Electromagnetic Induction – JEE Main PhysicsPractice Questions & PYQs

When a circular coil rotates within a uniform magnetic field, the continuous change in magnetic flux through its cross-sectional area induces an electromotive force. The peak voltage generated during this motion is proportional to the number of turns in the coil, the area enclosed, the strength of the magnetic field, and the angular velocity of the rotation.

The maximum induced voltage is determined by the equation:
$V_{\max} = N A B \omega$
Since the coil completes one revolution per second, its angular velocity is $\omega = 2 \pi \text{ rad/s}$. By substituting the given parameters-1000 turns, an area of $1 \text{ m}^2$, a magnetic field of 0.07 T, and an angular velocity of $2 \pi \text{ rad/s}$-into the formula, we perform the calculation $1000 \times 1 \times 0.07 \times 2 \pi$. This product yields approximately 440 V.

Faraday's law of electromagnetic induction states that the induced electromotive force in a circuit is the negative rate of change of magnetic flux. Given the flux expression $\phi = \frac{2}{3}(9-t^2)$, the magnetic flux reaches zero when the time t is 3 seconds. By differentiating the flux with respect to time, we find the magnitude of the induced electromotive force is $e = \frac{4t}{3}$.

The heat produced in the coil follows Joule's law, where the instantaneous power is the square of the electromotive force divided by the resistance of 8 ohms. Integrating this power over the time interval from 0 to 3 seconds gives the total heat energy as follows:
$\text{Heat} = \int_{0}^{3} \frac{e^2}{8} dt = \int_{0}^{3} \frac{(4t/3)^2}{8} dt$
Simplifying the integrand leads to $\int_{0}^{3} \frac{2t^2}{9} dt$, which evaluates to 2 Joules of total heat produced in the coil.

The induced electromotive force in a conducting loop arises from the rate of change of magnetic flux linked with the surface area, governed by Faraday's Law. Because the circular loop lies in the XY plane, its area vector is oriented along the Z-axis, expressed as $\vec{A} = \pi r^2 \hat{k}$. Given the radius is 1 meter, the area vector simplifies to $\vec{A} = \pi \hat{k}$. The magnetic flux through the loop is calculated using the dot product of the magnetic field $\vec{B} = 3t^3 \hat{j} + 3t^2 \hat{k}$ and the area vector $\vec{A}$. Since the dot product of perpendicular unit vectors is zero, only the component of the magnetic field parallel to the area vector-the Z-component-contributes to the flux, yielding $\Phi = 3t^2 \pi$.

Applying Faraday's Law, the induced EMF is the absolute value of the time derivative of this flux, specifically $\varepsilon = |\frac{d\Phi}{dt}|$. Differentiating $3t^2 \pi$ with respect to time results in $6t \pi$. Substituting the specified time of $t = 2$ seconds into this expression provides an induced EMF of $12 \pi$ volts, which confirms that the value of $n$ is 12.

The induced motional electromotive force in a conductor moving through a magnetic field is given by the expression $\epsilon = B l v_0$. This EMF drives a current through the loop, and the magnitude of this current depends on the total resistance of the circuit. The resistor network connected to the square loop simplifies to an equivalent external resistance of 4 ohms. When combined with the loop's own internal resistance, the total resistance of the circuit becomes 5 ohms.

Using Ohm's law in the form $I = \epsilon / R_{\text{eq}}$, we can relate the known parameters to the speed of the loop. We have a target current $I$ of 0.002 A, a magnetic field $B$ of 5 T, a loop side $l$ of 0.2 m, and a total resistance $R_{\text{eq}}$ of 5 ohms. Substituting these values into the expression yields:

$0.002 = \frac{5 \times v_0 \times 0.2}{5}$

Solving this equation for $v_0$ results in a speed of 0.01 m/s, which is equivalent to 1 cm/s.

Inductors store magnetic energy, where the current relates to the applied voltage through the expression $V = L \frac{di}{dt}$. With the supply starting at $t = 0$, the current at $t = 4$ s is found by integrating $\frac{di}{dt} = \frac{3t}{L}$, which leads to the integral $I = \int_{0}^{4} \frac{3t}{2} dt$. Evaluating this integral gives $I = 12$ A. The energy stored in the inductor is calculated using $E = \frac{1}{2} L I^2$. Substituting $L = 2$ H and $I = 12$ A into this formula, the calculation becomes $E = \frac{1}{2} \times 2 \times 12^2$, resulting in 144 J.

