Electromagnetic Induction – JEE Main PhysicsPractice Questions & PYQs

Faraday's law of electromagnetic induction states that the induced charge $Q$ flowing through a closed circuit is proportional to the change in magnetic flux $\Delta \Phi$ divided by the resistance $R$ of the loop, defined by the relation:

$Q = \frac{\Delta \Phi}{R}$

Because the magnetic flux is the product of the uniform magnetic field $B(t)$ and the area $A$ of the loop, the total change in flux during the interval from $t = 0$ to $t = 10 \, ms$ depends on the change in the magnetic field. With the area $A = 3.5 \times 10^{-3} \, m^2$ and the field $B(t) = 0.4 \sin(50\pi t)$, the initial field at $t = 0$ is $B_i = 0$. The final field at $t = 10 \, ms$ ($0.01 \, s$) is $B_f = 0.4 \sin(50\pi \times 0.01) = 0.4 \sin(\pi/2) = 0.4 \, T$. The change in flux is calculated as $\Delta \Phi = A(B_f - B_i) = 3.5 \times 10^{-3} \times (0.4 - 0) = 1.4 \times 10^{-3} \, Wb$. Substituting this flux change and the resistance $R = 10 \, \Omega$ into the charge equation yields $Q = \frac{1.4 \times 10^{-3}}{10} = 1.4 \times 10^{-4} \, C$, which is $0.14 \, mC$.

The self-inductance of a solenoid is fundamentally governed by the magnetic flux generated per unit current, which depends directly on the solenoid's physical dimensions. Using the relationship $\text{Inductance} = \frac{\mu N^2 A}{L}$, where $\mu$ is the magnetic permeability, $N$ is the total number of turns, $A$ is the cross-sectional area, and $L$ is the length, we can isolate the variables in question. Because $N$ and $A$ are held constant, the expression demonstrates that the inductance is inversely proportional to the length of the solenoid, resulting in the relationship $\text{Inductance} \propto \frac{1}{L}$.

The behavior of coupled coils is defined by the principle of mutual inductance, which asserts that the magnetic flux linked with one coil due to a current in another is proportional to that current, governed by a reciprocal relationship. Specifically, the relationship is expressed as $\phi_Q = M i_P$ and $\phi_P = M i_Q$, where the mutual inductance $M$ remains constant for the pair.

Initially, given that a current of 3 A produces a flux $\phi_Q = 10^{-3} \text{ Wb}$, we determine $M$ using the equation $M = \frac{10^{-3}}{3}$. Applying this constant to the second scenario where a current of 2 A flows through coil Q, the flux through coil P is determined by the equation $\phi_P = \frac{10^{-3}}{3} \times 2$. Evaluating this expression yields the result $\phi_P = 6.67 \times 10^{-4} \text{ Wb}$.

When a conducting object moves through a uniform magnetic field, charge carriers experience a Lorentz force that induces a potential difference across the conductor, a phenomenon known as motional EMF. This potential difference is calculated using the relationship $\epsilon = \vec{l} \cdot (\vec{v} \times \vec{B})$, where $\vec{l}$ represents the length vector between the faces of interest. With a velocity of 6 m/s in the positive y-direction ($\hat{j}$) and a magnetic field of 0.1 T in the positive z-direction ($\hat{k}$), the cross product $\vec{v} \times \vec{B}$ yields a field value of $0.6 \hat{i}$ V/m. Since the faces are perpendicular to the x-axis, the relevant length is 2 cm or 0.02 m in the x-direction. The induced EMF is the dot product of this length and the induced electric field, calculated as $\epsilon = 0.02 \times 0.6 = 0.012$ V, which is equivalent to 12 mV.

