Electromagnetic Induction – JEE Main PhysicsPractice Questions & PYQs

The motional electromotive force induced in a conductor moving through a magnetic field is defined by $E = Blv$, where $l$ represents the length of the conductor segment currently cutting the field lines. Since the bar starts at a vertex and travels along diverging rails, the length $l$ is directly proportional to the distance $vt$ traveled from the vertex, allowing us to write $l = kvt$ for some geometric constant $k$. Substituting this into the EMF formula yields $E = B(kvt)v$, which simplifies to $E = (Bkv^2)t$. Because $B$, $k$, and $v$ are all constant, the induced EMF is proportional to $t^1$, indicating that $n$ must be 1.

Self-inductance, characterized by the relation $L = \mu_r \mu_0 n^2 Al$, behaves as the electromagnetic equivalent of mass because it quantifies a system's inertia against changes in current. Since the expression depends on geometric parameters like the cross-sectional area and length of the coil as well as the magnetic permeability of the surrounding medium, the inductance is intrinsically tied to both the physical structure and material properties of the inductor. As a consequence of Lenz's law, the self-induced electromotive force opposes any current fluctuation, necessitating that work be performed by the circuit to build up the current flow.

The behavior of the circuit is governed by Kirchhoff's voltage law, which states that the induced EMF in the inductor must balance the resistive potential drop and the source EMF within the loop. The potential induced by the changing current is calculated as $L \frac{di}{dt} = 3 \times 8 = 24$ V. Using the relation $L \frac{di}{dt} = iR + \epsilon$ with the resistance $R = 12$ $\Omega$ and source EMF $\epsilon = 12$ V, we obtain the equation $24 = 12i + 12$. Solving this for the current $i$ yields a value of $i = 1$ A.

When a rectangular loop moves out of a uniform magnetic field at a constant speed, the magnetic flux $\phi$ passing through it changes based on the length of the loop $x$ still inside the field, represented as $\phi = Bbx$. Applying Faraday's law of induction, the magnitude of the induced electromotive force is given by $|E| = |\frac{d\phi}{dt}|$, which simplifies to $|E| = Bbv$ since the velocity $v$ is the rate of change of position with respect to time. Because the magnetic field strength $B$, the width of the loop $b$, and the speed $v$ are all constant, the induced EMF remains at a steady value throughout the interval the loop is partially inside the field, appearing as a constant horizontal line on the plot.

When a conducting disc rotates within a uniform magnetic field directed perpendicular to its surface, the moving charges within the material experience a magnetic force that separates them radially, establishing a potential difference between the center and the outer rim. This phenomenon is a specific application of motional electromotive force where the induced voltage depends on the strength of the magnetic field $B$, the angular velocity $\omega$ of the disc, and the square of its radius $R$.

The magnitude of this potential difference is determined by the relationship $E = \frac{1}{2} B \omega R^2$ Substituting the provided values of 0.4 T for the magnetic field, 10 \pi rad s^-1 for the angular velocity, and 0.2 m for the radius, the calculation is represented by $E = \frac{1}{2} \times 0.4 \times 10 \times 3.14 \times (0.2)^2$ After performing the arithmetic, the resulting potential difference is 0.2512 V.

The total induced electromotive force in a coil is determined by the combined effects of its own changing current and the changing magnetic flux originating from neighboring circuits, as dictated by Faraday's Law and Lenz's Law. The self-induction component, $-L_1 \frac{dl_1}{dt}$, accounts for the reaction of the coil to its own internal current variations, while the mutual induction component, $-M_{12} \frac{dl_2}{dt}$, accounts for the influence of the changing current in the second coil. Since Lenz's Law dictates that these induced effects oppose the variations that create them, the net induced electromotive force is expressed as $\epsilon _1 = -L_1 \frac{dl_1}{dt} - M_{12} \frac{dl_2}{dt}$.

The magnetic flux through a coil is determined by the relation $\phi = N B A \cos \theta$, where $\theta$ is the angle between the magnetic field $\vec{B}$ and the normal vector to the coil. When the magnetic field is parallel to the plane of the coil, the normal vector is perpendicular to the field, making $\theta = 90^\circ$ and resulting in a magnetic flux of $\phi = N B A \cos 90^\circ = 0$. Because the induced electromotive force is defined as the negative time derivative of the flux, or $\epsilon = N B A \omega \sin \theta$, the value of the emf at this instant reaches its maximum magnitude of $\epsilon = N B A \omega$.

