Electromagnetic Induction Previous Year Questions — JEE Main Physics
28 past-year Electromagnetic Induction questions from JEE Main (Physics) — page 2.
Q1287 (2025)
A coil of area A and N turns is rotating with angular velocity $\omega$ in a uniform magnetic field $\vec{B}$ about an axis perpendicular to $\vec{B}$ . Magnetic flux $\varphi$ and induced emf $\varepsilon$ across it, at an instant when $\vec{B}$ is parallel to the plane of coil, are:
- $\varphi = \mathrm{AB},\varepsilon = 0$
- $\varphi = 0, \varepsilon = \mathrm{NAB}\omega$
- $\varphi = 0, \varepsilon = 0$
- $\varphi = \mathrm{AB},\varepsilon = \mathrm{NAB}\omega$
### Related Formula
$$\phi = BAN \cos(\omega t)$$
$$\varepsilon = BAN\omega \sin(\omega t)$$
### Core Logic
{{SOL_IMG1}}
When the magnetic field vector $\vec{B}$ lines up parallel to the plane of the coil, the norm area vector stands perpendicular to $\vec{B}$, yielding $\omega t = \frac{\pi}{2}$. Thus :
$$\phi = BAN \cos\left(\frac{\pi}{2}\right) = 0$$
$$\varepsilon = BAN\omega \sin\left(\frac{\pi}{2}\right) = NAB\omega$$
### Pattern Recognition
Flux is zero when the field lines are parallel to the coil surface, but the rate of change of flux (and thus emf) peaks to its absolute maximum.
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q1372 (2024)
A rectangular loop of sides $12\mathrm{~cm}$ and $5\mathrm{~cm}$, with its sides parallel to the x-axis and y-axis respectively moves with a velocity of $5\mathrm{~cm/s}$ in the positive x-axis direction, in a space containing a variable magnetic field in the positive z-direction. The field has a gradient of $10^{-3}\mathrm{~T/cm}$ along the negative x-direction and it is decreasing with time at the rate of $10^{-3}\mathrm{~T/s}$. If the resistance of the loop is $6\mathrm{~m\Omega}$, the power dissipated by the loop as heat is $x \times 10^{-9}\mathrm{~W}$. The value of $x$ is:
### Related Formula
Total induced EMF in a moving loop within a time-varying spatial field:
$$\varepsilon_{\text{net}} = \varepsilon_{\text{motional}} + \varepsilon_{\text{time}}$$
$$\varepsilon_{\text{motional}} = l \cdot v \cdot \Delta B = l \cdot v \cdot \left(\frac{dB}{dx} \cdot \Delta x\right)$$
$$\varepsilon_{\text{time}} = A \cdot \frac{dB}{dt}$$
### Core Logic
Loop dimensions: $l = 5\mathrm{~cm} = 0.05\mathrm{~m}$, $\Delta x = 12\mathrm{~cm} = 0.12\mathrm{~m}$.
Velocity $v = 5\mathrm{~cm/s} = 0.05\mathrm{~m/s}$.
Spatial gradient $\frac{dB}{dx} = 10^{-3}\mathrm{~T/cm} = 0.1\mathrm{~T/m}$.
Time decay rate $\frac{dB}{dt} = 10^{-3}\mathrm{~T/s}$.
Calculate the motional component across the leading edges:
$$\varepsilon_{\text{motional}} = 300 \times 10^{-7}\mathrm{~V}$$
Calculate the time-varying field induction across the loop area:
$$A = 12 \times 5 = 60\mathrm{~cm}^2 = 60 \times 10^{-4}\mathrm{~m}^2$$
$$\varepsilon_{\text{time}} = A \cdot \frac{dB}{dt} = 60 \times 10^{-4} \times 10^{-3} = 60 \times 10^{-7}\mathrm{~V}$$
Both changes induce current in the same direction according to Lenz's law:
$$\varepsilon_{\text{net}} = 300 \times 10^{-7} + 60 \times 10^{-7} = 360 \times 10^{-7}\mathrm{~V}$$
### Step 1: Calculate Dissipated Power
Given loop resistance $R = 6\mathrm{~m\Omega} = 6 \times 10^{-3}\mathrm{~\Omega}$:
$$P = \frac{\varepsilon_{\text{net}}^2}{R} = \frac{(360 \times 10^{-7})^2}{6 \times 10^{-3}} = \frac{129600 \times 10^{-14}}{6 \times 10^{-3}}$$
$$P = 21600 \times 10^{-11} = 216 \times 10^{-9}\mathrm{~W}$$
Therefore, $x = 216$.