Inductance in long solenoids is fundamentally governed by the magnetic flux linkage, which is directly proportional to the winding density, cross-sectional area, and permeability of the core. Because the mutual inductance $M$ between these two co-axial coils is defined by $M = \mu_0 n_1 n_2 \pi r_1^2 l$ and the self-inductance $L_1$ of the inner coil is $L_1 = \mu_0 n_1^2 \pi r_1^2 l$, where $l$ represents the common length of the solenoids, we can evaluate their relationship through a simple ratio. When dividing $M$ by $L_1$, the shared physical constants $\mu_0$, $\pi$, $r_1^2$, and $l$ cancel out entirely, leaving the ratio as $n_2/n_1$.

The magnetic field at the center of the large circular coil is given by $B = \frac{\mu_0 I}{2R}$. Since the smaller coil has an area $A = \pi r^2$ and rotates about its diameter with an angular velocity $\omega$, the angle between the area vector and the magnetic field is $\theta = \omega t$, making the magnetic flux passing through it $\phi = B A \cos(\omega t) = \frac{\mu_0 I \pi r^2}{2R} \cos(\omega t)$. Applying Faraday's law, the induced electromotive force is $\varepsilon = -\frac{d\phi}{dt} = -\frac{d}{dt} \left( \frac{\mu_0 I \pi r^2}{2R} \cos(\omega t) \right)$, which simplifies to $\varepsilon = \frac{\mu_0 I}{2R} \omega \pi r^2 \sin(\omega t)$.

The magnetic flux $\Phi$ linked with a coil of $n$ turns in a uniform magnetic field $B$ is given by $\Phi = nBA \cos(\theta)$, where $\theta$ represents the angle between the area vector and the magnetic field. As the coil rotates with angular velocity $\omega$, the angle changes with time such that $\theta = \omega t$, allowing the induced electromotive force to be determined by Faraday's law of induction:
$e = -\frac{d\Phi}{dt} = -\frac{d}{dt}(nBA \cos(\omega t)) = nBA\omega \sin(\omega t)$
The maximum value of the induced e.m.f. is reached when $\sin(\omega t) = 1$, which equates to $nBA\omega$.

The terminal speed is reached when the gravitational force component pulling the rod down the incline, $mg \sin \theta$, is perfectly balanced by the upward magnetic force $F_m$ acting on the rod. The magnetic force is defined as $F_m = i l B$, where $i$ is the induced current, and the induced electromotive force in the rod is $\varepsilon = B v l$, resulting in an induced current $i = \frac{B v l}{R}$. Equating the gravitational component to the magnetic force gives $mg \sin \theta = \left(\frac{B v l}{R}\right) l B$, which simplifies to $mg \sin \theta = \frac{B^2 l^2 v}{R}$. Solving for the velocity $v$ yields the terminal speed $v = \frac{mg R \sin \theta}{B^2 l^2}$.

The phenomenon of mutual induction dictates that the electromotive force induced in a circuit is proportional to the rate of change of current in an adjacent circuit, expressed by the relationship $\varepsilon = M \frac{dI}{dt}$. The magnetic field at the center of the outer loop of radius $b$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2b}$, and assuming the inner loop of radius $a$ is sufficiently small to experience a uniform field, the magnetic flux through it is $\Phi = B(\pi a^2) = \frac{\mu_0 \pi a^2}{2b} I$. By comparing this to the definition of mutual inductance, $\Phi = MI$, we identify the mutual inductance as $M = \frac{\mu_0 \pi a^2}{2b}$. Given that the current varies as $I = I_0 \cos(\omega t)$, its derivative with respect to time is $-I_0 \omega \sin(\omega t)$, which leads to an induced electromotive force of $\varepsilon = M \frac{dI}{dt} = \frac{\mu_0 \pi a^2}{2b} I_0 \omega \sin(\omega t)$, equivalent to the expression $\frac{\pi \mu_0 I_0}{2} \frac{a^2}{b} \omega \sin(\omega t)$.

