Electromagnetic Induction – JEE Main PhysicsPractice Questions & PYQs

The phenomenon of self-inductance describes how a coil resists changes in current by inducing a reverse voltage, governed by the formula $V = L \left| \frac{di}{dt} \right|$. This relation implies that the magnitude of the induced electromotive force is directly proportional to the product of the inductance and the rate of current decay.

Using the rearranged equation
$L = \frac{V \Delta t}{\Delta i}$
we substitute the values $V = 100 \text{ V}$, $\Delta i = 0.25 \text{ A}$, and $\Delta t = 0.025 \times 10^{-3} \text{ s}$. Performing this calculation yields $L = \frac{100 \times 0.025 \times 10^{-3}}{0.25} = 0.01 \text{ H}$, which is equivalent to $10 \text{ mH}$.

The generation of induced electromotive force in a rotating coil relies on Faraday's law, where a change in magnetic flux linkage creates a potential difference. When a circular coil rotates within a magnetic field at a constant angular velocity, the magnetic flux passing through it changes sinusoidally. The induced EMF at any instant corresponds to the time derivative of this flux, and its peak magnitude occurs when the coil plane is parallel to the magnetic field lines.

The angular frequency is determined by the duration for a half-turn; because the coil completes half a rotation in 0.2 s, the full period is 0.4 s, resulting in an angular velocity $\omega = 2\pi/T$, which equals $\frac{2\pi}{0.4}$ rad/s. The area of the coil is determined by its radius of 0.1 m, yielding an area $A = \pi(0.1)^2$ square meters. Maximum EMF arises from the product of the magnetic field strength, the area, and the angular frequency:
$\text{EMF}_{\max} = BA\omega = BA\left(\frac{2\pi}{T}\right)$
Substituting the values 3.0 x 10^-5 T for the magnetic field, 0.01$\pi$ square meters for the area, and 0.4 s for the period leads to the following expression:
$\text{EMF}_{\max} = (3.0 \times 10^{-5}) \times (0.01\pi) \times \left(\frac{2\pi}{0.4}\right) = \frac{6\pi^2}{4} \times 10^{-6} \text{ Volts}$
Approximating $\pi^2$ as 10 simplifies the result to 15 x 10^-6 Volts, which equals 15 microvolts.

Rotating an elliptical loop with area $A = \pi a b$ inside a uniform magnetic field $B$ induces an electromotive force due to the changing magnetic flux linkage, which is governed by Faraday's law of induction. As the loop rotates with angular frequency $\omega$, the induced EMF $\epsilon$ varies sinusoidally in time, creating an instantaneous power dissipation across the resistor defined by the relationship $P = \frac{\epsilon^2}{R}$.

The average power loss is found by calculating the time average of this instantaneous power over one full cycle of rotation. Since the square of a sinusoidal function averages to $\frac{1}{2}$ over a complete period, the average power dissipation simplifies to the term $\frac{(B (\pi a b) \omega)^2}{2R}$. Expanding this expression leads directly to the final result of $\frac{\pi^2 a^2 b^2 B^2 \omega^2}{2R}$.

To find the potential difference between two points in a circuit branch, apply Kirchhoff's Voltage Law by calculating the sum of potential changes along the path from the first point to the second. As the current of 1 A flows through the inductor while decreasing at a rate of $10^2$ A/s, the induced electromotive force acts in a way that contributes a potential drop of 5 V. By traversing the branch from P to Q, we account for this inductor voltage, a 30 V source, and the potential difference across the resistor, where the current produces a drop of $2 \times 1$ V. These potentials sum up according to the equation $V_P - 5 - 30 + 2 \times 1 = V_Q$, which results in a potential difference of $V_P - V_Q = 33.00$ V.

Faraday's law of induction dictates that a changing magnetic field through a conductive loop creates an induced electromotive force proportional to the rate of change of the magnetic flux. Since the magnetic field is uniform and perpendicular to the square loop, the flux is the product of the field and the loop area, meaning the induced EMF is $|\epsilon| = a^2 \frac{dB}{dt}$.

The current flowing through the loop depends on the wire's resistance, determined by its resistivity $\rho$, total length $L$, and cross-sectional area $A = \pi r^2$. For a square loop with a total wire length of 30 cm, the side length is $a = 0.075 \text{ m}$, while a wire diameter of 4 mm gives a radius of $r = 0.002 \text{ m}$. Calculating resistance gives $R = \frac{\rho L}{\pi r^2}$, which leads to the current formula $i = \frac{a^2 (dB/dt) \pi r^2}{\rho L}$. Using the values of $a = 0.075 \text{ m}$, $dB/dt = 0.032 \text{ Ts}^{-1}$, $\rho = 1.23 \times 10^{-8} \ \Omega\text{m}$, and $L = 0.3 \text{ m}$ yields $i \approx 0.61 \text{ A}$.

