Electromagnetic Induction Previous Year Questions — JEE Main Physics
28 past-year Electromagnetic Induction questions from JEE Main (Physics).
Q8 (2025)
A solenoid having area $A$ and length $l$ is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is :
- $\frac{B^2 A l}{\mu_0}$
- $\frac{B^2 A l}{2 \mu_0}$
- $B^2 A l$
- $\frac{B^2 A l}{4 \mu_0}$
### Related Formula
1. Magnetic Energy Density (energy per unit volume):
$$u_m = \frac{B^2}{2 \mu} = \frac{B^2}{2 \mu_r \mu_0}$$
2. Total Energy stored:
$$U = u_m \times V = u_m \times (A l)$$
### Core Logic
We are given:
- Relative permeability $\mu_r = 2$
- Permeability of the filled core medium $\mu = \mu_r \mu_0 = 2 \mu_0$
Substitute $\mu_r = 2$ into the energy density expression:
$$u_m = \frac{B^2}{2 (2 \mu_0)} = \frac{B^2}{4 \mu_0}$$
Multiply by the total volume of the solenoid ($V = A l$):
$$U = u_m \cdot V = \frac{B^2}{4 \mu_0} A l$$
### Pattern Recognition
Sees: Magnetic energy stored in a solenoid with medium relative permeability.
Trap: Placing the relative permeability $\mu_r$ in the numerator of the formula instead of the denominator.
Shortcut: Magnetic energy density is always inversely proportional to permeability. With medium $\mu = 2\mu_0$, energy density is halved compared to free space, giving $\frac{B^2}{4\mu_0}$. Multiply by volume $Al$ to get $\frac{B^2 Al}{4\mu_0}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Class 12 Physics: Magnetism and Matter
Q925 (2025)
Conductor wire ABCDE with each arm $10\mathrm{~cm}$ in length is placed in magnetic field of $\frac{1}{\sqrt{2}}\mathrm{~Tesla}$, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10\mathrm{~cm/s}$, induced emf between points A and E is ________ mV.
{{IMG1}}
### Related Formula
Motional electromotive force formula:
$$\varepsilon = B v l_{\text{eff}}$$
where $l_{\text{eff}}$ is the perpendicular component of the straight-line displacement vector connecting the endpoints $A$ and $E$ ($l_{AE}$) relative to velocity $\vec{v}$.
### Core Logic
In a uniform magnetic field, the motional EMF induced in any arbitrary conductor wire depends solely on the straight-line displacement vector connecting its endpoints, rather than the detailed path:
$$\vec{\varepsilon} = (\vec{v} \times \vec{B}) \cdot \vec{l}_{\text{eff}}$$
{{SOL_IMG1}}
### Step 1: Compute Effective Length
From the geometry of the symmetric wire segments oriented at $45^{\circ}$ to the horizontal:
$$l_{\text{eff}} = 2 \times (10 \sin 45^{\circ}) = 2 \times 10 \times \frac{1}{\sqrt{2}} = 10\sqrt{2}\text{ cm} = 0.1\sqrt{2}\text{ m}$$
### Step 2: Calculate Induced EMF
Substitute the given parameters into the motional EMF expression:
$$\varepsilon = B v l_{\text{eff}}$$
$$\varepsilon = \left(\frac{1}{\sqrt{2}}\right) \times (0.1\text{ m/s}) \times (0.1\sqrt{2}\text{ m})$$
$$\varepsilon = \frac{1}{\sqrt{2}} \times 0.1 \times 0.1\sqrt{2} = 0.01\text{ V} = 10\text{ mV}$$
### Pattern Recognition
In a uniform magnetic field, motional EMF is path-independent. Replace any zig-zag or curved conductor with an equivalent straight line joining the two endpoints perpendicular to the velocity vector.
### Evaluation Rubric / Model Answer
10
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q1 (2025)
A uniform magnetic field of $0.4 \mathrm{T}$ acts perpendicular to a circular copper disc $20 \mathrm{~cm}$ in radius. The disc is having a uniform angular velocity of $10 \pi$ rad $\mathrm{s}^{-1}$ about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? ( $\pi = 3.14$ )
- $\text{0.0628 V}$
- $\text{0.5024 V}$
- $\text{0.2512 V}$
- $\text{0.1256 V}$
### Related Formula
The induced electromotive force (EMF) developed between the center and the rim of a rotating disc in a perpendicular magnetic field is given by:
$$E = \frac{1}{2} B \omega R^2$$
### Core Logic
Given parameters from the problem statement [cite: 655, 657, 658]:
* Magnetic field, $B = 0.4 \text{ T}$
* Radius of the disc, $R = 20 \text{ cm} = 0.2 \text{ m}$
* Angular velocity, \omega = 10\pi \text{ rad s}^{-1}
Substituting the values into the governing formula:
$$E = \frac{1}{2} \times 0.4 \times (10 \times 3.14) \times (0.2)^2$$
$$E = 0.2 \times 31.4 \times 0.04$$
$$E = 0.2512 \text{ V}$$
### Step 1: Evaluation
The potential difference developed between the axis of the disc and the rim is precisely $0.2512 \text{ V}$.
