Electromagnetic Induction – JEE Main PhysicsPractice Questions & PYQs

The movement of a conductor through a magnetic field induces an electromotive force because the magnetic field exerts a force on the charge carriers within the wire. When the wire is positioned at a right angle to the magnetic field, the induced emf is determined by the product of the magnetic field strength $B$, the length of the wire $l$, and its velocity $v$. By substituting the given parameters of 2 T, 1 m, and 8 m/s into the formula, the calculation yields the magnitude of the potential difference:

$\varepsilon = B l v = 2 \times 1 \times 8 = 16 \text{ V}$

The induced electromotive force in a coil is determined by the rate of change of current, which is described by the relation $\varepsilon = L \frac{di}{dt}$. Before the switch is opened, the steady current flowing through the circuit is calculated using Ohm's Law as $I = 12/6 = 2$ A. When the switch is opened, the current drops from $2$ A to $0$ A within $1$ ms, which is $10^{-3}$ s. By substituting these values into the formula to find the inductance, we have:

$20 = L \frac{2}{10^{-3}}$

Solving this equation leads to $L = 10$ mH.

The induced electromotive force in a moving metal rod is described by the principle of motional EMF, which establishes that moving a conductor through a uniform magnetic field creates a potential difference based on the equation $\epsilon = Blv$. Since the rod length must be in standard SI units, we convert 10 cm to 0.1 m and substitute the known parameters into the expression, giving $0.08 = 0.4 \times 0.1 \times v$. Multiplying the magnetic field and length simplifies the equation to $0.08 = 0.04 \times v$. Dividing both sides to isolate the variable $v$ confirms that the rod moves with a velocity of 2 m/s.

When a conducting rod rotates within a uniform magnetic field directed normal to its plane of rotation, motional emf is generated due to the Lorentz force acting on free charge carriers. Because the linear velocity of any segment along the rod increases linearly with its distance from the axis of rotation, we must consider a tiny element of length $dx$ at a distance $x$ from the fixed end. The linear velocity of this segment is $v = \omega x$, making the induced potential difference across the element $d\varepsilon = B v dx$, which simplifies to $B \omega x dx$. Integrating this infinitesimal contribution from the pivot point at $x = 0$ to the tip of the rod at $x = L$ yields the total induced emf as $\varepsilon = \int_{0}^{L} B \omega x dx = \frac{1}{2} B L^2 \omega$.

A conducting rod rotating perpendicular to a uniform magnetic field develops an induced potential difference because the individual charges within the rod experience a magnetic force as they move, which is expressed by the formula $\varepsilon = \frac{1}{2} B \omega l^2$. First, convert all physical quantities into standard SI units: the length $l = 0.2 \text{ m}$, the magnetic field $B = 0.2 \text{ T}$, and the angular frequency $\omega = 2\pi \times (\frac{210}{60}) \text{ rad/s}$. Simplifying the angular frequency yields $\omega = 2 \times \frac{22}{7} \times 3.5 = 22 \text{ rad/s}$. Inserting these parameters into the expression gives $\varepsilon = \frac{1}{2} \times 0.2 \times 22 \times (0.2)^2$, which evaluates to $0.088 \text{ V}$. Converting this result into millivolts provides a final value of $88 \text{ mV}$.

When an object falls through the Earth's magnetic field, its material properties dictate how it interacts with the surrounding environment. An insulating ball experiences no electromagnetic interference and falls with the standard gravitational acceleration $g$. Conversely, as a metallic ball drops, it moves through the Earth's magnetic flux, causing a change in flux through its conducting material. This process induces eddy currents within the metal, which, according to Lenz's law, create an opposing magnetic force that acts against the downward motion, causing the insulating ball to descend more quickly and arrive at the earth's surface first.

Current sensitivity of a galvanometer is defined by the ratio $\frac{\theta}{I} = \frac{NBA}{K}$, where $N$ represents the number of turns, $B$ the magnetic field, $A$ the area, and $K$ the torsional constant. Because this sensitivity is directly proportional to $N$, doubling the number of turns results in a doubling of the current sensitivity, confirming the first statement. Conversely, voltage sensitivity is defined as $\frac{\theta}{V} = \frac{\theta}{IR} = \frac{NBA}{KR}$, where $R$ denotes the coil resistance; since increasing the turns $N$ also increases the wire length and thus the resistance $R$ proportionally, the factor of $N$ and $R$ cancels out, meaning the voltage sensitivity remains constant, which invalidates the second statement.

According to Faraday's law of induction, the magnitude of the induced electromotive force in a closed loop is proportional to the rate of change of magnetic flux passing through it. Since the plate possesses a constant area $A$ and is oriented normally to the magnetic field $B$, the magnetic flux $\Phi$ is simply the product $BA$, meaning the induced EMF is calculated as $e = A \left| \frac{dB}{dt} \right|$.

