Electromagnetic Induction – JEE Main PhysicsPractice Questions & PYQs

The induced EMF is given by Faraday's Law, $|\varepsilon| = \frac{B |\Delta A|}{\Delta t}$. The initial area of the circular loop with radius $r = 0.07$ m is $A_i = \pi r^2 = \frac{22}{7} \times (0.07)^2 = 0.0154$ m$^2$. The constant perimeter of the loop is $P = 2\pi r = 2 \times \frac{22}{7} \times 0.07 = 0.44$ m, resulting in a square side length of $a = \frac{0.44}{4} = 0.11$ m and a final area of $A_f = a^2 = (0.11)^2 = 0.0121$ m$^2$. Substituting these values into the EMF equation yields $|\varepsilon| = \frac{0.2 \times |0.0121 - 0.0154|}{0.5} = 0.00132$ V. Converting this to millivolts results in $1.32$ mV.

The energy density $u$ inside a solenoid is given by the formula $u = \frac{B^2}{2\mu_0}$, where $B = \frac{\mu_0 N I}{l}$ is the magnetic field. Substituting $B$ into the energy density formula yields $u = \frac{\mu_0 N^2 I^2}{2 l^2}$.

The maximum current in the inductor is $I_{\text{max}} = 10 / 10 = 1$ A. When the current reaches $\frac{1}{e}$ of its maximum value, the instantaneous current is $I = \frac{1}{e}$ A. Substituting this into the energy density expression, we have $u = \frac{\mu_0 N^2}{2 l^2} (\frac{1}{e})^2 = \frac{\mu_0 N^2}{2 l^2 e^2}$.

Equating this to the given expression $\alpha \pi \times \frac{1}{e^2}$, we obtain $\alpha \pi = \frac{\mu_0 N^2}{2 l^2}$. Substituting $\mu_0 = 4\pi \times 10^{-7}$ T m/A, $N = 10^4$, and assuming a unit length $l = 1$ m, the equation becomes $\alpha \pi = \frac{4\pi \times 10^{-7} \times (10^4)^2}{2 \times 1^2} = 20\pi$. The value of $\alpha$ is 20.

When a conducting rod rotates within a magnetic field, each infinitesimal segment of the rod behaves like a tiny battery generating a motional electromotive force because it moves perpendicular to the field lines. For a segment of length $dr$ located at a distance $r$ from the origin, the tangential velocity is $v = \omega r$. Given that the magnetic field varies radially as $B(r) = B_{\emptyset} e^{-\lambda r}$, the induced emf for this segment is given by $d\epsilon = B(r) v dr$, which simplifies to $d\epsilon = B_{\emptyset} \omega r e^{-\lambda r} dr$.

To find the total emf, we integrate this expression over the entire length of the rod from $0$ to $L$. Solving the integral $\int_0^L r e^{-\lambda r} dr$ using integration by parts results in the antiderivative $-e^{-\lambda r} (\frac{r}{\lambda} + \frac{1}{\lambda^2})$. Evaluating this at the boundaries $L$ and $0$ yields the final expression:

$\epsilon = B_{\emptyset} \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) \right]$

At the instant the switch is turned ON ($t = 0$), an inductor opposes any sudden change in current and behaves as an open circuit, meaning no current flows through the branches containing inductors. Replacing the inductors with open circuits removes their corresponding branches from the network. With a source voltage of $9$ V and the remaining $9\ \Omega$ resistor in the primary path, the ammeter reading is calculated using Ohm's law as $I = \frac{V}{R} = \frac{9}{9} = 1$ A.

Faraday's law of induction dictates that an electromotive force arises in a conducting loop whenever the magnetic flux threading through it varies with time. Because the circular loop is oriented perpendicular to the magnetic field lines, the magnetic flux $\Phi$ is simply the product of the field magnitude $B$ and the loop's cross-sectional area $\pi r^2$. Given that the current $i$ is defined by the ratio of the induced electromotive force to the resistance $R$, the relationship is expressed as:

$i = \frac{\pi r^2}{R} \left(\frac{dB}{dt}\right)$

The rate of change of the magnetic field, $\frac{dB}{dt}$, is obtained by differentiating the expression $\vec{B} = 2t^2 + 2t + 3$ with respect to time, which yields $4t + 2$. At the specified time of $t = 3 \text{ s}$, this rate of change is $4(3) + 2 = 14 \text{ T/s}$. With a loop radius of $20 \text{ cm}$, or $0.2 \text{ m}$, the area $A$ is $\pi(1/5)^2$, or $\pi/25 \text{ m}^2$. Substituting these values and the resistance of $2 \text{ }\Omega$ into the current equation results in $i = \frac{\pi/25}{2} \cdot 14$. Applying the value $\pi = 22/7$, the calculation becomes $i = \frac{22/7 \cdot 1/25}{2} \cdot 14$, which simplifies to $44/50 \text{ A}$. Consequently, the value of $\alpha$ is 44.