When an inductor and a resistor are connected in series with a battery, the current does not rise instantaneously to its steady-state value because the inductor opposes the change in current. The instantaneous current is described by the expression $I(t) = \frac{\epsilon}{R}(1 - e^{-Rt/L})$, where the time constant $t_c$ is equal to $L/R$. To determine the total charge that flows from the battery during the interval from $t = 0$ to $t = t_c$, we must integrate the current function with respect to time over this specific duration.

The total charge is given by the integral $q = \int_{0}^{t_c} \frac{\epsilon}{R}(1 - e^{-t/t_c}) dt$. Performing the integration yields $q = \frac{\epsilon}{R} \left[ t + t_c e^{-t/t_c} \right]_{0}^{t_c}$. Evaluating this at the specified limits, the lower bound at $t=0$ yields $t_c$ from the exponential term, while the upper bound at $t=t_c$ provides $t_c$ from the first term and $t_c e^{-1}$ from the second. The resulting expression simplifies to $q = \frac{\epsilon}{R} (t_c e^{-1})$. Substituting back the relation $t_c = \frac{L}{R}$ allows us to arrive at the final result of $\frac{\epsilon L}{eR^2}$.

Magnetic flux is defined as the dot product of the magnetic field vector and the area vector of a surface, expressed as $\phi = \vec{B} \cdot \vec{A}$. When a loop is composed of multiple planar sections, the total flux is obtained by calculating the sum of the flux through each component, which is equivalent to finding the dot product of the magnetic field with the net area vector of the loop.

The loop in this problem consists of two square faces: face ABCD, which lies in the $xy$-plane and has an area vector of $\vec{A}_{\text{ABCD}} = 25\hat{k}$, and face ADEF, which lies in the $yz$-plane and has an area vector of $\vec{A}_{\text{ADEF}} = 25\hat{i}$. These combine to form a net area vector of $\vec{A}_{\text{net}} = 25\hat{i} + 25\hat{k}$. Calculating the flux with the magnetic field $\vec{B} = 3\hat{i} + 4\hat{k}$ yields the following result:

$\phi = \vec{B} \cdot \vec{A}_{\text{net}} = (3\hat{i} + 4\hat{k}) \cdot (25\hat{i} + 25\hat{k}) = 75 + 100 = 175 \text{ Wb}$

Magnetic field lines generated by a current-carrying loop always form continuous, closed loops, which is a fundamental consequence of the non-existence of magnetic monopoles. Because these lines must complete their path, every magnetic field line that penetrates the interior area of the coil must necessarily return through the surrounding infinite plane. Since these field lines cross the interior region and the exterior plane in opposite directions relative to the surface normal, their associated fluxes are equal in magnitude but opposite in sign, yielding the relation $\phi_i = -\phi_0$.

Faraday's law of induction states that an induced electromotive force arises from the rate of change of magnetic flux through a conductor. Because the loop rotates in a uniform magnetic field, the flux is given by $\phi = BA \cos(\omega t)$, which makes the magnitude of the induced EMF $|e| = |BA \omega \sin(\omega t)|$. The maximum EMF magnitude occurs when the sine component is at its peak, $\omega t = \pi / 2$, whereas the minimum magnitude occurs when the flux change is momentarily zero, $\omega t = \pi$.

With a rotation period of 10 s, the angular frequency is $\omega = 2\pi / 10 = \pi / 5$. For the maximum EMF, we set $\omega t = \pi / 2$, resulting in $(\pi / 5) t = \pi / 2$, which solves to $t = 2.5$ s. For the minimum EMF, we set $\omega t = \pi$, which results in $(\pi / 5) t = \pi$, solving to $t = 5$ s.

Faraday's law of induction dictates that an electromotive force is generated in a coil whenever the magnetic flux threading through it changes, defined by the relationship $\varepsilon = - \frac{d\Phi}{dt}$. As the bar magnet enters the coil, the increasing magnetic flux generates a current that results in a positive deflection on the galvanometer. Once the magnet is fully positioned inside the coil, the magnetic flux remains constant and stops changing, which causes the induced voltage to drop to zero. Upon the magnet exiting the coil, the flux changes in the opposite sense, inducing a current that results in a negative deflection on the galvanometer.