In a cyclotron, a charged particle gains kinetic energy every time it passes through the accelerating potential gap. Since the proton crosses this gap twice per revolution, the total energy gained after $n$ revolutions is given by the expression $2 n q V$. According to the principle of conservation of energy, this total work done on the proton equals the kinetic energy it acquires, expressed as $\frac{1}{2} m_p v^2$. With the target velocity set to one-sixth the speed of light, we determine the speed as $v = \frac{3 \times 10^8}{6} = 5 \times 10^7 \text{ m/s}$. Equating the energy expressions gives the relation $2 n q V = \frac{1}{2} m_p v^2$. Substituting the values $q = 1.6 \times 10^{-19} \text{ C}$, $V = 12000 \text{ V}$, $m_p = 1.67 \times 10^{-27} \text{ kg}$, and $v = 5 \times 10^7 \text{ m/s}$ into the equation, we get $n \times 2 \times 1.6 \times 10^{-19} \times 12000 = \frac{1}{2} \times 1.67 \times 10^{-27} \times (5 \times 10^7)^2$. This simplifies to $n \times 3.84 \times 10^{-15} = 2.0875 \times 10^{-12}$. Solving for $n$ yields approximately 543.62, and since the number of revolutions must be an integer, the proton completes 543 revolutions to reach the required speed.

Mutual inductance, denoted as $M$, represents the ratio of the magnetic flux $\phi$ linked through a circuit to the current $I$ flowing in another, defined by the relationship $\phi = MI$. Since the side length of the larger square loop $b$ is significantly greater than the side length of the inner square loop $a$, the magnetic field generated by the larger loop is approximately uniform over the entire area of the smaller loop. The magnetic field at the center of the larger loop is the sum of the fields produced by its four identical sides, where the distance from the center to the midpoint of each side is $b/2$. Using the law of Biot-Savart, the field from a single side is $B_{side} = \frac{\mu_0}{4\pi} \frac{I}{b/2} (\sin 45^\circ + \sin 45^\circ)$.

Summing these four contributions gives a total magnetic field $B = 4 \left( \frac{\mu_0}{4\pi} \frac{2I}{b} \right) \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right)$, which simplifies to $B = \frac{\mu_0}{4\pi} \frac{8 \sqrt{2} I}{b}$. The magnetic flux linked through the inner loop is the product of this uniform field and the area of the smaller loop $a^2$, resulting in $\phi = \left( \frac{\mu_0}{4\pi} \frac{8 \sqrt{2} I}{b} \right) a^2$. Dividing this total flux by the current $I$ reveals the coefficient of mutual inductance to be $M = \frac{\mu_0}{4\pi} \frac{8 \sqrt{2} a^2}{b}$.

When a coil rotates at a constant speed within a uniform magnetic field, the changing magnetic flux through its surface induces an alternating voltage. The peak value of this electromotive force is determined by the formula $\varepsilon = N B A \omega$, where $N$ is the number of turns, $B$ is the magnitude of the magnetic field, $A$ represents the area of the coil, and $\omega$ is the angular speed. Since the area of the circular coil is calculated as $A = \pi r^2$, the radius must first be converted from 8.0 cm to 0.08 m to maintain consistency with standard units.

Substituting the given parameters into the equation yields $\varepsilon = 20 \times 3.0 \times 10^{-2} \times 50 \times \pi \times 0.08^2$. This calculation results in approximately $60.3 \times 10^{-2}$ V. Rounding this final value to the nearest integer provides 60.

The magnetic energy stored in a solenoid is proportional to the square of the current, expressed by the formula $U = \frac{1}{2} L I^2$. Since the magnetic energy is required to reach $25\%$ of its maximum value $U_0 = \frac{1}{2} L I_0^2$, the relationship $I^2 = 0.25 I_0^2$ must hold, which means the instantaneous current must be exactly half of its maximum value, or $I = \frac{I_0}{2}$.

