units: the loop side length of $2.0 \text{ cm}$ leads to an area $A = (0.02 \text{ m})^2 = 4 \times 10^{-4} \text{ m}^2$, and the turns density $n = 50 \text{ turns/cm} = 5000 \text{ turns/m}$. The peak magnetic field $B_0$ produced by the current amplitude $I_0 = 2.5 \text{ A}$ is:
$B_0 = \mu_0 n I_0 = (4\pi \times 10^{-7}) (5000) (2.5) = 5\pi \times 10^{-3} \text{ T}$
Given the angular frequency $\omega = 700 \text{ rad s}^{-1}$, the amplitude of the induced emf $\varepsilon_0$ is calculated using the relationship $\varepsilon_0 = B_0 A \omega$:
$\varepsilon_0 = (5\pi \times 10^{-3}) (4 \times 10^{-4}) (700)$
Substituting $\pi = \frac{22}{7}$ into this expression results in:
$\varepsilon_0 = 5 \times \frac{22}{7} \times 4 \times 700 \times 10^{-7} = 44 \times 10^{-4} \text{ V}$
Comparing this to the provided form $x \times 10^{-4} \text{ V}$, the value of $x$ is 44.

The potential difference between two points in a circuit is determined by applying Kirchhoff's Voltage Law while traversing the branch containing the resistor, the inductor, and the voltage source. Starting at point A with potential $V_A$, we account for the voltage drops across the resistor, the inductor, and the battery to reach point B with potential $V_B$, following the equation:
$V_A - IR - L\frac{dI}{dt} - 12 = V_B$
Substituting the given values of resistance $R = 2\,\Omega$, current $I = 12\,\text{A}$, inductance $L = 6\,\text{H}$, and the rate of change of current $\frac{dI}{dt} = -1\,\text{A/s}$ into the equation gives:
$V_A - (2 \times 12) - 6(-1) - 12 = V_B$
Simplifying these terms leads to $V_A - 24 + 6 - 12 = V_B$, which rearranges to $V_A - V_B = 24 - 6 + 12 = 30\,\text{V}$. Thus, the potential difference $V_{AB}$ at this instant is 30 V.

The magnetic flux through a coil is mathematically defined as the dot product of the magnetic field vector and the area vector, expressed by the equation $\phi = \vec{B} \cdot \vec{A} = BA \cos \theta$. Since the magnitude of the flux depends directly on the strength of the magnetic field $B$, the surface area $A$ enclosed by the coil, and the angle $\theta$ between the magnetic field and the area vector, any variation in these parameters will change the flux. Consequently, altering the magnitude of the magnetic field, changing the effective area of the coil, or rotating the coil to change the angle between the magnetic field and the plane of the coil will each result in a change in the magnetic flux.

Faraday's law of induction states that the total charge flowing through a circuit equals the net change in magnetic flux divided by the circuit's resistance, expressed as $Q = \frac{N A \Delta B}{R}$ for a coil of $N$ turns. Because the magnetic field intensity changes from 1.5 T to 1.5 T in the opposite direction, the net change in the field is $\Delta B = 1.5 - (-1.5) = 3 {\text{T}}$. The total charge flowing through the circuit is determined by using the values $N = 100$, $A = 24 \times 10^{-4} {\text{m}}^2$, and $R = 12 \Omega$:
$Q = \frac{100 \times 24 \times 10^{-4} \times 3}{12}$
Solving this equation yields $0.06 {\text{C}}$, which corresponds to a total charge flow of $60 {\text{mC}}$.

When a conducting object, such as a metallic cube, moves through a magnetic field, the free electrons inside the conductor experience a magnetic Lorentz force. This force causes charge separation between the faces perpendicular to the direction of motion and the magnetic field, generating an internal electric field that results in an induced potential difference. Equilibrium is reached when the electric force on these charges balances the magnetic force, a relationship that leads to the formula $\Delta V = vBL$. Substituting a velocity of 2 ms^-1, a magnetic field of 0.5 T, and a side length of 0.15 m into this formula gives:

$\Delta V = 2 \times 0.5 \times 0.15 = 0.15$

This result is in volts, and converting it to millivolts gives 150 mV.