The induced electromotive force ($e$) in a conductor moving with velocity ($v$) through a uniform magnetic field ($B$) is determined by the straight-line displacement between its endpoints, specifically the component of this displacement perpendicular to the velocity vector. This is expressed by the formula
$e = B v L_{\text{eff}}$
where $L_{\text{eff}}$ is the effective vertical length of the conductor. Given the velocity $v = 10 \text{ cm/s} = 0.1 \text{ m/s}$, the magnetic field $B = 1/\sqrt{2} \text{ T}$, and the effective vertical displacement $L_{\text{eff}} = 10\sqrt{2} \text{ cm} = 0.1\sqrt{2} \text{ m}$, substituting these values into the equation yields
$e = \frac{1}{\sqrt{2}} \times 0.1 \times 0.1\sqrt{2}$
The $\sqrt{2}$ terms cancel, resulting in $e = 0.1 \times 0.1 = 0.01 \text{ V}$, which is equivalent to $10 \text{ mV}$.

Maxwell's equations provide the foundational framework for electromagnetism by relating electric and magnetic fields to their respective sources and temporal changes. The Ampere-Maxwell law, $\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$, explains how magnetic fields are generated by electric currents and changing electric flux, whereas Faraday's law, $\oint \vec{E} \cdot d\vec{l} = \frac{d\phi_B}{dt}$, describes the generation of electric fields from changing magnetic flux. Additionally, Gauss's law for electricity, $\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$, quantifies electric flux through a closed surface in terms of enclosed charge, while Gauss's law for magnetism, $\oint \vec{B} \cdot d\vec{A} = 0$, signifies the absence of isolated magnetic poles.

The mutual inductance of this system is governed by the magnetic flux threading the smaller inner loop due to the current circulating in the larger outer loop. Since the inner loop is small, we can treat the magnetic field produced by the outer loop as uniform across the entire area of the inner loop. Applying the Biot-Savart law to the larger square loop of side $L$ carrying current $i$, the total magnetic field $B$ at the center is the sum of contributions from all four sides, given by $B = \frac{2\sqrt{2}\mu_0 i}{\pi L}$.

The magnetic flux $\phi$ through the inner loop is the product of this field and the area $\ell^2$, leading to $\phi = \frac{2\sqrt{2}\mu_0 i \ell^2}{\pi L}$. Mutual inductance $M$ is defined as the ratio $\phi/i$, which yields $M = \frac{2\sqrt{2}\mu_0 \ell^2}{\pi L}$. Using the physical constant $\mu_0 = 4\pi \times 10^{-7}$ H/m and the given geometric condition where $\ell^2/L = 1$, the expression simplifies to $M = 2\sqrt{2} \times \frac{4\pi \times 10^{-7}}{\pi} = 8\sqrt{2} \times 10^{-7}$ H. By expressing the coefficient as a single square root, $8\sqrt{2} = \sqrt{64 \times 2} = \sqrt{128}$, the value of the mutual inductance becomes $\sqrt{128} \times 10^{-7}$ H, which establishes that $x = 128$.

When a conducting wire moves through a magnetic field, it experiences a motional electromotive force across its length due to the interaction of moving charges with the magnetic field, which is a consequence of the magnetic Lorentz force. For a straight wire moving perpendicularly to the Earth's horizontal magnetic field, the induced EMF is determined by the equation $e = Bvl$, where $B$ is the magnetic field intensity, $v$ is the velocity of the wire, and $l$ is its length. Substituting the given values of $B = 0.60 \times 10^{-4} \text{ Wbm}^{-2}$, $v = 10 \text{ ms}^{-1}$, and $l = 5 \text{ m}$ into the formula, we perform the following calculation:

$e = 0.60 \times 10^{-4} \times 10 \times 5 = 3 \times 10^{-3} \text{ V}$

This result confirms that the missing numerical value to be filled in is 3.