### Pattern Recognition
When a loop moves through a field that changes in both space and time, the total induced EMF is the sum of the motional EMF ($v\frac{\partial B}{\partial x}$) and the transformer EMF ($A\frac{\partial B}{\partial t}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q52 (2024)
A horizontal straight wire $5\text{ m}$ long extending from east to west falling freely at right angle to horizontal component of earth's magnetic field $0.60 \times 10^{-4}\text{ Wb m}^{-2}$. The instantaneous value of emf induced in the wire when its velocity is $10\text{ ms}^{-1}$ is $x \times 10^{-3}\text{ V}$. The value of $x$ is:
### Related Formula
The motional electromotive force (emf) induced in a conductor of length $L$ moving with velocity $v$ perpendicular to a magnetic field $B$ is:
$e = B v L$
### Core Logic
Given parameters:
* Length of wire, $L = 5\text{ m}$
* Horizontal magnetic field component, $B_H = 0.60 \times 10^{-4}\text{ Wb m}^{-2}$
* Velocity of fall, $v = 10\text{ ms}^{-1}$
### Step 1: Calculate the Induced EMF
Substitute the parameters directly into the motional emf formula:
$e = B_H v L$
$$e = (0.60 \times 10^{-4}\text{ Wb m}^{-2}) \times (10\text{ ms}^{-1}) \times (5\text{ m})$$
$$e = 3 \times 10^{-3}\text{ V}$$
Comparing this to $x \times 10^{-3}\text{ V}$, we find:
$x = 3$
### Pattern Recognition
Motional EMF is directly the product of field, velocity, and length ($e = B v L$) when they are mutually perpendicular. A simple multiplication is all that is required here.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q42 (2024)
A rectangular loop of length $2.5\text{ m}$ and width $2\text{ m}$ is placed at $60^{\circ}$ to a magnetic field of $4\text{ T}$. The loop is removed from the field in $10\text{ sec}$. The average emf induced in the loop during this time is:
- $-2\text{ V}$
- $+2\text{ V}$
- $+1\text{ V}$
- $-1\text{ V}$
### Related Formula
$$\text{emf} = -\frac{\Delta\phi}{\Delta t}$$
### Core Logic
Initial magnetic flux is given by $\phi_i = B A \cos\theta$, where $A = 2.5 \times 2 = 5\text{ m}^2$, $B = 4\text{ T}$, and $\theta = 60^{\circ}$ (angle aligned with the axis mapping context rules in the problem source text).
$$\phi_i = 4 \times 5 \times \cos(60^{\circ}) = 20 \times 0.5 = 10\text{ Wb}$$
Final flux after removal $\phi_f = 0$.
### Step 1: Compute Induced EMF
$$\text{emf} = -\frac{\phi_f - \phi_i}{\Delta t} = -\frac{0 - 10}{10} = +1\text{ V}$$
### Pattern Recognition
Removal fields yield positive flux variants under canonical sign configurations due to the absolute reduction profile mapped by Lenz/Faraday relationships.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q53 (2024)
Two coils have mutual inductance $0.002\text{ H}$. The current changes in the first coil according to the relation $i = i_{0}\sin\omega t$, where $i_{0} = 5\text{ A}$ and \omega = $50\pi\text{ rad/s}$. The maximum value of emf in the second coil is $\frac{\pi}{\alpha}\text{ V}$. The value of $\alpha$ is ______.
### Related Formula
$$\text{emf} = -M \frac{di}{dt}$$
### Core Logic
Differentiating the current expression with respect to time:
$$\frac{di}{dt} = \frac{d}{dt}(i_0 \sin\omega t) = i_0 \omega \cos\omega t$$
Hence, the expression for induced emf is:
$$\text{emf} = -M i_0 \omega \cos\omega t$$
### Step 1: Isolate Maximum value
The peak magnitude occurs when $\cos\omega t = 1$:
$$\text{emf}_{\text{max}} = M i_0 \omega$$
Substitute given numerical values:
$$\text{emf}_{\text{max}} = 0.002 \times 5 \times 50\pi = 0.5\pi = \frac{\pi}{2}\text{ V}$$
### Step 2: Match with target variable
Comparing $\frac{\pi}{2}$ with $\frac{\pi}{\alpha}$ gives:
$\alpha = 2$
### Pattern Recognition
Harmonic driving functions induce derivative cosine arrays whose peak amplitudes scale strictly as the product of primary parameters: $M \cdot i_0 \cdot \omega$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