Motional electromotive force arises when a conducting rod moves through a magnetic field, resulting in a potential difference across its length. A long straight wire carrying current $I$ produces a magnetic field at a radial distance $r$ given by $B = \frac{\mu_0 I}{2\pi r}$. As the connector of length $l$ slides with velocity $v$ perpendicular to these magnetic field lines, it experiences an induced EMF of magnitude $\mathcal{E} = Bvl$.

By substituting the magnetic field expression into the formula for motional EMF, the potential difference becomes $\mathcal{E} = \frac{\mu_0 I v l}{2\pi r}$. Since the loop has a resistance $R$, Ohm's Law allows us to determine the induced current $i$ by dividing the total induced EMF by the resistance. This calculation results in the final expression:

$i = \frac{\mu_0 I v l}{2\pi R r}$

Faraday's law of electromagnetic induction states that an electromotive force (EMF) is generated when the magnetic flux through a coil changes over time. Because coil $C_1$ is positioned concentrically inside coil $C_2$, it experiences a magnetic field $B = \frac{\mu_0 N_2 I}{2 R_2}$ created by the current in the outer coil. The total magnetic flux linkage $\phi$ is the product of the number of turns $N_1$, the magnetic field $B$, and the area $A_1 = \pi R_1^2$, resulting in the induced EMF expression $\varepsilon = \frac{d\phi}{dt} = \frac{\mu_0 N_1 N_2 \pi R_1^2}{2 R_2} \frac{dI}{dt}$.

The derivative of the current function $I(t) = 5t^2 - 2t + 3$ is $\frac{dI}{dt} = 10t - 2$, which evaluates to $8 \text{ A/s}$ at $t = 1 \text{ s}$. Substituting the values $N_1 = 500$, $N_2 = 200$, $R_1 = 1 \text{ cm}$, and $R_2 = 20 \text{ cm}$ into the formula yields an induced voltage of $0.8 \text{ mV}$. Equating this result to the given expression $\frac{4}{x} \text{ mV}$ leads to $x = 5.00$.

Electromagnetic induction follows Faraday's law, where the induced electromotive force in a loop is directly proportional to the time derivative of the magnetic flux passing through it. Given the magnetic field of the solenoid depends on the current $I(t) = I_0(t - t^2)$, the magnetic flux threading the ring is proportional to this time-dependent term, specifically $\phi = (\pi R^2 \mu_0 n I_0)(t - t^2)$. Applying the derivative, the induced EMF becomes $V_R = -\frac{d\phi}{dt} = -\frac{d}{dt} [(\pi R^2 \mu_0 n I_0)(t - t^2)]$, which simplifies to $V_R = (\pi R^2 \mu_0 n I_0)(2t - 1)$.

Because the induced current $I_R$ is obtained by dividing the induced EMF by the resistance of the ring, it shares the same temporal dependence as the EMF, governed by the term $(2t - 1)$. At the specific instant $t = 0.5$, this term equals zero, causing both the induced EMF and the induced current to vanish. For any time interval before $0.5$, the expression $(2t - 1)$ results in a negative value, while for any time after $0.5$, it becomes positive, indicating that the polarity of the induced EMF and the direction of the induced current must reverse as $t$ passes through $0.5$.

Faraday's Law of Induction establishes that the induced electromotive force in a closed loop is directly proportional to the rate of change of magnetic flux through it. When the magnetic field magnitude changes over time while the area remains constant, we can calculate the induced EMF using the relationship $\varepsilon = A \frac{dB}{dt}$.

In this problem, the area of the loop is 56 cm^2, and the magnetic field changes at a rate of 100 Gauss per second. Converting these to standard SI units, we have an area of 56 × 10^{-4} m^2 and a rate of change of 100 × 10^{-4} T/s. Substituting these values into our equation gives $\varepsilon = (100 \times 10^{-4} \text{ T/s}) \times (56 \times 10^{-4} \text{ m}^2) = 56 \times 10^{-6} \text{ V}$. Therefore, the induced EMF in the loop is 56 µV.

The energy stored within an inductor is proportional to the square of the current flowing through it, described by the expression $U = \frac{1}{2}Li^2$. When the current in the circuit changes over time, a self-induced electromotive force is generated in opposition to this change, governed by the relation $\varepsilon = L \frac{di}{dt}$. By evaluating the relationship between the induced voltage and the rate of current change, one can determine the inductance of the coil and subsequently calculate the total change in magnetic potential energy.