Mutual inductance is governed by the relation $\Phi = Mi$, where the magnetic flux $\Phi$ through the smaller coil is directly proportional to the current $i$ in the larger coil. Because the circular loop is placed inside a much larger square loop, the magnetic field produced by the square loop is approximately uniform across the entire area of the circle. This total magnetic field is the sum of the contributions from the four straight segments of the square, with each segment acting as a finite conductor at a perpendicular distance of $L/2$ from the center.

The magnetic field contribution from a single side of the square is $B_{side} = \frac{\mu_0 i}{4 \pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i \sqrt{2}}{2 \pi L}$. Summing this contribution for all four sides gives a total magnetic field of $B = \frac{2 \sqrt{2} \mu_0 i}{\pi L}$. The total magnetic flux linking the circular loop of area $\pi R^2$ is $\Phi = \frac{2 \sqrt{2} \mu_0 i R^2}{L}$. By dividing this flux by the current $i$, the mutual inductance is $M = \frac{2 \sqrt{2} \mu_0 R^2}{L}$.

Faraday's law of induction provides the foundation for this problem, as it relates the induced electromotive force to the rate of change of magnetic flux through a loop. When the radius of a circular conducting loop increases within a uniform magnetic field, the enclosed area grows, causing the magnetic flux to change even though the field strength remains constant. Because the magnetic field is perpendicular to the plane of the loop, the flux is given by $\Phi = B \pi r^2$, and the magnitude of the induced emf is found by differentiating this expression with respect to time, resulting in $\varepsilon = B \pi (2r) \frac{dr}{dt}$. Substituting the values $B = 0.4 \text{ T}$, $r = 0.02 \text{ m}$, and $\frac{dr}{dt} = 0.001 \text{ m/s}$ into the derivative formula yields $\varepsilon = 0.4 \times \pi \times 2 \times 0.02 \times 0.001$. Evaluating this product results in $\varepsilon = 16 \pi \times 10^{-6} \text{ V}$, and considering $\pi \approx 3.14$, the calculation gives an induced emf of $50 \mu\text{V}$.

According to Faraday's law of electromagnetic induction, the induced electromotive force in a loop is determined by the rate of change of magnetic flux through it. Since the loop is positioned perpendicular to a uniform magnetic field $B=0.8 {\text{T}}$, the magnetic flux $\Phi$ is calculated as the product of the field and the area of the circular loop, denoted by $\Phi = B \pi r^2$. Differentiating this expression with respect to time provides the relationship for the induced EMF:

$\mathcal{E} = \frac{d}{dt}(B \pi r^2) = 2 \pi r B \frac{dr}{dt}$

Substituting the given values into this formula, where the radius $r=10 {\text{cm}} = 0.1 {\text{m}}$ and the constant rate of shrinkage $\frac{dr}{dt} = 2 {\text{cm/s}} = 0.02 {\text{m/s}}$, the expression becomes $\mathcal{E} = 2 \times \pi \times 0.1 \times 0.8 \times 0.02$. This calculation yields approximately $0.01005 {\text{V}}$, which is equal to $10.05 {\text{mV}}$. Rounding this result to the nearest integer gives $10 {\text{mV}}$.

When a galvanometer coil oscillates within a magnetic field, the movement of the metallic core induces eddy currents within the conductive material due to electromagnetic induction. According to Lenz's law, these induced currents create an opposing magnetic torque that acts directly against the motion of the coil. This electromagnetic damping effect effectively extracts kinetic energy from the system, which serves to bring the coil to rest as quickly as possible.