The total electromotive force induced in the rectangular loop is the superposition of two components: the motional EMF arising from the motion of the loop through a spatially non-uniform magnetic field, and the transformer EMF induced by the temporal change in the magnetic field magnitude. The vertical sides of the loop, which have a length of $5 {\text{cm}}$, move with a velocity of $5 {\text{cm/s}}$ perpendicular to the magnetic field. Because the magnetic field has a spatial gradient of $10^{-3} {\text{T/cm}}$ along the x-axis, the difference in magnetic flux density across the $12 {\text{cm}}$ length of the loop's horizontal sides is $(10^{-3} {\text{T/cm}}) \times 12 {\text{cm}} = 1.2 \times 10^{-2} {\text{T}}$. The motional EMF generated is determined by $\epsilon_{\text{motional}} = B_{\text{effective}} l v = (1.2 \times 10^{-2} {\text{T}}) \times (5 \times 10^{-2} {\text{m}}) \times (5 \times 10^{-2} {\text{m/s}})$, which simplifies to $300 \times 10^{-7} {\text{V}}$.

Simultaneously, the magnetic field is decreasing over time, which induces an additional EMF according to Faraday's law of induction. Given the loop area is $12 {\text{cm}} \times 5 {\text{cm}} = 60 {\text{cm}}^2$, or $60 \times 10^{-4} {\text{m}}^2$, and the rate of change of the magnetic field is $10^{-3} {\text{T/s}}$, the induced transformer EMF is calculated as $\epsilon_{\text{time}} = A \left| \frac{dB}{dt} \right| = (60 \times 10^{-4} {\text{m}}^2) \times (10^{-3} {\text{T/s}}) = 60 \times 10^{-7} {\text{V}}$. Combining these two effects, the total net EMF acting on the loop is $\epsilon_{\text{net}} = 300 \times 10^{-7} {\text{V}} + 60 \times 10^{-7} {\text{V}} = 360 \times 10^{-7} {\text{V}}$. With a resistance of $6 {\text{m}} \Omega$, or $6 \times 10^{-3} \Omega$, the power dissipated by the loop is determined by the formula $P = \frac{\epsilon_{\text{net}}^2}{R}$. Substituting the values, we get:

$P = \frac{(360 \times 10^{-7} {\text{V}})^2}{6 \times 10^{-3} \Omega} = \frac{129600 \times 10^{-14}}{6 \times 10^{-3}} {\text{W}} = 216 \times 10^{-9} {\text{W}}$

When a coil rotates within a uniform magnetic field, the changing magnetic flux induces an electromotive force described by Faraday's law of induction. This rotation generates a sinusoidal voltage whose maximum magnitude is determined by the formula $\epsilon_{\max} = NAB\omega$ where $N$ is the number of turns, $A$ is the area of the coil, $B$ is the magnetic field strength, and $\omega$ is the angular frequency of rotation. Given that the coil has $N=200$ turns, an area of $A=0.20 {\text{m}}^2$, and is placed \in a magnetic field $B=0.01 {\text{T}}$, we first calculate the angular frequency. Since the coil rotates at half a revolution per second, $\omega = 2\pi f = 2\pi(0.5) = \pi {\text{ rad/s}}$. Substituting these values into the equation for maximum voltage, we obtain $\epsilon_{\max} = 200 × 0.20 × 0.01 × \pi$, which simplifies to $0.4\pi$ or $\frac{2\pi}{5}$ volts. By comparing this result with the expression $\frac{2\pi}{\beta}$, we find that $\beta$ equals 5.

Faraday's Law of Induction dictates that an induced electromotive force is generated by a rate of change in magnetic flux, which depends on the magnetic field, the area of the loop, and the number of turns. Because the coil is placed perpendicular to the magnetic field, the magnetic flux through each individual turn is simply the product of the field magnitude and the cross-sectional area of the coil.

To find the area, we use the provided diameter of $0.02 \text{ m}$, which gives a radius of $0.01 \text{ m}$, resulting in an area of $\pi(0.01)^2 \text{ m}^2$. The magnetic field changes from $5000 \text{ T}$ to $3000 \text{ T}$, creating a change in the field of $2000 \text{ T}$. Consequently, the change in magnetic flux experienced by the coil is calculated as $\Delta\phi = 2000 \times \pi(0.01)^2 = 0.2\pi \text{ Wb}$.

We relate these parameters using the induced emf equation, $\varepsilon = N \frac{\Delta\phi}{\Delta t}$, where $N$ represents the number of turns. Substituting the given emf of $22 \text{ V}$, the flux change of $0.2\pi \text{ Wb}$, and the time interval of $2 \text{ s}$ into the formula yields the following:
$22 = N \left( \frac{0.2\pi}{2} \right)$
Simplifying this expression results in $22 = N \times 0.1\pi$. Solving for the number of turns by dividing $22$ by $0.1\pi$ gives $N = \frac{220}{\pi}$, which equals $70$.

