JEE Main Electromagnetic Induction PYQ

1. (JEE Main 2024 (Online) 6th April Evening Shift)

In a coil, the current changes from $- 2 A$ to $+ 2 A$ in $0.2 s$ and induces an emf of $0.1 V$ . The self inductance of the coil is :

A. 4 mH

B. 2.5 mH

C. 1 mH

D. 5 mH

Correct Option is (D)

To find the self-inductance of the coil, we can use the formula for induced electromotive force (emf), which is given by Faraday's law of electromagnetic induction as it applies to self-induction:

$emf = - L \frac{Δ I}{Δ t}$

Where:

$emf$ = induced voltage in volts (V)
$L$ = self-inductance of the coil in henries (H)
$Δ I$ = change in current in amperes (A)
$Δ t$ = time interval in seconds (s) over which the current change occurs

Here, the problem gives us the following data:

$emf = 0.1 V$
$Δ I = 2 A - (- 2 A) = 4 A$
$Δ t = 0.2 s$

Substituting the given values into the formula, we get:

$0.1 = - L \frac{4}{0.2}$

Solving for $L$ , the equation becomes:

$0.1 = - L \cdot 20$

Therefore:

$L = - \frac{0.1}{20}$

Calculating the value of $L$ :

$L = - 0.005 H$

Or, expressing $L$ in millihenries (mH):

$L = - 5 mH$

However, considering the absolute value (since inductance is a magnitude and cannot be negative in this context):

$L = 5 mH$

Thus, the self-inductance of the coil is 5 mH, which corresponds to Option D.

2. (JEE Main 2024 (Online) 5th April Evening Shift)

The current in an inductor is given by $I = (3 t + 8)$ where $t$ is in second. The magnitude of induced emf produced in the inductor is $12 mV$ . The self-inductance of the inductor _________ $mH$ .

Correct answer is 4

The induced emf ( $ε$ ) in an inductor is given by Faraday's law of electromagnetic induction, which in its differential form for an inductor can be expressed as:

$ε = L \frac{d I}{d t}$

where:

$ε$ is the induced emf in the inductor,
$L$ is the inductance of the inductor,
$\frac{d I}{d t}$ is the rate of change of current through the inductor.

Given that the current $I = (3 t + 8)$ , where $t$ is in seconds, we can find the rate of change of current by differentiating $I$ with respect to $t$ .

$\frac{d I}{d t} = \frac{d}{d t} (3 t + 8) = 3$

The given magnitude of induced emf is $12 mV = 12 \times 10^{- 3} V$ (since $1 mV = 10^{- 3} V$ ).

Now, plug these values into the formula to find $L$ :

$12 \times 10^{- 3} = L \cdot 3$

Solving for $L$ gives:

$L = \frac{12 \times 10^{- 3}}{3} = 4 \times 10^{- 3} H = 4 mH$

Therefore, the self-inductance of the inductor is 4 mH.

3. (JEE Main 2023 (Online) 15th April Morning Shift)

$12 V$ battery connected to a coil of resistance $6 Ω$ through a switch, drives a constant current in the circuit. The switch is opened in $1 ms$ . The emf induced across the coil is $20 V$ . The inductance of the coil is :

A. 5 mH

B. 8 mH

C. 10 mH

D. 12 mH

Correct Option is (C)

When the switch is closed, the circuit is a simple DC circuit and the current in the circuit is given by Ohm's law:

$I = \frac{V}{R} = \frac{12 V}{6 Ω} = 2 A$ .

When the switch is opened, the current in the circuit drops to zero instantaneously.

However, the magnetic field generated by the current in the coil does not disappear immediately, and it continues to produce a back EMF that opposes the change in current.

This back EMF induces a voltage across the coil that can be calculated using Faraday's law of induction: $E = - L \frac{Δ I}{Δ t}$ , where $E$ is the induced voltage, $L$ is the inductance of the coil, and $Δ I / Δ t$ is the rate of change of current in the coil.

In this case, we know that the induced voltage is $20 V$ and the rate of change of current is

$Δ I / Δ t = - 2 A / (1 ms) = - 2 \times 10^{3} A/s$ .

Substituting these values into the equation above, we get: $20 V = - L \times (- 2 \times 10^{3} A/s)$ .

Solving for $L$ , we get: $L = \frac{20 V}{2 \times 10^{3} A/s} = 0.01 H$ .

Therefore, the inductance of the coil is $0.01 H$ , or 10 mH.

4. (JEE Main 2023 (Online) 30th January Morning Shift)

JEE Main 2023 (Online) 30th January Morning Shift Physics - Electromagnetic Induction Question 39 English

As per the given figure, if $\frac{dI}{dt} = - 1 A / s$ then the value of $V_{AB}$ at this instant will be ____________ $V$ .

Correct answer is 30

From the circuit :

$V_{A} - i R - \frac{L d i}{d t} - 12 = V_{B}$

$\Rightarrow V_{A} - V_{B} = 2 \times 12 + 6 (- 1) + 12$ volts

$= 30$ volts

5. (JEE Main 2023 (Online) 24th January Evening Shift)

Three identical resistors with resistance R = 12 $Ω$ and two identical inductors with self inductance L = 5 mH are connected to an ideal battery with emf of 12 V as shown in figure. The current through the battery long after the switch has been closed will be _____________ A.

JEE Main 2023 (Online) 24th January Evening Shift Physics - Electromagnetic Induction Question 34 English

Correct answer is 3

After long time, inductors are shorted.

Effective circuit becomes

JEE Main 2023 (Online) 24th January Evening Shift Physics - Electromagnetic Induction Question 34 English Explanation

Current through battery $= \frac{V}{R_{eq}} = \frac{12 V}{4 Ω} = 3 A$

where $R_{eq} = 3$ resistors in parallel.