Given that the area of the metallic plate is $4 \text{ m}^2$, we determine the rate of change of the magnetic field by finding the slope of the $B-t$ graph between $t = 2 \text{ s}$ and $t = 4 \text{ s}$. During this time interval, the magnetic field rises linearly from $4 \text{ T}$ to $8 \text{ T}$, which corresponds to a slope of $\frac{8 - 4}{4 - 2} = 2 \text{ T/s}$. Multiplying the area of $4 \text{ m}^2$ by this rate of $2 \text{ T/s}$ yields an induced EMF magnitude of 8.

When a conductive loop moves through a uniform magnetic field, the change in magnetic flux induces an electromotive force, which generates an electric current. Because this current flows through the resistance of the loop, a magnetic force is exerted on the loop that opposes its motion. The work done to pull the loop out of the field at a constant speed is equal to the energy dissipated as heat by this induced current throughout the process.

For a square loop with an area of $25 \text{ cm}^2$, the length of each side is $l = 0.05 \text{ m}$. Pulling the loop completely out of the magnetic field in $1.0 \text{ s}$ requires moving it at a constant velocity of $v = \frac{0.05 \text{ m}}{1.0 \text{ s}} = 0.05 \text{ m/s}$. The motional electromotive force induced across the loop is given by
$\varepsilon = Blv = 40 \times 0.05 \times 0.05 = 0.1 \text{ V}$
This EMF drives an induced current of $I = \frac{0.1 \text{ V}}{10 \text{ \Omega}} = 0.01 \text{ A}$. The magnetic force resisting the motion is calculated as $F = IlB = 0.01 \text{ A} \times 0.05 \text{ m} \times 40 \text{ T} = 0.02 \text{ N}$. Consequently, the total work done to extract the loop against this force over a distance of $0.05 \text{ m}$ is $W = F \times l = 0.02 \text{ N} \times 0.05 \text{ m} = 1.0 \times 10^{-3} \text{ J}$.

Faraday's law of electromagnetic induction establishes that the magnitude of the induced electromotive force in a coil is determined by the rate at which the magnetic flux changes over time. Starting with the provided flux relationship $\phi = 5t^3 + 4t^2 + 2t - 5$, we differentiate the expression with respect to time to derive the formula for the induced emf:
$|\epsilon| = \frac{d\phi}{dt} = 15t^2 + 8t + 2$
By substituting the specific time of $t = 2$ into this expression, we calculate the instantaneous voltage generated within the coil:
$|\epsilon| = 15(2)^2 + 8(2) + 2 = 78$
This results in an induced potential of $78$ V. To find the current, we use Ohm's law by dividing the induced voltage by the coil's resistance of $5$ $\Omega$:
$I = \frac{78}{5} = 15.6$
The induced current through the coil is $15.6$ A.

Faraday's Law of electromagnetic induction dictates that the magnitude of an induced electromotive force in a circuit is the absolute rate of change of magnetic flux. Calculating the time derivative of the flux function $\phi = 8t^{2}-9t+5$ provides the expression for the EMF magnitude as $|\frac{d\phi}{dt}| = |16t - 9|$. Substituting the time 0.25 s into this expression results in an induced EMF of 5 V. Applying Ohm's Law in the form $i = \frac{\varepsilon}{R}$, the induced current is found by dividing 5 V by 20 ohms, which equals 0.25 A. Converting this result to milliamperes by multiplying by 1000 yields 250 mA.

An inductor opposes any instantaneous change in current, meaning it acts as an open circuit the moment the switch is closed. Because the branch containing the inductor is effectively blocked, the circuit simplifies to a series loop consisting of the 6 V battery and two resistors of 2 ohms and 4 ohms. The current $I$ flowing through the battery is determined by the total resistance of this series path according to Ohm's Law:

$I = \frac{6}{2+4} = 1$

The resulting current is 1 ampere.

Mutual inductance quantifies the magnetic coupling between two coils and is defined by the relationship $\phi = M i$, where $\phi$ is the magnetic flux through the smaller loop due to a current $i$ in the larger loop. Because the small loop of side $l$ is placed at the center of a much larger loop of side $L$ ($L \gg l$), we can approximate the magnetic field $B$ as uniform across the area of the smaller loop.

To find this field $B$ at the center, we apply the Biot-Savart law to the four sides of the large loop. Each side, at a perpendicular distance of $L/2$ from the center, contributes a field magnitude calculated as $B_{side} = \frac{\mu_0 i}{4 \pi (L/2)} (\sin 45^\circ + \sin 45^\circ)$. Summing these contributions from all four sides yields $B = 4 \times \frac{\mu_0 i}{2 \pi L} \times \sqrt{2}$, which simplifies to $B = \frac{2 \sqrt{2} \mu_0 i}{\pi L}$. The total flux $\phi$ through the smaller loop is then the product of this magnetic field and the area of the small loop, $\phi = B \cdot l^2 = \frac{2 \sqrt{2} \mu_0 i l^2}{\pi L}$. By definition, dividing this flux by the current $i$ gives the mutual inductance $M = \frac{2 \sqrt{2} \mu_0 l^2}{\pi L}$.