The interaction between two coils is characterized by mutual inductance, where a changing current in the first coil induces an electromotive force (EMF) in the second coil according to Faraday's law, expressed as
$e = -M \frac{di}{dt}$
Given that the current varies as $i = i_0 \sin(\omega t)$, the rate of change is determined by the derivative $di/dt = i_0 \omega \cos(\omega t)$. Consequently, the magnitude of the induced EMF is $M i_0 \omega \cos(\omega t)$, and its maximum value occurs when the cosine term equals one, resulting in $EMF_{\max} = M i_0 \omega$. By substituting the given parameters $M = 0.002$ H, $i_0 = 5$ A, and $\omega = 50\pi$ rad/s, the maximum EMF evaluates to $0.002 \times 5 \times 50\pi = 0.5\pi$ V. This matches the form $\pi / \alpha$ V, identifying that $\alpha$ is 2.

When a conducting rod rotates within a magnetic field, the induced emf across its length is determined by the vertical component of the magnetic field acting perpendicular to the rotational plane. The vertical component is found using the relation $B_V = B \sin(\theta)$, which yields $B_V = 0.5 \times 10^{-4} \times \sin(30^\circ) = 0.25 \times 10^{-4} \text{ T}$. Converting the rotational speed of 1200 rpm to angular velocity results in $\omega = 40\pi \text{ rad/s}$. With a blade length of 0.8 m, the emf is determined by the formula $\varepsilon = \frac{1}{2} B_V \omega L^2$. Substituting these values into the equation gives $\varepsilon = \frac{1}{2} \times 0.25 \times 10^{-4} \times 40\pi \times (0.8)^2$, which simplifies to $\varepsilon = 32\pi \times 10^{-5} \text{ V}$, confirming that $N = 32$.

The magnitude of the induced electromotive force in an inductor is determined by the product of its self-inductance and the absolute value of the rate of change of current, expressed as $|\epsilon| = L \left| \frac{dI}{dt} \right|$. Differentiating the given current function $I = 3t + 8$ with respect to time yields a rate of change of 3 A/s. Substituting the induced EMF of 12 mV into the equation results in $12 = L \times 3$, which reveals that the self-inductance of the inductor is 4 mH.

Mutual inductance represents the magnetic coupling between two circuits, defined by the relationship where the magnetic flux $\phi$ linked through the inner loop is proportional to the current $i$ flowing in the outer loop, such that $\phi = M i$. Since the inner loop with radius $a$ sits within the nearly uniform magnetic field $B = \frac{\mu_0 i}{2b}$ produced at the center of the outer loop of radius $b$, the total flux passing through the inner area $A = \pi a^2$ is $\phi = B \cdot A = \frac{\mu_0 i}{2b} \cdot \pi a^2$. Dividing this flux by the current $i$ to isolate the proportionality constant yields the mutual inductance $M = \frac{\mu_0 \pi a^2}{2b}$.

A changing current within a coil induces an electromotive force, a process governed by the relationship $\mathcal{E} = -L \frac{di}{dt}$. The magnitude of the induced voltage is related to the inductance by the equation

$L = \frac{|\mathcal{E}|}{\left| \frac{\Delta i}{\Delta t} \right|}$

The total current change of 4 A is the result of shifting from -2 A to 2 A. Dividing this change by the time interval of 0.2 s yields a rate of change of 20 A/s. Inserting these values into the inductance equation results in $L = \frac{0.1}{20}$, which equals 0.005 H, or 5 mH.

The efficiency of a transformer represents the proportion of the electrical power supplied to the primary coil that successfully transfers to the secondary coil as useful power output. Since efficiency is defined as the ratio of output power to input power, determining the actual power available in the secondary circuit involves multiplying the input power of 4000 W by the efficiency of 0.8, which results in 3200 W.

Because the output power is also equal to the product of the secondary voltage and the secondary current, we use the relationship $P_{out} = V_s I_s$. By substituting the known secondary voltage of 240 V and the calculated output power of 3200 W, we solve for the current using $I_s = \frac{3200}{240}$, which results in a secondary current of 13.33 A.