The relationship between induced charge $q$, change in magnetic flux $\Delta\phi$, and resistance $R$ is defined by $q = \frac{\Delta\phi}{R}$. The total charge passed through the coil corresponds to the area enclosed by the current-time graph. Because the graph depicts a triangle, the total charge is calculated as $\frac{1}{2} \times 10 \times 0.5$, resulting in 2.5 Coulombs. Rearranging the initial relationship to $\Delta\phi = q \times R$ and substituting 2.5 Coulombs for the charge and 100 ohms for the resistance, the total change in flux is calculated to be 250 Wb.

Stable equilibrium of a current-carrying loop in a magnetic field occurs when the potential energy $U = -\vec{m} \cdot \vec{B}$ is minimized. This condition is satisfied when the magnetic dipole moment $\vec{m}$ is aligned parallel to the magnetic field vector $\vec{B}$, which corresponds to an angle of $\theta = 0^\circ$ between the two vectors. Therefore, the stable orientation is achieved when the loop is positioned such that its area vector, determined by the right-hand rule, points in the same direction as the magnetic field.

When a magnetic field changes over time through a conducting loop, it induces an electromotive force (EMF) proportional to the rate of change of magnetic flux, as defined by Faraday's law. The magnetic flux $\Phi$ is the product of the field $B(t) = B_0 e^{-t/\tau}$ and the area $A = \pi r^2$, and differentiating this with respect to time gives the induced EMF magnitude $|\varepsilon| = \frac{\pi r^2 B_0}{\tau} e^{-t/\tau}$. The power dissipated as heat in the resistance $R$ follows the relation $P = \varepsilon^2 / R$, so to find the total heat energy, we integrate this power over time from zero to infinity.

$H = \int_{0}^{\infty } \frac{\varepsilon^2}{R} dt = \frac{\pi^2 r^4 B_0^2}{\tau^2 R} \int_{0}^{\infty } e^{-2t/\tau} dt$

Evaluating the exponential integral $\int_{0}^{\infty } e^{-2t/\tau} dt$ results in $\frac{\tau}{2}$, which, when multiplied by the preceding constant terms, simplifies to $\frac{\pi^2 r^4 B_0^2}{2\tau R}$.

When the metallic sheet moves through the uniform magnetic field, free electrons within the conductor experience a magnetic Lorentz force that drives them to one side, establishing a charge separation that builds an opposing internal electric field until equilibrium is reached. In this steady state, the magnetic force $qvB$ is exactly balanced by the internal electric force $qE$, which requires the internal field to satisfy $E = vB$. For a thin conducting sheet, the electric field generated by the induced surface charge densities $\sigma_1$ and $\sigma_2$ is related to the charge density by $E = \frac{\sigma}{\epsilon_0}$. Equating these two expressions for $E$ yields a surface charge density magnitude of $\sigma = \epsilon_0 v B$, which, corresponding to the observed charge distribution, results in $\sigma_1 = \epsilon_0 v B$ and $\sigma_2 = -\epsilon_0 v B$.

An ideal solenoid produces a magnetic field that is uniform and confined strictly within its interior, meaning there is no external magnetic field interacting with the environment outside the coil. Due to the perfect cylindrical symmetry of the coaxial arrangement, the magnetic interactions between the two coils do not result in a net directional force on either structure. Consequently, the net magnetic force exerted by the outer solenoid on the inner one, $\vec{F}_1$, and the force exerted by the inner solenoid on the outer one, $\vec{F}_2$, are both zero, resulting in $\vec{F}_1 = \vec{F}_2 = 0$.

Motional EMF arises when a conductor moves through a magnetic field, and for a rotating rod, we calculate the total voltage by summing the contributions of infinitesimal segments. Because the rod is attached to a string of length $2l$, the rod occupies the region from $2l$ to $3l$ relative to the pivot. Each element of length $dx$ at a distance $x$ moves with a linear speed of $v = \omega x$, creating an induced EMF of $de = B(\omega x) dx$.

