Electromagnetic Induction – JEE Main PhysicsPractice Questions & PYQs

When an inductor is connected in series with a resistor and a battery, the current does not rise to its maximum value instantaneously because the inductor generates a back electromotive force that opposes the change. The current follows an exponential growth curve dependent on the ratio of resistance to inductance. The maximum steady-state current is calculated as:

$I_0 = \frac{E}{R}$

Using the 5 V battery and 5 Ohm resistance, the maximum current is 1 A. The current at any time $t$ is described by the relation:

$I = I_0 (1 - e^{-\frac{R}{L} t})$

By substituting the inductance of 10 H, resistance of 5 Ohm, and time of 2 s into the exponent, the term $\frac{R}{L} t$ becomes $\frac{5}{10} \times 2$, which simplifies to 1. Consequently, the current flowing in the circuit at 2 s is $1 - e^{-1}$ A.

When the battery is disconnected while the inductor and resistor remain in a closed loop, the current does not drop to zero immediately due to the inductor's tendency to oppose changes in magnetic flux. Instead, the energy stored in the inductor discharges through the resistor, causing the current to decay exponentially according to the equation $i = i_0 e^{-Rt/L}$.

Before disconnection, the circuit reached a steady state where the inductor acted as a simple conductor, resulting in an initial current $i_0 = E/R = 100/100 = 1$ A. The rate at which this current decays is governed by the time constant $\tau = L/R$, which calculates to $100 \times 10^{-3}/100 = 1$ ms. Because the time elapsed is $1$ ms, which equals exactly one time constant, the expression for the current becomes:

$i = 1 e^{-1/1} = 1 e^{-1} = \frac{1}{e} \text{ A}$

Consequently, the current flowing in the circuit at that instant is $1/e$ A.

Mutual inductance represents the efficiency of magnetic coupling between circuits and is defined as the ratio of magnetic flux to the current flowing through a conductor. By substituting the dimensional components of magnetic flux, $[M L^2 T^{-1} Q^{-1}]$, and current, $[Q T^{-1}]$, into the expression $\frac{\phi}{I}$, the time terms cancel out to leave the dimension $M L^2 Q^{-2}$. This specific dimensional combination characterizes the Henry, which serves as the standard SI unit for measuring both self and mutual inductance.

Faraday's law of electromagnetic induction states that the induced electromotive force in a coil is equal to the negative time rate of change of the magnetic flux. Given the expression for flux $\phi = 10t^2 - 50t + 250$, the induced EMF is determined by calculating $e = -\frac{d\phi}{dt}$. This operation leads to the expression $e = -(20t - 50)$. At $t = 3$ seconds, the induced EMF becomes $e = -(20(3) - 50)$, which simplifies to an induced EMF of $-10$ V.

The induced electromotive force in a moving conductor within a magnetic field arises from the rate at which the conductor cuts through magnetic field lines, which is determined by the velocity of the conductor relative to the field. As both conducting U-tubes approach one another at a constant speed $v$, the speed at which they close the distance between them is the sum of their individual speeds, resulting in a relative velocity of $v + v = 2v$. Substituting this value into the standard motional EMF formula yields the final induced voltage:

$\varepsilon = B \cdot l \cdot (2v) = 2Blv$

When a circuit contains an inductor, the current does not rise to its maximum value instantaneously because the induced electromotive force opposes the change. Instead, it follows an exponential growth pattern governed by the following equation:
$i = i_0(1 - e^{-\frac{Rt}{L}})$
where $i_0$ is the steady-state current and the ratio $\frac{L}{R}$ represents the time constant of the circuit.

To determine when the current reaches half its steady-state value, we set $i = \frac{i_0}{2}$. Substituting this into the growth equation results in the relation $\frac{1}{2} = 1 - e^{-\frac{Rt}{L}}$, which simplifies to $e^{-\frac{Rt}{L}} = \frac{1}{2}$. By taking the natural logarithm of both sides, we find that $\frac{Rt}{L} = \ln(2)$, allowing us to solve for time as $t = \frac{L}{R} \ln(2)$.

Given that the inductance is 300 mH or 0.3 H and the resistance is 2 ohms, we plug these values into our derived expression. The calculation $t = \frac{0.3}{2} \times 0.69$ leads to $0.15 \times 0.69$, which equals approximately 0.1 s.