The magnetic field inside a long solenoid is given by $B = \mu_0 n I$, where $n = 500 \; \text{m}^{-1}$ is the number of turns per unit length and $I = 10 \sin(\omega t) \; \text{A}$ is the current. The magnetic flux $\Phi$ through the circular conducting loop of radius $r = 0.01 \; \text{m}$ is $\Phi = B \pi r^2 = \mu_0 n \pi r^2 (10 \sin(\omega t))$, and the induced electromotive force $\varepsilon$ is determined by Faraday's law of induction, $\varepsilon = - \frac{d\Phi}{dt} = -\mu_0 n \pi r^2 (10 \omega \cos(\omega t))$.

The induced current in the loop is given by $i = \frac{\varepsilon}{R} = -\frac{\mu_0 n \pi r^2 (10 \omega)}{R} \cos(\omega t)$, where $R = 10 \; \Omega$ and $\omega = 1000 \; \text{rad/s}$. The peak current $I_0$ is $\frac{\mu_0 n \pi r^2 (10 \omega)}{R}$, meaning the root-mean-square current is $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{\mu_0 n \pi r^2 (10 \omega)}{R \sqrt{2}}$. Equating this expression to the given form $\frac{\alpha}{\sqrt{2}} \mu \text{A} = \frac{\alpha}{\sqrt{2}} \times 10^{-6} \; \text{A}$ leads to $\alpha = \frac{\mu_0 n \pi r^2 (10 \omega) \times 10^6}{R}$.

Substituting the values $\mu_0 = 4\pi \times 10^{-7} \; \text{T}\cdot\text{m/A}$, $n = 500 \; \text{m}^{-1}$, $r = 0.01 \; \text{m}$, $\omega = 1000 \; \text{rad/s}$, and $R = 10 \; \Omega$ yields $\alpha = \frac{4\pi \times 10^{-7} \cdot 500 \cdot \pi \cdot (0.01)^2 \cdot 10 \cdot 1000 \cdot 10^6}{10} = 20 \pi^2$. Using the approximation $\pi^2 \approx 9.87$, $\alpha \approx 197.4$, so the nearest integer value is 197.

The magnetic energy stored in an inductor is determined by the formula $U = \frac{1}{2}Li^2$. In this circuit configuration, the total current $I$ flows through the primary inductor, while the remaining current splits equally into two parallel branches, each carrying a current of $I/2$. The energy stored in the primary inductor is $\frac{1}{2}LI^2$, while the energy stored in each of the two parallel inductors is $\frac{1}{2}L(I/2)^2$, which simplifies to $\frac{1}{8}LI^2$.

The total magnetic energy in the circuit $U_t$ is the sum of these individual values, calculated as $U_t = \frac{1}{2}LI^2 + \frac{1}{8}LI^2 + \frac{1}{8}LI^2 = \frac{3}{4}LI^2$. Comparing this total energy to the energy stored specifically in the second inductor $U_l = \frac{1}{8}LI^2$ gives the ratio $U_t/U_l = \frac{3/4 LI^2}{1/8 LI^2}$, which results in 6.

Kirchhoff's Voltage Law provides the fundamental principle for understanding this circuit, stating that the sum of potential differences around any closed loop must equal zero. When an inductor with an inductance of 10 mH and a resistance of 100 $\Omega$ is connected to a battery with an EMF of 1.0 V, the supplied electromotive force is divided between the resistance and the inductor. This relationship is defined by $E = iR + V_L$, where $E$ is the battery voltage, $iR$ is the potential drop across the resistance, and $V_L$ is the instantaneous voltage across the inductor.