Given a self-induced electromotive force of 25 V and a current that increases from 10 A to 25 A over 1 s, the inductance is derived by rearranging the voltage equation:
$L = \frac{25}{15/1} = 5/3 \text{ Henry}$
With the inductance identified, the variation in energy as the current changes is calculated by finding the difference in energy between the initial and final states:
$\Delta E = \frac{1}{2} \times \frac{5}{3} \times (25^2 - 10^2)$
Computing this expression leads to $\frac{5}{6} \times 525$, which results in a total energy change of 437.5 J.

Faraday's law dictates that the induced current in the outer coil is proportional to the negative rate of change of magnetic flux, which is confined solely within the solenoid's radius $R$ because the magnetic field outside is negligible. Given the solenoid current $I(t)=k t e^{-\alpha t}$, the induced current follows the time derivative of this expression:

$I_{ind} \propto -\frac{d}{dt}(k t e^{-\alpha t}) = -k e^{-\alpha t}(1-\alpha t)$

This relationship results in a graph that starts at a negative value at $t=0$, crosses the horizontal axis at $t=1/\alpha$ where the derivative of the flux is zero, and then approaches zero asymptotically from the positive side as time increases.

When a conducting loop moves through a magnetic field, the motion of charge carriers within the conductor across magnetic field lines induces an electromotive force known as motional EMF. This potential difference is determined by the relationship $E = B l v$. As the loop is part of a larger circuit, the induced current flows according to Ohm's law, where the current is the ratio of this induced EMF to the total resistance of the entire path.

Substituting the given values, the induced EMF is calculated as $E = 1 \times 5 \times 10^{-2} \times 10^{-2} = 5 \times 10^{-4}$ volts. Since the network adds a resistance of 1.3 ohms to the loop's own 1.7 ohms, the total resistance in the circuit is 3.0 ohms. Dividing the induced EMF by this total resistance yields $I = \frac{5 \times 10^{-4}}{3.0} = 1.66 \times 10^{-4}$ amperes, which is approximately 170 microamperes.

The rate at which energy dissipates as heat through a resistor is given by $i^2 R$, while the rate at which energy is stored in the magnetic field of an inductor is defined by $L i \frac{di}{dt}$. Equating these two rates produces $i^2 R = L i \frac{di}{dt}$, which reduces to the simplified condition $i R = L \frac{di}{dt}$. In a series RL circuit, the growth of current over time is $i = i_0 (1 - e^{-t/\tau})$ and its rate of change is $\frac{di}{dt} = \frac{i_0}{\tau} e^{-t/\tau}$, where the time constant is $\tau = L/R$. Substituting these expressions into the balance condition yields $i_0 (1 - e^{-t/\tau}) R = L \frac{i_0}{\tau} e^{-t/\tau}$. Because $\tau = L/R$, this simplifies to $(1 - e^{-t/\tau}) R = R e^{-t/\tau}$, which further reduces to $1 - e^{-t/\tau} = e^{-t/\tau}$. This equality implies $2 e^{-t/\tau} = 1$, or $e^{t/\tau} = 2$, which solves to $t = \tau \ln 2$. With $L = 20$ H and $R = 10$ $\Omega$, the time constant is $\tau = 20/10 = 2$ s, giving the final value $2 \ln 2$.

When a conductive wire moves through a magnetic field such that its velocity, length, and the magnetic field vector are mutually perpendicular, an electromotive force is induced across its ends due to the magnetic Lorentz force acting on the charge carriers. This physical principle allows for the determination of the induced potential difference using the formula $e = Blv$. By substituting the provided values-a magnetic field strength of $0.3 \times 10^{-4} \, \text{Wb/m}^2$, a velocity of $5.0 \, \text{m/s}$, and a length of $10 \, \text{m}$-the resulting electromotive force is calculated as:
$e = (0.3 \times 10^{-4}) \times 5.0 \times 10 = 1.5 \times 10^{-3} \, \text{V}$

The self-inductance of a coil depends on the cross-sectional area of the frame and the total number of turns wound along its perimeter. The area of the equilateral triangle frame is proportional to the square of the side length, $a^2$, and since the number of turns per unit length is kept constant, the total number of turns $N$ is proportional to the perimeter of the frame, which scales linearly as $a$. Applying these relationships to the geometry of the coil, the self-inductance scales proportionally such that increasing the side length of the frame by a factor of 3 results in the self-inductance increasing by a factor of 3.