Mutual inductance arises when a changing current in one coil induces a magnetic flux through an adjacent coil. For two concentric coils, we can consider the outer coil with radius $a = 10 \text{ m}$ and $N_2 = 200$ turns as the source of a magnetic field, which is approximately uniform across the inner coil with radius $b = 0.01 \text{ m}$ and $N_1 = 10$ turns. The magnetic field at the center of the outer coil is given by the expression $B = \frac{\mu_0 N_2 I}{2a}$.

The flux linkage through the inner coil is calculated as $\phi = B N_1 (\pi b^2)$, which leads to the mutual inductance formula $M = \frac{\mu_0 N_1 N_2 \pi b^2}{2a}$. Substituting the values $N_1 = 10$, $N_2 = 200$, $b = 0.01 \text{ m}$, and $a = 10 \text{ m}$ with $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$, the mutual inductance is $M = \frac{4\pi \times 10^{-7} \times 10 \times 200 \times \pi \times (0.01)^2}{2 \times 10}$. Simplifying this expression yields $M = 4 \pi^2 \times 10^{-9} \text{ H}$. Given the approximation $\pi^2 = 10$, the result is $4 \times 10^{-8} \text{ H}$.

In a DC circuit, an ideal inductor eventually reaches a steady state where it behaves as a simple conducting wire with zero resistance. Given that the inductors effectively short-circuit their specific branches long after the switch is closed, the three $12 \Omega$ resistors are left connected in parallel across the $12 \text{V}$ battery. The equivalent resistance for these three resistors in parallel is $R_{eq} = \frac{12}{3} = 4 \Omega$. Applying Ohm's law, the steady current drawn from the battery is:
$I = \frac{12 \text{V}}{4 \Omega} = 3 \text{A}$

Faraday's law of induction states that the magnitude of the induced electromotive force in a closed loop is equal to the rate of change of magnetic flux, given by the equation $\varepsilon = A \frac{dB}{dt}$. The radius of the circular loop is $r = \frac{10}{\sqrt{\pi}} \text{ cm}$, which converts to $r = \frac{0.1}{\sqrt{\pi}} \text{ m}$. Using this radius, the area is calculated as $A = \pi r^2 = \pi \left(\frac{0.01}{\pi}\right) = 0.01 \text{ m}^2$. Since the magnetic field decreases from $0.5 \text{ T}$ to $0 \text{ T}$ over $0.5 \text{ s}$ at a steady rate, the field changes at a rate of $\frac{dB}{dt} = \frac{0.5 \text{ T}}{0.5 \text{ s}} = 1 \text{ T/s}$. Substituting these values into the Faraday's law equation yields an induced electromotive force of $\varepsilon = 0.01 \text{ m}^2 \times 1 \text{ T/s} = 0.01 \text{ V}$, which equals $10 \text{ mV}$.

When a bar magnet falls through a metallic pipe, the continuously changing magnetic flux linked with the pipe's interior induces eddy currents in the metal according to Faraday's Law of induction. These circulating currents generate a magnetic field that opposes the motion of the falling magnet as dictated by Lenz's Law, resulting in a retarding force that significantly slows its descent. Because a non-magnetic bar does not induce these eddy currents, it experiences no such opposing force and falls much faster under the influence of gravity alone. Thus, the presence of eddy currents is the direct cause of the increased time required for the magnetic bar to travel through the pipe.

When a conducting rod moves through a uniform magnetic field, it generates a motional electromotive force $e = Bvl$ that drives a current $I = e / R$ through the circuit, resulting in an opposing magnetic force $F = IlB$. Combining these principles yields the general formula $F = \frac{B^2 l^2 v}{R}$. By substituting the given values $B = 0.15 \text{ T}$, $l = 1 \text{ m}$, $v = 4 \text{ m/s}$, and $R = 5 \Omega$, the expression becomes $F = \frac{(0.15)^2 \times (1)^2 \times 4}{5}$. This calculation results in $F = 18 \times 10^{-3} \text{ N}$.