By rearranging this to solve for the inductor voltage, we obtain $V_L = E - iR$. Substituting the given values, the expression becomes $V_L = 1.0 - 100i$. When the current is 2 mA, or 0.002 A, the voltage across the inductor is $V_L = 1.0 - 100 \times 0.002 = 0.8$ V. Similarly, when the current reaches 4 mA, or 0.004 A, the inductor voltage is $V_L = 1.0 - 100 \times 0.004 = 0.6$ V. Calculating the ratio of these two instantaneous voltages, 0.8 divided by 0.6, results in 4/3.

The motional electromotive force $\varepsilon$ induced in a conducting rod of length $l$ moving with velocity $v$ through a magnetic field $B$ is $\varepsilon = Blv$. Substituting the provided parameters results in $\varepsilon = 0.10 \times 1 \times 1.5 = 0.15 \text{ V}$. The induced current $I$ circulating through the resistance $R$ follows Ohm's law, $I = \frac{\varepsilon}{R} = \frac{0.15}{2} = 0.075 \text{ A}$. Since the rod moves at a constant velocity, the external applied force must counteract the magnetic force $F = IlB$, which is $F = 0.075 \times 1 \times 0.10 = 7.5 \times 10^{-3} \text{ N}$.

The terminal speed is achieved when the acceleration of the rod becomes zero, meaning the gravitational force $mg$ is exactly balanced by the upward magnetic force $F_m$ acting on the rod. The motional electromotive force $\varepsilon$ induced in the rod as it moves with speed $v$ is given by $\varepsilon = Blv$, where $l$ is the separation between the rails. Using Ohm's law, the induced current $I$ flowing through the loop is $I = \frac{\varepsilon}{R} = \frac{Blv}{R}$. The magnetic force $F_m$ acting on the segment of the rod of length $l$ within the rails is $F_m = IlB = \left(\frac{Blv}{R}\right)lB = \frac{B^2 l^2 v}{R}$. Setting the magnetic force equal to the gravitational force at terminal speed $v_t$, $mg = \frac{B^2 l^2 v_t}{R}$, leads to $v_t = \frac{mgR}{B^2 l^2}$.

According to Lenz's Law, the induced current in a coil generates a magnetic field that opposes any change in the net magnetic flux threading it. An anti-clockwise current in $C_1$ creates a magnetic field directed to the right through $C_2$, while a clockwise current in $C_3$ creates a magnetic field directed to the left. When $C_1$ moves towards $C_2$ and $C_3$ moves away, the rightward magnetic flux through $C_2$ increases and the leftward magnetic flux decreases, causing a net increase in rightward flux which induces a clockwise current in $C_2$ to generate an opposing magnetic field pointing to the left.

The principle of electromagnetic induction dictates that a time-varying magnetic flux through a stationary loop generates an induced electromotive force. This flux is determined by the relation $\Phi = BA \cos \theta$, where $A$ is the area of the loop and $\theta$ is the angle between the magnetic field vector and the normal to the loop surface. For a square loop with a side length of 0.02 m, the area is calculated as $A = (0.02)^2 = 4 \times 10^{-4}$ m².

Given the magnetic field $B = 0.4 \sin(300t)$, the induced emf is found by calculating the magnitude of the rate of change of magnetic flux, $\varepsilon = |d\Phi/dt| = |A \cos \theta \frac{dB}{dt}|$. Differentiating the magnetic field with respect to time yields $\frac{dB}{dt} = 0.4 \times 300 \cos(300t)$. Substituting the known parameters, the instantaneous emf is expressed as $\varepsilon = 4 \times 10^{-4} \times \cos(60^\circ) \times 0.4 \times 300 \cos(300t)$.

The maximum induced emf is attained when the oscillating term, represented by the cosine function, reaches its peak amplitude of one. This simplifies the expression for the maximum emf to $4 \times 10^{-4} \times \cos(60^\circ) \times 0.4 \times 300$. Using the value $\cos(60^\circ) = 0.5$, the calculation results in $4 \times 10^{-4} \times 0.5 \times 0.4 \times 300$, which evaluates to $240 \times 10^{-4}$ V. Converting this value into millivolts gives 24 mV.