Substituting this condition into the standard current growth equation for an $LR$ circuit, $I = I_0(1 - e^{-Rt/L})$, we obtain the expression $I_0(1 - e^{-Rt/L}) = \frac{I_0}{2}$. Simplifying this by dividing both sides by $I_0$ results in $1 - e^{-Rt/L} = \frac{1}{2}$, which further rearranges to $e^{-Rt/L} = \frac{1}{2}$. Applying the natural logarithm to both sides leads to $-\frac{Rt}{L} = \ln(\frac{1}{2})$, or equivalently $-\frac{Rt}{L} = -\ln 2$, confirming that the time taken is $t = \frac{L}{R} \ln 2$.

Motional electromotive force develops in a conductive loop when it moves through a magnetic field, specifically when the charge carriers within the wire segments experience a magnetic Lorentz force. Since the loop moves with velocity $\vec{v} = v_0 \hat{i}$ in a magnetic field $B = B_0 (\frac{x}{a}) \hat{k}$, the two sides of the loop parallel to the X-axis experience no EMF because the induced force acts perpendicular to these conductors. Consequently, the net EMF is determined solely by the potential difference generated across the two vertical sides of length $d$.

At the leading edge, positioned at $x+d$, the induced EMF is $\mathcal{E}_1 = B_0 \frac{(x+d)}{a} v_0 d$, while the trailing edge at position $x$ induces an EMF of $\mathcal{E}_2 = B_0 \frac{x}{a} v_0 d$. The net EMF is the difference between these potentials:
$\mathcal{E} = B_0 \frac{(x+d)}{a} v_0 d - B_0 \frac{x}{a} v_0 d = \frac{B_0 v_0 d^2}{a}$

Faraday's law of induction establishes that the induced electromotive force in a circuit is proportional to the negative rate of change of magnetic flux through it. The magnetic flux is calculated by multiplying the magnetic field by the area of the coil, yielding $\Phi = B \cdot \pi R^2 = 4 \times 10^{-3} (1 - t/100)$ Weber. Differentiating this expression with respect to time provides the induced emf, which evaluates to $\varepsilon = 4 \times 10^{-5}$ V. The magnetic field is completely switched off at 100 seconds, marking the duration over which dissipation occurs. Calculating the total energy using $E = \frac{\varepsilon^2}{R} \cdot t$ with a resistance of 2 $\mu\Omega$ gives $E = \frac{(4 \times 10^{-5})^2}{2 \times 10^{-6}} \cdot 100 = 0.08$ Joules. This energy dissipation equates to 80 mJ.

At the instant the switch is closed, inductors behave as open circuits because they oppose any sudden change in current, which effectively isolates the branches containing them. The circuit simplifies to a single path containing the battery and two of the resistors in series, resulting in an equivalent resistance of $R_{\text{eq}} = 2 \Omega + 2 \Omega = 4 \Omega$. Applying Ohm's law, the initial current $i$ is determined by the equation $i = \frac{9 \text{ V}}{4 \Omega} = 2.25 \text{ A}$.

An RL circuit's behavior is determined by the ratio of its inductance to its resistance, which establishes its characteristic time constant. The magnetic energy stored in an inductor is governed by $U_m = \frac{1}{2} L i^2$. By substituting the given energy of 64 J and current of 8 A, the relationship becomes
$64 = \frac{1}{2} \times L \times 64$
which simplifies to an inductance of 2 H.

The power dissipated by the coil is due to its internal resistance, calculated using $P = i^2 R$. Inserting the power of 640 W and the current of 8 A gives the equation
$640 = 64 \times R$
indicating that the resistance is 10 ohms. Since the time constant is defined as $\tau = \frac{L}{R}$, dividing the inductance by the resistance yields
$\tau = \frac{2}{10} = 0.2 \text{ s}$

When a conducting rod moves through a magnetic field, the changing area of the loop induces an electromotive force based on Faraday's law, where $e = -\frac{d\phi}{dt}$. As the arm travels from $x = 0$ to $x = b$, the area enclosed by the circuit increases linearly, leading to a corresponding linear change in magnetic flux, followed by a decrease upon the return journey from $x = b$ to $0$; this pattern identifies the graph for flux. Since the induced electromotive force is directly proportional to the time rate of change of this flux, it remains constant while the arm traverses the field and vanishes while the arm is outside the field, which describes the behavior of the electromotive force graph. Because power dissipated is calculated as $P = \frac{e^2}{R}$, the power follows a consistent magnitude while the rod is within the magnetic field, mapping correctly to the remaining graph.