The electrical power dissipated in a short-circuited coil is determined by the relationship $P = \mathcal{E}^2 / R$, where $\mathcal{E}$ is the induced electromotive force and $R$ is the electrical resistance of the wire. According to Faraday's law of induction, the induced EMF is directly proportional to the number of turns $N$ in the coil, meaning that halving the number of turns results in a new EMF of $\mathcal{E}' = \mathcal{E} / 2$.

The resistance of the wire is governed by the formula $R = \rho (L / A)$, where $L$ is the total length of the wire and $A$ is its cross-sectional area. Reducing the number of turns by half correspondingly halves the total wire length $L$ to $L' = L / 2$. Furthermore, doubling the radius of the wire increases the cross-sectional area $A$ by a factor of $4$ since $A = \pi r^2$, leading to a new resistance of $R' = \rho (L/2) / (4A) = R / 8$.

Substituting these new values into the power equation yields $P' = (\mathcal{E} / 2)^2 / (R / 8)$. Simplifying this expression gives $P' = (\mathcal{E}^2 / 4) / (R / 8) = 2(\mathcal{E}^2 / R) = 2P$. As a result, the electrical power dissipated by the coil is doubled.

Magnetic potential energy stored in an inductor is calculated using the relationship $U = \frac{1}{2} L I^2$, where $L$ is the self-inductance and $I$ is the current. Because the total energy depends on the magnitude of the current, the provided time-varying function simply indicates that the current reaches a maximum of 2 A. With an inductance of 2.0 H and a final current of 2 A, the energy calculation becomes $U = \frac{1}{2} \times 2.0 \times 2^2$, resulting in 4 J.

When a conducting rod rotates within a magnetic field, the magnetic force acting on the free charge carriers generates a potential difference across its ends. This induced electromotive force depends on the magnetic field strength, the angular velocity of the rod, and its length, as described by the equation $Emf = \frac{1}{2} B \omega l^2$. Substituting the provided values-a magnetic field of 0.2 \times 10^-4 T, an angular velocity of 5 rad s^-1, and a length of 1 m-the equation becomes $Emf = \frac{1}{2} \times (0.2 \times 10^{-4}) \times 5 \times (1)^2$. Calculating this expression results in 0.5 × 10^-4 V, which is equivalent to 50 μV.

When two inductive coils are connected in series, the equivalent self-inductance depends not only on the individual inductances $L_1$ and $L_2$ but also on their magnetic coupling represented by the mutual inductance $M$. Because these coils are linked in opposition, the mutual magnetic flux generated by one coil acts against the flux generated by the other, effectively reducing the total inductance of the circuit. Since this mutual influence is reciprocal-meaning each coil induces a voltage in the other-the mutual inductance term is subtracted twice from the sum of the individual inductances to produce an equivalent self-inductance of $L = L_1 + L_2 - 2M$.

The average electromotive force induced in an inductor due to a change in current is governed by Faraday's law, which relates the voltage to the inductance and the rate of current decay through the equation $e_{avg} = L \frac{\Delta I}{\Delta t}$ Before the switch is opened, the coil carries a steady current determined by Ohm's Law, $I = V / R$. Given a battery of 20 V and a resistance of 10 ohms, the initial current is 2 A. When the switch opens, the current drops from 2 A to 0 A over an interval of 100 microseconds. By substituting the inductance of 20 mH, the change \in current of 2 A, and the time duration of 100 microseconds into the formula, the average induced EMF is $e_{avg} = \frac{20 \times 10^{-3} \times 2}{100 \times 10^{-6}} = 400 \text{ V}$

When a conducting rod moves through a magnetic field, the motion perpendicular to the field lines induces an electromotive force across the conductor. In this scenario, the rod is positioned North-South and moves toward the East, meaning it cuts across the horizontal component of the Earth's magnetic field. Since the rod length is perpendicular to both the velocity vector and the horizontal magnetic field component, we calculate the induced EMF using the motional EMF formula $E = B_H l v$ where $l$ is the length of the rod, $v$ is the velocity, and $B_H$ is the horizontal component of the Earth's magnetic field. Converting the length of 20 cm to 0.2 m and substituting the given values, the induced emf is 4 x 10^-3 T multiplied by 0.2 m and 20 m/s, which results in 0.016 V or 16 mV.

The magnetic induction $B$ at the center of a circular coil is determined by the relationship $B = \frac{N \mu_0 I}{2R}$, where $N$ represents the number of turns and $R$ is the radius of the coil. Since the wire is simply unwound and reshaped without changing its total length, the product of the number of turns and the radius remains constant, meaning $N_1 R_1 = N_2 R_2$. Given the initial turns $N_1 = 2$ and final turns $N_2 = 5$, the radius changes such that $R_2 = \frac{2}{5} R_1$. Applying these values to the ratio of the magnetic fields, the calculation proceeds as follows:

$\frac{B_2}{B_1} = \frac{N_2}{N_1} \times \frac{R_1}{R_2} = \frac{5}{2} \times \frac{R_1}{(2/5)R_1} = \frac{5}{2} \times \frac{5}{2} = \frac{25}{4}$