Dimensional analysis expresses physical quantities in terms of the fundamental dimensions mass ($M$), length ($L$), time ($T$), and current ($A$). Magnetic induction ($B$) is derived from the Lorentz force formula $F = BIL$, yielding dimensions $[MT^{-2}A^{-1}]$, which corresponds to III. Magnetic flux ($\Phi$) is the product of magnetic induction and area ($L^2$), resulting in dimensions $[ML^2T^{-2}A^{-1}]$, which matches IV. Magnetic permeability ($\mu$) is determined from the force per unit length between two parallel conductors, $F = (\mu I_1 I_2 L) / (2\pi d)$, providing dimensions $[MLT^{-2}A^{-2}]$, which corresponds to I. Finally, self-inductance ($L$) is defined by the electromotive force equation $\varepsilon = L(dI/dt)$, resulting in dimensions $[ML^2T^{-2}A^{-2}]$, which matches II.

The induced EMF $\varepsilon$ in a conducting rod of length $l$ rotating with angular velocity $\omega$ within a uniform magnetic field $B$ perpendicular to the plane of motion is given by $\varepsilon = \frac{1}{2} B \omega l^2$. To determine the maximum angular velocity $\omega$ achieved at the lowest point of the oscillation, apply the principle of conservation of mechanical energy by equating the initial gravitational potential energy $mgl(1 - \cos \theta)$ to the rotational kinetic energy $\frac{1}{2} I \omega^2$. Since a simple pendulum behaves as a point mass $m$ attached to a string of length $l$, the moment of inertia is $I = ml^2$, which simplifies the energy equation to $\omega = \sqrt{\frac{2g(1 - \cos \theta)}{l}}$.

Substituting the values $\theta = 60^\circ$, $g = 10 \, \text{m/s}^2$, and $l = 0.1 \, \text{m}$ yields a maximum angular velocity of $\omega = \sqrt{\frac{2 \times 10 \times (1 - 0.5)}{0.1}} = 10 \, \text{rad/s}$. Applying this angular velocity and the magnetic field strength $B = 2 \, \text{T}$ to the EMF formula results in $\varepsilon = \frac{1}{2} \times 2 \times 10 \times (0.1)^2 = 0.1 \, \text{V}$. Expressed in millivolts, this value is $100 \, \text{mV}$.

The induced electromotive force (e.m.f.) in a solenoid arises due to the change in magnetic flux linked with it, which occurs when the current through the coil changes over time. We calculate this magnitude using the relationship $\varepsilon = L \frac{di}{dt}$, where $L$ is the self-inductance of the solenoid. The self-inductance of a solenoid is given by the formula $L = \mu_0 n^2 A l$, where $n$ represents the number of turns per unit length.

Converting the provided parameters into SI units, we have the number of turns per meter as $n = 1000 \, m^{-1}$, the cross-sectional area as $A = 5 \times 10^{-4} \, m^2$, and the length as $l = 0.3 \, m$. Substituting these values along with the permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \, H/m$, into the inductance formula results in $L = (4\pi \times 10^{-7}) \times (10^3)^2 \times (5 \times 10^{-4}) \times 0.3$, which simplifies to $L = 6\pi \times 10^{-5} \, H$.

Given that the current changes from 2 A to 4 A over 3.14 seconds, the rate of change is $\frac{di}{dt} = \frac{2}{3.14} \, A/s$. Using the approximation that 3.14 is equivalent to $\pi$, this rate simplifies to $\frac{2}{\pi} \, A/s$. Multiplying the inductance by this rate of change yields $\varepsilon = (6\pi \times 10^{-5}) \times (\frac{2}{\pi}) = 12 \times 10^{-5} \, V$. By comparing this result to the given expression $\alpha \times 10^{-5} \, V$, we identify that $\alpha = 12$.

The mutual inductance between two loops describes how effectively a current in one loop induces magnetic flux through the other. Since the circular loop has a radius much smaller than the square's side length, the magnetic field generated by the square loop can be treated as uniform over the area of the inner circle. The magnetic field at the center of a square loop of side length $L$ is $B = \frac{2\sqrt{2}\mu_0 I}{\pi L}$. By multiplying this field by the cross-sectional area of the circular loop, $\pi R^2$, we determine the total magnetic flux linkage $\Phi = \frac{2\sqrt{2}\mu_0 I R^2}{L}$. Dividing this magnetic flux by the current $I$ yields the mutual inductance $M = \frac{2\sqrt{2}\mu_0 R^2}{L}$.