### Pattern Recognition
For any rotating conductor of length $R$ or a continuous disc rotating about its center in a perpendicular magnetic field, the induced EMF is mathematically equivalent to a single radial rod sweeping the area, leading directly to the formula $\frac{1}{2}B\omega R^2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q21 (2025)
A conducting \bar moves on two conducting rails as shown in the figure. A constant magnetic field B \exists into the page. The \bar starts to move from the vertex at time $t = 0$ with a constant velocity. If the induced EMF is $E \propto t^n$ , then value of $n$ is
{{IMG1}}
### Related Formula
The motional EMF induced across a moving conductor of instantaneous length $\ell$ inside a perpendicular uniform magnetic field is given by:
$$E = B \cdot \ell \cdot v$$
### Core Logic
Let the V-shaped guide rails form an \angle, so that the instantaneous length $\ell$ of the conducting \bar grows linearly with its horizontal position distance $x$ from the vertex [cite: 782, 791]:
$\ell \propto x$
Since the \bar moves with a constant velocity $v$, its displacement position at any time $t$ is :
$$x = v \cdot t \implies \ell \propto v \cdot t$$
Substituting this time-dependent length into the induced EMF expression :
$$E = B \cdot \ell \cdot v \implies E \propto B \cdot (v \cdot t) \cdot v \implies E \propto t^1$$
Comparing this to the given relation $E \propto t^n$ gives the exponent[cite: 188, 791]:
$n = 1$
### Step 1: Geometric Analysis
The expanding circuit loop configuration across time is shown below:
{{SOL_IMG1}}
### Pattern Recognition
For \parallel rails, the length $\ell$ remains constant, meaning induced EMF is independent of time ($E \propto t^0$). For V-shaped divergent rails, the effective length increases linearly with distance, making the induced EMF directly proportional to time ($E \propto t^1$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q1282 (2025)
Consider $\mathrm{I}_1$ and $\mathrm{I}_2$ are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If $\mathrm{L}_1 =$ self inductance of coil 1, $\mathrm{M}_{12} =$ mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be
- $\varepsilon_{1} = -\mathrm{L}_{1}\frac{\mathrm{dI}_{1}}{\mathrm{d}t} +\mathrm{M}_{12}\frac{\mathrm{dI}_{2}}{\mathrm{d}t}$
- $\varepsilon_{1} = -\mathrm{L}_{1}\frac{\mathrm{dI}_{1}}{\mathrm{d}t} -\mathrm{M}_{12}\frac{\mathrm{dI}_{1}}{\mathrm{d}t}$
- $\varepsilon_{1} = -\mathrm{L}_{1}\frac{\mathrm{dI}_{1}}{\mathrm{d}t} -\mathrm{M}_{12}\frac{\mathrm{dI}_{2}}{\mathrm{d}t}$
- $\varepsilon_{1} = -\mathrm{L}_{1}\frac{\mathrm{dI}_{2}}{\mathrm{dt}} -\mathrm{M}_{12}\frac{\mathrm{dI}_{1}}{\mathrm{dt}}$
### Related Formula
$$\phi_1 = L_1 I_1 + M_{12} I_2$$
$$\varepsilon_1 = -\frac{\mathrm{d}\phi_1}{\mathrm{d}t}$$
### Core Logic
The total flux linked with coil 1 is due to its own current $I_1$ and the mutual influence of current $I_2$ in the neighboring coil :
$$\phi_1 = L_1 I_1 + M_{12} I_2$$
Differentiating with respect to time according to Faraday\'s Law yields :
$$\varepsilon_1 = -L_1 \frac{\mathrm{d}I_1}{\mathrm{d}t} - M_{12} \frac{\mathrm{d}I_2}{\mathrm{d}t}$$
### Pattern Recognition
Total induced emf sums both self-induction and mutual induction effects additively with standard Lenz law negative signs.
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