According to Faraday's law of electromagnetic induction, the induced electromotive force in a coil is directly proportional to the number of turns and the rate of change of magnetic flux. Given the magnetic flux changes from $W_1$ to $W_2$ over a duration $t$, the rate of change is $\frac{W_2 - W_1}{t}$, which results in an induced electromotive force of $-n \frac{(W_2 - W_1)}{t}$. Because the coil with resistance $R$ and the galvanometer with resistance $4R$ are connected in series, the total resistance of the circuit is the sum of these individual resistances, totaling $5R$. Dividing the induced electromotive force by this total resistance determines the magnitude of the induced current in the circuit as $i = -\frac{n (W_2 - W_1)}{5 R t}$

When a conducting rod rotates through a magnetic field perpendicular to its plane of rotation, the motion of its internal charge carriers creates an induced potential difference between its ends. Since the linear speed of each segment of the rod increases linearly with its distance from the pivot, the total induced electromotive force is determined by integrating this motional effect across the length of the conductor, leading to the relationship $\varepsilon = \frac{1}{2} B \omega l^2$.

By substituting the known parameters-a magnetic field $B = 0.2 \times 10^{-4}$ T, angular velocity $\omega = 5$ rad/s, and length $l = 1$ m-into this expression, we have:
$\varepsilon = \frac{1}{2} \times 0.2 \times 10^{-4} \times 5 \times 1^2$

Simplifying this calculation results in an induced EMF of $\varepsilon = 0.5 \times 10^{-4}$ V. Converting this value into scientific notation gives $5 \times 10^{-5}$ V, which corresponds to $50$ $\mu$V.

The rotation of a wire loop within a magnetic field induces a time-varying magnetic flux through its area, defined by $\phi = B A \cos(\omega t)$. Based on Faraday's Law, this changing flux generates an induced electromotive force that drives a sinusoidal current $i = \frac{\omega B A}{R} \sin(\omega t)$ through the circuit, where $A$ denotes the area of the semicircle.

Calculating the instantaneous power dissipated through the resistance $R$ gives $P = i^2 R = \frac{(\omega B A)^2}{R} \sin^2(\omega t)$. Since the average value of $\sin^2(\omega t)$ over one complete cycle is $1/2$, the mean power is expressed as $P_{avg} = \frac{(\omega B A)^2}{2 R}$. Substituting the area of a semicircle, $A = \frac{\pi r^2}{2}$, leads to the final calculation:
$P_{avg} = \frac{(\omega B \pi r^2)^2}{8 R}$

Mutual inductance, represented by the symbol $M$, serves as a measure of the geometric coupling between two circuits, indicating how effectively magnetic flux generated by one coil links with the windings of the other. Since this flux linkage is determined by the spatial arrangement and angular alignment of the magnetic field lines, $M$ changes whenever the distance between the coils or their relative orientation is altered. Properties such as the current intensity or the wire composition do not affect this relationship, as it is strictly a function of physical geometry.

When the current flowing through an inductor changes, it generates an induced electromotive force that resists this change according to the principle of self-induction. The relationship between the induced EMF and the rate of current change is defined by the formula $|e| = L \frac{\Delta I}{\Delta t}$, where $L$ represents the self-induction coefficient.

The current changes from 2 A to -2 A, meaning the total change in magnitude is 4 A over an interval of 0.05 s. Substituting the known values into the equation gives:
$8 = L \times \frac{4}{0.05}$
To find the self-induction coefficient, the equation is rearranged as:
$L = \frac{8 \times 0.05}{4}$
Performing this arithmetic yields a coefficient of self-induction of 0.1 H.

When a conducting loop moves through a magnetic field, an electromotive force is induced due to the magnetic Lorentz force acting on charge carriers within the moving conductor. As the loop of side length $L$ moves with uniform velocity $v$ perpendicular to the magnetic field $B$, only the segment traversing the flux gradient contributes to the potential difference. The magnitude of this induced EMF is determined by the product of the magnetic field strength, the velocity of the conductor, and the length of the segment cutting the field lines:

$e = B v L$

When inductors are connected in parallel, the reciprocal of the equivalent inductance is calculated by summing the reciprocals of the individual inductances. For these three inductors each rated at 3 H, the total inductance between terminals A and D is determined by the equation:

$\frac{1}{L_P} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3}$

Solving this gives $\frac{1}{L_P} = 1$, which confirms that the equivalent inductance is 1 H.