The induced electromotive force (EMF) across a wire of length $l$ moving with velocity $v$ perpendicular to a uniform magnetic field $B$ is defined by $\epsilon = B l v$. The velocity acquired by the wire falling from rest through a vertical distance $h = 200$ m under gravity is $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 200} = 20\sqrt{10}$ m/s. With the magnetic field $B = 0.5$ Gauss $= 0.5 \times 10^{-4}$ T and length $l = 20$ m, the induced EMF is $\epsilon = (0.5 \times 10^{-4}) \times 20 \times 20\sqrt{10} = 200 \times 10^{-4} \sqrt{10}$ V. Converting this to millivolts yields $\epsilon = 200 \times 10^{-4} \sqrt{10} \times 10^3 = 20\sqrt{10}$ mV.

The induced electromotive force $\epsilon$ in a rotating conducting loop is given by the formula $\epsilon = BA\omega \sin(\theta)$, where $B$ is the magnetic field, $A$ is the area of the loop, $\omega$ is the angular speed, and $\theta$ is the angle of rotation. Substituting the provided values $\epsilon = 15.4 \times 10^{-3} \text{ V}$, $B = 0.5 \text{ T}$, $\omega = 100 \text{ rad/s}$, and $\theta = 30^\circ$ into the equation leads to $15.4 \times 10^{-3} = 0.5 \times A \times 100 \times \sin(30^\circ)$, which simplifies to $15.4 \times 10^{-3} = 25A$. Solving for the area $A$ results in $A = \frac{15.4 \times 10^{-3}}{25} = 6.16 \times 10^{-4} \text{ m}^2$. Using the formula for the area of a circular loop $A = \pi r^2$ with $\pi = \frac{22}{7}$, we find $r^2 = \frac{6.16 \times 10^{-4} \times 7}{22} = 1.96 \times 10^{-4} \text{ m}^2$. Taking the square root gives $r = 1.4 \times 10^{-2} \text{ m}$, which indicates that the radius is 14 mm.

The power dissipated in an induction coil follows the relation $P = \frac{\epsilon^2}{R}$, where the induced electromotive force is determined by Faraday's law as $\epsilon \propto NA$, and the resistance of the wire is related to its dimensions by $R = \rho \frac{\ell}{\pi r^2}$. By representing the total wire length in terms of the number of turns and the coil radius derived from its area where $R_{coil} = \sqrt{A/\pi}$, we find the resistance scales as $R \propto \frac{N \sqrt{A}}{r^2}$. Substituting these dependencies into the power formula results in the proportionality $P \propto N A^{3/2} r^2$.

When we transition to the second coil with parameters $2N$, $2A$, and $3r$, the new power $P'$ becomes proportional to $(2N)(2A)^{3/2}(3r)^2$. Expanding this expression gives $P' \propto (2N)(2\sqrt{2}A^{3/2})(9r^2)$, which simplifies to $P' = 36\sqrt{2} P$. By matching this result with the given form $P' = \sqrt{2} \alpha P$, it is evident that $\alpha = 36$.

Faraday's law of induction states that the induced electromotive force is $\epsilon = - \frac{d\phi}{dt}$, where the magnetic flux is $\phi = B A$. Given $B = \sin(100t)$ and $A = 1.0 \, \text{m}^2$, the flux is $\phi = \sin(100t)$, resulting in an EMF of $\epsilon = -100 \cos(100t) \, \text{V}$. The induced current in the loop with resistance $R = 100 \, \Omega$ is $I = \frac{\epsilon}{R} = -\cos(100t) \, \text{A}$, and the instantaneous power dissipated is $P = I^2 R = 100 \cos^2(100t) \, \text{W}$.

The total energy dissipated in one period $T = \frac{2\pi}{100} = \frac{\pi}{50} \, \text{s}$ is the integral of the instantaneous power over the interval $[0, T]$, which is $E = \int_0^{\pi/50} 100 \cos^2(100t) \, dt$. Applying the trigonometric identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$, the expression becomes $E = 50 \int_0^{\pi/50} (1 + \cos(200t)) \, dt = 50 \left[ t + \frac{\sin(200t)}{200} \right]_0^{\pi/50}$. Evaluating this definite integral yields $50 \left( \frac{\pi}{50} + 0 - (0 + 0) \right) = \pi \, \text{J}$.