According to Lenz's law, the induced current in a circuit always flows in a direction that opposes the change in magnetic flux responsible for its existence. Since the coil exhibits an anti-clockwise induced current, it acts to generate an outward-pointing magnetic field $B$, effectively creating a North pole on the outward face to reinforce the flux. Because this induced magnetic field works to counteract the current change, it reveals that the original outward-directed field must be decreasing over time, prompting the coil to attempt to restore the flux balance.

Faraday's law dictates that the magnitude of the induced electromotive force is determined by the rate at which the magnetic flux changes over time, calculated as $|\epsilon| = | \frac{d\phi_B}{dt} |$. Applying this derivative to the flux function $\phi_B(t) = 10t^2 + 20t$ produces an induced voltage of $|\epsilon| = 20t + 20$ in millivolts. Using Ohm's law, the current magnitude is found by dividing this induced voltage by the resistance of 2 $\Omega$, which simplifies the relationship to $|i| = 10t + 10$ in milliamperes. Substituting the specified time of 5 seconds into this equation results in a final current value of 60 mA.

According to Faraday's law of electromagnetic induction, the induced electromotive force in a loop is determined by the relation $\epsilon = -\frac{d\phi}{dt}$. As the magnet approaches the loop, the magnetic flux increases, resulting in a negative induced EMF. When the magnet reaches the center of the loop, the flux stops changing momentarily because the density of magnetic field lines reaches a maximum, causing the induced EMF to drop to zero. As the magnet continues through the loop, the flux decreases, reversing the sign of the rate of change and thereby generating an induced EMF of the opposite polarity.

Motional electromotive force arises when a conductor moves through a magnetic field, described by the relation $\epsilon = B l v \sin \theta$, where $l$ represents the effective length of the conductor perpendicular to both the magnetic field and the direction of velocity. As the circular ring moves into the region with a magnetic field of $1 \text{ T}$ and reaches the position at $t=1 \text{ s}$, the vertical diameter of the ring acts as a rod of length $2R = 2 \text{ m}$ cutting the flux lines perpendicularly. Given the velocity of $1 \text{ m/s}$, the induced EMF is calculated by substituting these parameters into the formula:

$\epsilon = 1 \text{ T} \times 2 \text{ m} \times 1 \text{ m/s} = 2 \text{ V}$

The motion of the conducting bar within the magnetic field generates a motional electromotive force defined by $\varepsilon = vBL$ across its ends. According to Lenz's law, the induced currents must flow in directions that counteract the change in magnetic flux caused by this motion through the loops formed by the rails and the resistors. Consequently, to effectively oppose the change in flux, $I_1$ flows in a clockwise direction, whereas $I_2$ flows in an anti-clockwise direction.

The induced EMF across the tips of the aeroplane's wings arises because the wings, acting as a conducting rod, move through the Earth's magnetic field. As the aeroplane travels horizontally, its wings cut through the vertical component of this magnetic field, which is defined by $B_{\perp} = B \sin \theta$, where $B$ is the total field intensity and $\theta$ is the angle of dip.

First, convert the aeroplane's speed into SI units, resulting in $50 \text{ m/s}$ since $180 \text{ km/h}$ divided by $3.6$ equals $50 \text{ m/s}$. Using the formula for motional EMF $\epsilon = B \sin \theta \cdot l \cdot v$, substitute the known parameters: $\epsilon = 2.5 \times 10^{-4} \times \sin 60^\circ \times 10 \times 50$. Evaluating this expression, where $\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$, yields $\epsilon = 2.5 \times 10^{-4} \times 500 \times \frac{\sqrt{3}}{2}$. This simplifies to $0.0625 \times 1.732$, resulting in $0.10825 \text{ V}$, which is equivalent to $108.25 \text{ mV}$.