Integrating this expression over the rod's span from $x = 2l$ to $x = 3l$ yields the total potential difference:

$e = \int_{2l}^{3l} B\omega x \, dx$

$e = B\omega \left[ \frac{x^2}{2} \right]_{2l}^{3l}$

$e = \frac{B\omega}{2} (9l^2 - 4l^2)$

$e = \frac{5B\omega l^2}{2}$

The concept of reciprocity in mutual induction simplifies this problem significantly, allowing us to determine the flux through the larger loop by calculating the flux that would be linked with the smaller loop if the current were flowing through the larger one. Because the mutual inductance between two circuits is a constant property depending only on their geometry and relative orientation, the magnetic flux $\phi = M I$ remains consistent regardless of which loop serves as the source of the magnetic field. Calculating the flux directly from the small loop is difficult due to its non-uniform field, so we treat the larger loop as the primary coil, assuming its magnetic field is approximately uniform across the cross-section of the smaller loop.

The magnetic field produced by the larger loop of radius $R$ at a point on its axis at distance $d$ from its center is defined by $B = \frac{\mu_0 I R^2}{2(R^2 + d^2)^{3/2}}$. By taking the product of this magnetic field and the area of the smaller loop, $A = \pi r^2$, we define the total magnetic flux linkage. Using the provided values of $R = 0.2$ m, $d = 0.15$ m, $r = 0.003$ m, $I = 2.0$ A, and the magnetic permeability of free space $\mu_0 = 4\pi \times 10^{-7}$ T$\cdot$m/A, the calculation for the flux $\phi$ follows the expression:

$\phi = \frac{\mu_0 I R^2 \pi r^2}{2(R^2 + d^2)^{3/2}}$

Substituting the physical dimensions and current into this equation, the denominator term $(R^2 + d^2)^{3/2}$ evaluates to $0.015625$. Combining these factors, the magnetic flux linked with the loops is calculated as $9.1 \times 10^{-11}$ weber.

This phenomenon is rooted in Lenz's law, which is a direct consequence of the law of conservation of energy. As the coil oscillates, the periodic change in magnetic flux $\phi$ through the nearby aluminum plate induces circulating eddy currents within the metal. These currents generate an opposing magnetic field that acts against the motion of the coil, creating a resistive force known as electromagnetic damping that gradually extracts kinetic energy from the system until the oscillations cease.

The motional electromotive force is generated when a conductor moves through a magnetic field, as the magnetic Lorentz force separates charge carriers to create a potential difference across the conductor. Since the boat's vertical aerial moves horizontally through the Earth's horizontal magnetic field, the vectors of velocity, field, and length are mutually perpendicular, and the induced EMF can be determined using the relationship $E = B l v$. Substituting the specified values into this equation gives:

$E = (5.0 \times 10^{-5}) \times 2 \times 1.50$

This calculation results in an EMF of 15 x 10^-5 V, which simplifies to an induced EMF of 0.15 mV.

The motion of a conductor with length $l$ at velocity $v$ within a magnetic field $B$ induces an electromotive force calculated as $\epsilon = Blv$. This induced voltage acts as the power source for the entire loop circuit. Because the geometry of the loop is symmetric, the total induced current splits evenly between the two parallel paths, confirming the relationship $I_1 = I_2$.

Based on the circuit configuration, the current in each individual branch is determined to be $I_1 = I_2 = \frac{Blv}{3R}$. Since the total current output is the sum of these two parallel branches, represented by the relationship $I = I_1 + I_2$, combining these components results in a total current of $I = \frac{2Blv}{3R}$.

The mutual inductance of two coaxial solenoids depends on the magnetic permeability of the core, the number of turns in each coil, the cross-sectional area, and the length, described by the formula $M = \frac{\mu_0 N_1 N_2 A}{l}$. First, express the dimensions in SI units by converting 10 cm^2 to 0.001 m^2 and 20 cm to 0.2 m. Substituting the values where $N_1 = 300$, $N_2 = 400$, $\mu_0 = 4\pi \times 10^{-7}$, $A = 0.001$ m^2, and $l = 0.2$ m into the equation gives $M = 2.4\pi \times 10^{-4}$ H.